diff options
| author | hardythe1 | 2015-07-03 12:23:43 +0530 |
|---|---|---|
| committer | hardythe1 | 2015-07-03 12:23:43 +0530 |
| commit | 5a86a20b9de487553d4ef88719fb0fd76a5dd6a7 (patch) | |
| tree | db67ac5738a18b921d9a8cf6e86f402703f30bdf /Engineering_Physics | |
| parent | 37d315828bbfc0f5cabee669d2b9dd8cd17b5154 (diff) | |
| download | Python-Textbook-Companions-5a86a20b9de487553d4ef88719fb0fd76a5dd6a7.tar.gz Python-Textbook-Companions-5a86a20b9de487553d4ef88719fb0fd76a5dd6a7.tar.bz2 Python-Textbook-Companions-5a86a20b9de487553d4ef88719fb0fd76a5dd6a7.zip | |
add/remove books
Diffstat (limited to 'Engineering_Physics')
50 files changed, 0 insertions, 16748 deletions
diff --git a/Engineering_Physics/Chapter1.ipynb b/Engineering_Physics/Chapter1.ipynb deleted file mode 100755 index aec285c1..00000000 --- a/Engineering_Physics/Chapter1.ipynb +++ /dev/null @@ -1,463 +0,0 @@ -{
- "metadata": {
- "name": "",
- "signature": "sha256:305dcc155810a92329b40a851ec0d721020f53211b825d9c1f3b0e3bb9312285"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "heading",
- "level": 1,
- "metadata": {},
- "source": [
- "1: Acoustics"
- ]
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example number 1.1, Page number 12"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#importing modules\n",
- "import math\n",
- "from __future__ import division\n",
- "\n",
- "#Variable declaration\n",
- "p=50; #sound waves with output power(W)\n",
- "r=4; #Distance(m)\n",
- "\n",
- "#Calculation\n",
- "I=p/(4*math.pi*r**2) #Intensity(W/m**2)\n",
- "\n",
- "#Result\n",
- "print \"Intensity of sound at a distance of 4m from the source is\",round(I,2),\"W/m**2\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Intensity of sound at a distance of 4m from the source is 0.25 W/m**2\n"
- ]
- }
- ],
- "prompt_number": 2
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example number 1.2, Page number 12"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#importing modules\n",
- "import math\n",
- "from __future__ import division\n",
- "\n",
- "#Variable declaration\n",
- "Io=10**-12; #Initial intensity of sound(W/m**2)\n",
- "d=50; #number of decibels given by 10log(Io/I1)\n",
- "P=70; #Output power(W)\n",
- "\n",
- "#Calculation\n",
- "I1=(10**5)*Io; #Intensity(W/m**2)\n",
- "r=math.sqrt(P/(4*math.pi*I1)); #distance(m)\n",
- "\n",
- "#Result\n",
- "print \"The distance at which sound reduces to a level of 50dB is\",round(r/10**3,2),\"*10**3 m\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The distance at which sound reduces to a level of 50dB is 7.46 *10**3 m\n"
- ]
- }
- ],
- "prompt_number": 5
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example number 1.3, Page number 12"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#importing modules\n",
- "import math\n",
- "from __future__ import division\n",
- "\n",
- "#Variable declaration\n",
- "v=8000; #volume of the hall(m**3)\n",
- "T=1.5; #Reverberation time(sec)\n",
- "\n",
- "#Calculation\n",
- "A=(0.161*v)/T; #Total absorption time(m**2 sabine)\n",
- "\n",
- "#Result\n",
- "print \"The total reverberation in the hall is\",round(A,2),\"m**2 sabine\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The total reverberation in the hall is 858.67 m**2 sabine\n"
- ]
- }
- ],
- "prompt_number": 7
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example number 1.4, Page number 12"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#importing modules\n",
- "import math\n",
- "from __future__ import division\n",
- "\n",
- "#Variable declaration\n",
- "l=25; #length of the hall(m)\n",
- "b=20; #breadth of the hall(m)\n",
- "h=10; #height of the hall(m)\n",
- "T=4; #Reverberation time(s)\n",
- "\n",
- "#Calculation\n",
- "V=l*b*h; #Volume of the hall(m**3)\n",
- "A=(0.161*V)/T; #Total absorption time(m**2 sabine)\n",
- "a=A/(2*((l*b)+(b*h)+(l*h))); #a is absorption co-efficient\n",
- "\n",
- "#Result\n",
- "print \"The average absorption co-efficients of surfaces is\",round(a,3)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The average absorption co-efficients of surfaces is 0.106\n"
- ]
- }
- ],
- "prompt_number": 9
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example number 1.5, Page number 12"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#importing modules\n",
- "import math\n",
- "from __future__ import division\n",
- "\n",
- "#Variable declaration\n",
- "Y=77*(10**10); #Youngs modulus for quartz(dyne/cm**2)\n",
- "rho=2.6; #density of quartz(g/cm**3)\n",
- "t=0.4; #thickness(cm)\n",
- "\n",
- "#Calculation\n",
- "f=((1/(2*t))*math.sqrt(Y/rho))*10**-3; #frequency(kHz)\n",
- "\n",
- "#Result\n",
- "print \"The frequency of ultrasonic waves produced is\",int(f),\"kHz\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The frequency of ultrasonic waves produced is 680 kHz\n"
- ]
- }
- ],
- "prompt_number": 11
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example number 1.6, Page number 12"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#importing modules\n",
- "import math\n",
- "from __future__ import division\n",
- "\n",
- "#Variable declaration\n",
- "Y=81*10**10; #Young's modulus for barium titanate(dynes/cm**2)\n",
- "rho=5.51; #density of barium titanate(g/cm**3)\n",
- "f=900; #frequency of ultrasonic waves(kHZ)\n",
- "\n",
- "#Calculation\n",
- "t=((1/(2*f))*math.sqrt(Y/rho))*10**-2; #thickness of crystal(mm)\n",
- "\n",
- "#Result\n",
- "print \"The thickness of the crystal to produce ultrasonic waves is\",round(t,2),\"mm\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The thickness of the crystal to produce ultrasonic waves is 2.13 mm\n"
- ]
- }
- ],
- "prompt_number": 13
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example number 1.7, Page number 13"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#importing modules\n",
- "import math\n",
- "from __future__ import division\n",
- "\n",
- "#Variable declaration\n",
- "r=3; #distance(m)\n",
- "I=0.86; #Intensity of sound source(W/m**2)\n",
- "\n",
- "#Calculation\n",
- "P=4*math.pi*r**2*I; #Power of the sound source(W)\n",
- "\n",
- "#Result\n",
- "print \"The output power of the sound source is\",round(P,2),\"W\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The output power of the sound source is 97.26 W\n"
- ]
- }
- ],
- "prompt_number": 15
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example number 1.8, Page number 13"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#importing modules\n",
- "import math\n",
- "from __future__ import division\n",
- "\n",
- "#Variable declaration\n",
- "d=60; #Number of decibels given by 10*log(I/I0)\n",
- "I0=10**-12; #Initial intensity of sound(W/m**2)\n",
- "I=10**-6; #since 10log(I/I0)=60\n",
- "r=200; #distance(m)\n",
- "\n",
- "#Calculation\n",
- "P=4*math.pi*r**2*I; #output power of the sound source(W)\n",
- "\n",
- "#Result\n",
- "print \"The output power of the sound source is\",round(P,1),\"W\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The output power of the sound source is 0.5 W\n"
- ]
- }
- ],
- "prompt_number": 17
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example number 1.9, Page number 13"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#importing modules\n",
- "import math\n",
- "from __future__ import division\n",
- "\n",
- "#Variable declaration\n",
- "V=9250; #volume of the hall(m**3)\n",
- "A=900; #Total absorption(m**2 sabine)\n",
- "\n",
- "#Calculation\n",
- "T=(0.161*V)/A; #Reverberation time(s)\n",
- "\n",
- "#Result\n",
- "print \"The reverberation time in a hall is\",round(T,2),\"s\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The reverberation time in a hall is 1.65 s\n"
- ]
- }
- ],
- "prompt_number": 19
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example number 1.10, Page number 13"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#importing modules\n",
- "import math\n",
- "from __future__ import division\n",
- "\n",
- "#Variable declaration\n",
- "f1=400; #Initial frequency(kHZ)\n",
- "f2=500; #New frequency(kHZ)\n",
- "t1=3; #initial thickness of the crystal(mm)\n",
- "\n",
- "#Calculation\n",
- "t2=(f1*t1)/f2; #required thickness(mm)\n",
- "\n",
- "#Result\n",
- "print \"The required thickness of the crystal is\",t2,\"mm\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The required thickness of the crystal is 2.4 mm\n"
- ]
- }
- ],
- "prompt_number": 20
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example number 1.11, Page number 13"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#importing modules\n",
- "import math\n",
- "from __future__ import division\n",
- "\n",
- "#Variable declaration\n",
- "t1=2; #Initial thickness(mm)\n",
- "t2=2.8; #New thickness(mm)\n",
- "\n",
- "#Calculation\n",
- "F=t1/t2; #ratio of new to old frequencies\n",
- "\n",
- "#Result\n",
- "print \"The ratio of new to old frequencies is\",round(F,3),"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The ratio of new to old frequencies is 0.714\n"
- ]
- }
- ],
- "prompt_number": 23
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Engineering_Physics/Chapter10.ipynb b/Engineering_Physics/Chapter10.ipynb deleted file mode 100755 index 7a9d784a..00000000 --- a/Engineering_Physics/Chapter10.ipynb +++ /dev/null @@ -1,493 +0,0 @@ -{
- "metadata": {
- "name": "",
- "signature": "sha256:9dafdb7acb5e988ab3a5ace98a3f2deebed0e1d539e288cbefca9baaaeda9388"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "heading",
- "level": 1,
- "metadata": {},
- "source": [
- "10: Quantum Mechanics"
- ]
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example number 10.1, Page number 196"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#importing modules\n",
- "import math\n",
- "from __future__ import division\n",
- "\n",
- "#Variable declaration\n",
- "v=10**7; #speed of electron(m/s)\n",
- "h=6.626*10**-34; #plancks constant\n",
- "m=9.1*10**-31; #mass of electron(kg)\n",
- "\n",
- "#Calculation \n",
- "lamda=h/(m*v); #de Broglie wavelength(m)\n",
- "\n",
- "#Result\n",
- "print \"The de Broglie wavelength is\",round(lamda*10**11,2),\"*10**-11 m\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The de Broglie wavelength is 7.28 *10**-11 m\n"
- ]
- }
- ],
- "prompt_number": 3
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example number 10.2, Page number 196"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#importing modules\n",
- "import math\n",
- "from __future__ import division\n",
- "\n",
- "#Variable declaration\n",
- "h=6.626*10**-34; #plancks constant\n",
- "lamda=0.3; #de Broglie wavelength(nm)\n",
- "#For electron\n",
- "me=9.1*10**-31; #mass of electron(kg)\n",
- "#For proton\n",
- "mp=1.672*10**-27; #mass of proton(kg)\n",
- "\n",
- "#Calculation \n",
- "p=h/(lamda*10**-9); #uncertainity in determining momentum(kg m/s)\n",
- "K1=p**2/(2*me); #kinetic energy of electron(J)\n",
- "K2=p**2/(2*mp); #kinetic energy of proton(J)\n",
- "\n",
- "#Result\n",
- "print \"The kinetic energy of electron is\",round(K1*10**18,1),\"*10**-18 J\"\n",
- "print \"The kinetic energy of proton is\",round(K2*10**21,2),\"*10**-21 J\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The kinetic energy of electron is 2.7 *10**-18 J\n",
- "The kinetic energy of proton is 1.46 *10**-21 J\n"
- ]
- }
- ],
- "prompt_number": 5
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example number 10.3, Page number 196"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#importing modules\n",
- "import math\n",
- "from __future__ import division\n",
- "\n",
- "#Variable declaration\n",
- "#K=p^2/(lambda^2*2*m) where K is kinetic energy\n",
- "h=6.626*10**-34; #plancks constant\n",
- "lamda=10**-14; #de Broglie wavelength(m)\n",
- "m=9.1*10**-31; #mass of electron(kg)\n",
- "e=1.6*10**-19;\n",
- "\n",
- "#Calculation \n",
- "K=(h**2/((lamda**2)*2*m*e))*10**-9; \n",
- "\n",
- "#Result\n",
- "print \"The kinetic energy is\",int(K),\"GeV\"\n",
- "print \"It is not possible to confine the electron to a nucleus.\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The kinetic energy is 15 GeV\n",
- "It is not possible to confine the electron to a nucleus.\n"
- ]
- }
- ],
- "prompt_number": 9
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example number 10.4, Page number 197"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#importing modules\n",
- "import math\n",
- "from __future__ import division\n",
- "\n",
- "#Variable declaration\n",
- "m=9.1*10**-31; #mass of electron(kg)\n",
- "v=6*10**3; #speed of electron(m/s)\n",
- "h=6.626*10**-34; #plancks constant\n",
- "a=0.00005; \n",
- "\n",
- "#Calculation \n",
- "p=m*v; #uncertainity in momentum(kg m/s)\n",
- "deltap=a*p; #uncertainity in p\n",
- "deltax=(h/(4*math.pi*deltap))*10**3 #uncertainity in position(mm)\n",
- "\n",
- "#Result\n",
- "print \"The uncertainity in position is\",round(deltax,3),\"mm\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The uncertainity in position is 0.193 mm\n"
- ]
- }
- ],
- "prompt_number": 11
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example number 10.5, Page number 197"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#importing modules\n",
- "import math\n",
- "from __future__ import division\n",
- "\n",
- "#Variable declaration\n",
- "L=3*10**-5; #diameter of the sphere(nm)\n",
- "h=6.626*10**-34; #plancks constant\n",
- "m=1.67*10**-27; #mass of the particle(kg)\n",
- "n=1;\n",
- "e=1.6*10**-19;\n",
- "\n",
- "#Calculation \n",
- "E1=((h**2)*(n**2))/(8*m*(L**2)*e)*10**12 #first energy level(MeV)\n",
- "E2=E1*2**2; #second energy level(MeV)\n",
- "\n",
- "#Result\n",
- "print \"The first energy level is\",round(E1,3),\"MeV\"\n",
- "print \"The second energy level is\",round(E2,4),\"MeV\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The first energy level is 0.228 MeV\n",
- "The second energy level is 0.9128 MeV\n"
- ]
- }
- ],
- "prompt_number": 14
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example number 10.6, Page number 197"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#importing modules\n",
- "import math\n",
- "from __future__ import division\n",
- "\n",
- "#Variable declaration\n",
- "h=6.626*10**-34; #plancks constant\n",
- "a=2*10**12; #angular frequency(rad/s)\n",
- "e=1.6*10**-19;\n",
- "\n",
- "#Calculation \n",
- "E0=(0.5*(h/(2*math.pi*e))*a)*10**3; #ground state energy(MeV)\n",
- "E1=(1.5*(h/(2*math.pi*e))*a)*10**3; #first excited state energy(MeV)\n",
- "\n",
- "#Result\n",
- "print \"The ground state energy is\",round(E0,3),\"MeV\" \n",
- "print \"The first excited state energy is\",round(E1,3),\"MeV\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The ground state energy is 0.659 MeV\n",
- "The first excited state energy is 1.977 MeV\n"
- ]
- }
- ],
- "prompt_number": 16
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example number 10.7, Page number 197"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#importing modules\n",
- "import math\n",
- "from __future__ import division\n",
- "\n",
- "#Variable declaration\n",
- "h=6.626*10**-34; #plancks constant\n",
- "E=85; #Energy(keV)\n",
- "c=3*10**8; #speed of light(m/s)\n",
- "e=1.6*10**-19;\n",
- "\n",
- "#Calculation \n",
- "lamda=(h*c)/(E*10**3*e); #de Broglie wavelength(m)\n",
- "m=9.1*10**-31; #mass of electron(kg)\n",
- "K=((h**2)/((lamda**2)*2*m*e)); #kinetic energy of electron(keV)\n",
- "\n",
- "#Result\n",
- "print \"The kinetic energy of the electron is\",round(K*10**-3,2),\"keV\"\n",
- "print \"answer in the book varies due to rounding off errors\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The kinetic energy of the electron is 7.06 keV\n",
- "answer in the book varies due to rounding off errors\n"
- ]
- }
- ],
- "prompt_number": 21
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example number 10.8, Page number 198"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#importing modules\n",
- "import math\n",
- "from __future__ import division\n",
- "\n",
- "#Variable declaration\n",
- "lamda=0.08; #de Briglie wavelength(nm)\n",
- "m=9.1*10**-31; #mass of electron(kg)\n",
- "h=6.626*10**-34; #plancks constant\n",
- "\n",
- "#Calculation \n",
- "v=h/(m*lamda*10**-9); #velocity of the electron(m/s)\n",
- "\n",
- "#Result\n",
- "print \"The velocity of the electron is\",round(v/10**6,1),\"*10**6 m/s\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The velocity of the electron is 9.1 *10**6 m/s\n"
- ]
- }
- ],
- "prompt_number": 28
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example number 10.9, Page number 198"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#importing modules\n",
- "import math\n",
- "from __future__ import division\n",
- "\n",
- "#Variable declaration\n",
- "h=6.626*10**-34; #plancks constant\n",
- "lamda=589*10**-9; #wavelength(m)\n",
- "m=9.1*10**-31; #mass of electron(kg)\n",
- "e=1.6*10**-19;\n",
- "\n",
- "#Calculation \n",
- "V=((h**2)/((lamda**2)*2*m*e))*10**6; #potential diference(micro V)\n",
- "\n",
- "#Result\n",
- "print \"The potential difference through which an electron should be accelerated is\",round(V,2),\"micro V\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The potential difference through which an electron should be accelerated is 4.35 micro V\n"
- ]
- }
- ],
- "prompt_number": 31
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example number 10.10, Page number 198"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#importing modules\n",
- "import math\n",
- "from __future__ import division\n",
- "\n",
- "#Variable declaration\n",
- "deltax=0.92*10**-9; #uncertainity in position(m)\n",
- "m=9.1*10**-31; #mass of electron(kg)\n",
- "h=6.626*10**-34; #plancks constant\n",
- "\n",
- "#Calculation \n",
- "deltav=h/(4*math.pi*m*deltax); #uncertainity in velocity(m/s)\n",
- "\n",
- "#Result\n",
- "print \"The uncertainity in velocity is\",round(deltav/10**4,1),\"*10**4 m/s\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The uncertainity in velocity is 6.3 *10**4 m/s\n"
- ]
- }
- ],
- "prompt_number": 33
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example number 10.11, Page number 198"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#importing modules\n",
- "import math\n",
- "from __future__ import division\n",
- "\n",
- "#Variable declaration\n",
- "h=6.626*10**-34; #plancks constant\n",
- "n=3; #for second excited state\n",
- "m=1.67*10**-27; #mass of proton(kg)\n",
- "E=0.5; #energy(MeV)\n",
- "e=1.6*10**-19;\n",
- "\n",
- "#Calculation \n",
- "L=((h*n)/math.sqrt(8*m*E*10**6*e))*10**15; #length of the box(fm)\n",
- "\n",
- "#Result\n",
- "print \"The length of the box for proton in its second excited state is\",round(L,1),\"fm\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The length of the box for proton in its second excited state is 60.8 fm\n"
- ]
- }
- ],
- "prompt_number": 35
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Engineering_Physics/Chapter11.ipynb b/Engineering_Physics/Chapter11.ipynb deleted file mode 100755 index 39762859..00000000 --- a/Engineering_Physics/Chapter11.ipynb +++ /dev/null @@ -1,620 +0,0 @@ -{
- "metadata": {
- "name": "",
- "signature": "sha256:1d5f4970753c62d94a2bd202867cc0f79046f1baac4b1a42721a5ae6844ad5f4"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "heading",
- "level": 1,
- "metadata": {},
- "source": [
- "11: Nuclear Radiations and Detectors"
- ]
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example number 11.1, Page number 227"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#importing modules\n",
- "import math\n",
- "from __future__ import division\n",
- "\n",
- "#Variable declaration\n",
- "r0=1.2; #radius(fm)\n",
- "A=7; #mass number \n",
- "\n",
- "#Calculation \n",
- "r=r0*A**(1/3);\t #radius of Li(fm) \n",
- "\n",
- "#Result\n",
- "print \"The radius of Li is\",round(r,4),\"fm\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The radius of Li is 2.2955 fm\n"
- ]
- }
- ],
- "prompt_number": 2
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example number 11.2, Page number 227"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#importing modules\n",
- "import math\n",
- "from __future__ import division\n",
- "\n",
- "#Variable declaration\n",
- "M=235.043945; #atomic mass of uranium(u)\n",
- "Z=92; #atomic number of uranium\n",
- "mp=1.007825; #mass of proton(kg)\n",
- "N=143; #no.of neutrons\n",
- "mn=1.008665; #mass of neutron(kg)\n",
- "A=235; #number of nucleons\n",
- "\n",
- "#Calculation \n",
- "B=(((Z*mp)+(N*mn)-(M))/A)*931.5; #Binding energy(MeV)\n",
- "\n",
- "#Result\n",
- "print \"The binding energy per nucleon is\",round(B,3),\"MeV\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The binding energy per nucleon is 7.591 MeV\n"
- ]
- }
- ],
- "prompt_number": 4
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example number 11.3, Page number 227"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#importing modules\n",
- "import math\n",
- "from __future__ import division\n",
- "\n",
- "#Variable declaration\n",
- "#After removing one neutron from Ca(A=43;Z=20) it becomes Ca(A=42;Z=20)\n",
- "M=41.958622; #mass of Ca(A=42;Z=20)(kg)\n",
- "mn=1.008665; #mass of neutron(kg)\n",
- "E=42.95877; #mass of Ca(A=43;Z=20)(kg)\n",
- "\n",
- "#Calculation \n",
- "C=M+mn;\n",
- "D=C-E;\n",
- "B=D*931.5; #Binding energy of neutron(MeV)\n",
- "\n",
- "#Result\n",
- "print \"The binding energy of neutron is\",round(B,4),\"MeV\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The binding energy of neutron is 7.9336 MeV\n"
- ]
- }
- ],
- "prompt_number": 7
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example number 11.4, Page number 227"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#importing modules\n",
- "import math\n",
- "from __future__ import division\n",
- "\n",
- "#Variable declaration\n",
- "mBe=9.012182; #Atomic mass of beryllium(u)\n",
- "mHe=4.002603; #Atomic mass of helium\n",
- "mn=1.008665; #mass of neutron(kg)\n",
- "mC=12.000000; #Atomic mass of carbon\n",
- "\n",
- "#Calculation \n",
- "Q=(mBe+mHe-mn-mC)*931.5 #energy balance of the reaction(MeV)\n",
- "\n",
- "#Result\n",
- "print \"The Q-value is\",round(Q,1),\"MeV\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The Q-value is 5.7 MeV\n"
- ]
- }
- ],
- "prompt_number": 9
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example number 11.5, Page number 227"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#importing modules\n",
- "import math\n",
- "from __future__ import division\n",
- "\n",
- "#Variable declaration\n",
- "mLi=7.016004; #mass of Lithium(A=7)(u)\n",
- "mH=1.007825; #mass of Hydrogen(A=1)(u)\n",
- "mHe=4.002603; #mass of helium(A=4)(u)\n",
- "p=0.5; #energy of proton(MeV)\n",
- "\n",
- "#Calculation \n",
- "Q=(mLi+mH-2*(mHe))*931.5 #energy balance of the reaction(MeV)\n",
- "#The energy of 2 alpha particles is equal to the Q-value + energy of proton.\n",
- "Ealpha=(Q+p)/2; #energy of each alpha particle(MeV)\n",
- "\n",
- "#Result\n",
- "print \"The Q-value of the reaction is\",round(Q,2),\"MeV\"\n",
- "print \"The energy of each alpha particle is\",round(Ealpha,3),\"MeV\"\n",
- "print \"answer for energy in the book varies due to rounding off errors\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The Q-value of the reaction is 17.35 MeV\n",
- "The energy of each alpha particle is 8.924 MeV\n",
- "answer for energy in the book varies due to rounding off errors\n"
- ]
- }
- ],
- "prompt_number": 14
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example number 11.6, Page number 228"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#importing modules\n",
- "import math\n",
- "from __future__ import division\n",
- "\n",
- "#Variable declaration\n",
- "wt=1000; #weight(gm)\n",
- "A=235; #mass number of uranium\n",
- "N=(6.02*10**23/A)*wt; #no.of nuclei in 1kg of uranium\n",
- "Q=208; #energy-balance of the reaction\n",
- "\n",
- "#Calculation \n",
- "E=N*Q; #Energy released(MeV)\n",
- "#1MeV=4.45*10^-20kWh\n",
- "E=E*4.45*10**-20;\n",
- "\n",
- "#Result\n",
- "print \"The energy released is\",round(E/10**7,3),\"*10**7 kWh\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The energy released is 2.371 *10**7 kWh\n"
- ]
- }
- ],
- "prompt_number": 16
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example number 11.7, Page number 228"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#importing modules\n",
- "import math\n",
- "from __future__ import division\n",
- "\n",
- "#Variable declaration\n",
- "wt=5000; #weight(gm)\n",
- "A=235; #mass number of uranium\n",
- "Ef=208; #Energy released per fission(MeV)\n",
- "\n",
- "#Calculation \n",
- "N=(6.02*10**23/A)*wt; #number of nuclei in 5 Kg\n",
- "E=N*Ef; #Energy(MeV)\n",
- "E=E*1.6*10**-13; #Energy(J)\n",
- "T=24*60*60; #time\n",
- "P=E/T; #power(MW)\n",
- "\n",
- "#Result\n",
- "print \"The power output of a nuclear reactor is\",round(P/10**6),\"MW\"\n",
- "print \"answer given in the book is wrong\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The power output of a nuclear reactor is 4934.0 MW\n",
- "answer given in the book is wrong\n"
- ]
- }
- ],
- "prompt_number": 19
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example number 11.8, Page number 228"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#importing modules\n",
- "import math\n",
- "from __future__ import division\n",
- "\n",
- "#Variable declaration\n",
- "A=235; #mass number of uranium\n",
- "p=1000; #amount of electric power produced(MW)\n",
- "e=0.32; #energy conversion efficiency of the plant\n",
- "f=200; #fission energy per event(MeV)\n",
- "\n",
- "#Calculation \n",
- "I=p/e; #Input nuclear energy(MW)\n",
- "TE=I*(10**6)*3600*24*365; #total energy(J)\n",
- "EF=f*(10**6)*1.6*10**-19; #Energy released per fission event(J)\n",
- "N=TE/EF; #Number of nuclei required\n",
- "M=N*A/(6.02*10**23); #corresponding mass(g)\n",
- "\n",
- "#Result\n",
- "print \"The amount of uranium required is\",round(M*10**-3,1),\"kg\"\n",
- "print \"answer in the book varies due to rounding off errors\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The amount of uranium required is 1202.2 kg\n",
- "answer in the book varies due to rounding off errors\n"
- ]
- }
- ],
- "prompt_number": 22
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example number 11.9, Page number 229"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#importing modules\n",
- "import math\n",
- "from __future__ import division\n",
- "\n",
- "#Variable declaration\n",
- "q=1.6*10**-19; #charge of the particle(c)\n",
- "B=1; #magnetic field(T)\n",
- "m=1.67*10**-27; #mass of proton(kg)\n",
- "r=0.5; #radius(m)\n",
- "\n",
- "#Calculation \n",
- "omega=(q*B)/m; #angular frequency(radian/s)\n",
- "v=(omega/(2*math.pi))*10**-8; #frequency(MHz)\n",
- "s=omega*r; #speed of proton(m/s)\n",
- "K=(m*(s**2))*(1/2)*6.27*10**12; #kinetic energy of protons emerging from cyclotron(MeV)\n",
- "\n",
- "#Result\n",
- "print \"The frequency of oscillator to accelerate protons is\",round(omega/10**8,2),\"*10**8 radian/s\"\n",
- "print \"The speed of proton is\",round(s/10**7,1),\"*10**7 m/s\"\n",
- "print \"The kinetic energy of protons emerging from the cyclotron is\",int(K),\"MeV\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The frequency of oscillator to accelerate protons is 0.96 *10**8 radian/s\n",
- "The speed of proton is 4.8 *10**7 m/s\n",
- "The kinetic energy of protons emerging from the cyclotron is 12 MeV\n"
- ]
- }
- ],
- "prompt_number": 28
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example number 11.10, Page number 229"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#importing modules\n",
- "import math\n",
- "from __future__ import division\n",
- "\n",
- "#Variable declaration\n",
- "rho=1.83*10**17; #average density of carbon nucleus(kg/m^3)\n",
- "m=12; #mass(u)\n",
- "e=1.66*10**-27;\n",
- "\n",
- "#Calculation \n",
- "r=(m*e/((4/3)*math.pi*rho))**(1/3)*10**15; #radius(fm)\n",
- "\n",
- "#Result\n",
- "print \"The radius is\",round(r,2),\"fm\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The radius is 2.96 fm\n"
- ]
- }
- ],
- "prompt_number": 30
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example number 11.11, Page number 229"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#importing modules\n",
- "import math\n",
- "from __future__ import division\n",
- "\n",
- "#Variable declaration\n",
- "q=1.6*10**-19; #charge of the particle(c)\n",
- "B=5; #magnetic field(T)\n",
- "m=9.1*10**-31; #mass of electron(kg)\n",
- "\n",
- "#Calculation \n",
- "v=(q*B)/(2*math.pi*m); #cyclotron frequency(Hz)\n",
- "\n",
- "#Result\n",
- "print \"cyclotron frequency of electron is\",round(v/10**11,1),\"*10**11 Hz\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "cyclotron frequency of electron is 1.4 *10**11 Hz\n"
- ]
- }
- ],
- "prompt_number": 32
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example number 11.12, Page number 229"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#importing modules\n",
- "import math\n",
- "from __future__ import division\n",
- "\n",
- "#Variable declaration\n",
- "k=1.5; #maximum kinetic energy(MeV)\n",
- "m=1.67*10**-27; #mass of proton(kg)\n",
- "q=1.6*10**-19; #charge of particle(c)\n",
- "r=0.35; #radius(m)\n",
- "\n",
- "#Calculation \n",
- "B=math.sqrt(k*10**6*q*2*m)/(q*r); #magnetic field(T)\n",
- "\n",
- "#Result\n",
- "print \"The mgnetic field is\",round(B,1),\"T\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The mgnetic field is 0.5 T\n"
- ]
- }
- ],
- "prompt_number": 34
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example number 11.13, Page number 229"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#importing modules\n",
- "import math\n",
- "from __future__ import division\n",
- "\n",
- "#Variable declaration\n",
- "m=1.67*10**-27; #mass of proton(kg)\n",
- "q=1.6*10**-19; #charge of particle(q)\n",
- "v=25; #cyclotron frequency(MHz)\n",
- "\n",
- "#Calculation \n",
- "B=(v*10**6*2*math.pi*m)/q; #magnetic field(T)\n",
- "\n",
- "#Result\n",
- "print \"The required magnetic field is\",round(B,4),\"T\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The required magnetic field is 1.6395 T\n"
- ]
- }
- ],
- "prompt_number": 37
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example number 11.14, Page number 229"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#importing modules\n",
- "import math\n",
- "from __future__ import division\n",
- "\n",
- "#Variable declaration\n",
- "v=20; #cyclotron frequency(MHz)\n",
- "B=1.3; #magnetic field(T)\n",
- "\n",
- "#Calculation \n",
- "d=2*math.pi*v*10**6/B; #charge to mass ratio of proton(C/kg)\n",
- "\n",
- "#Result\n",
- "print \"charge to mass ratio of proton is\",round(d/10**6,2),\"*10**6 C/kg\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "charge to mass ratio of proton is 96.66 *10**6 C/kg\n"
- ]
- }
- ],
- "prompt_number": 39
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Engineering_Physics/Chapter11_1.ipynb b/Engineering_Physics/Chapter11_1.ipynb deleted file mode 100755 index 39762859..00000000 --- a/Engineering_Physics/Chapter11_1.ipynb +++ /dev/null @@ -1,620 +0,0 @@ -{
- "metadata": {
- "name": "",
- "signature": "sha256:1d5f4970753c62d94a2bd202867cc0f79046f1baac4b1a42721a5ae6844ad5f4"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "heading",
- "level": 1,
- "metadata": {},
- "source": [
- "11: Nuclear Radiations and Detectors"
- ]
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example number 11.1, Page number 227"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#importing modules\n",
- "import math\n",
- "from __future__ import division\n",
- "\n",
- "#Variable declaration\n",
- "r0=1.2; #radius(fm)\n",
- "A=7; #mass number \n",
- "\n",
- "#Calculation \n",
- "r=r0*A**(1/3);\t #radius of Li(fm) \n",
- "\n",
- "#Result\n",
- "print \"The radius of Li is\",round(r,4),\"fm\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The radius of Li is 2.2955 fm\n"
- ]
- }
- ],
- "prompt_number": 2
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example number 11.2, Page number 227"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#importing modules\n",
- "import math\n",
- "from __future__ import division\n",
- "\n",
- "#Variable declaration\n",
- "M=235.043945; #atomic mass of uranium(u)\n",
- "Z=92; #atomic number of uranium\n",
- "mp=1.007825; #mass of proton(kg)\n",
- "N=143; #no.of neutrons\n",
- "mn=1.008665; #mass of neutron(kg)\n",
- "A=235; #number of nucleons\n",
- "\n",
- "#Calculation \n",
- "B=(((Z*mp)+(N*mn)-(M))/A)*931.5; #Binding energy(MeV)\n",
- "\n",
- "#Result\n",
- "print \"The binding energy per nucleon is\",round(B,3),\"MeV\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The binding energy per nucleon is 7.591 MeV\n"
- ]
- }
- ],
- "prompt_number": 4
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example number 11.3, Page number 227"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#importing modules\n",
- "import math\n",
- "from __future__ import division\n",
- "\n",
- "#Variable declaration\n",
- "#After removing one neutron from Ca(A=43;Z=20) it becomes Ca(A=42;Z=20)\n",
- "M=41.958622; #mass of Ca(A=42;Z=20)(kg)\n",
- "mn=1.008665; #mass of neutron(kg)\n",
- "E=42.95877; #mass of Ca(A=43;Z=20)(kg)\n",
- "\n",
- "#Calculation \n",
- "C=M+mn;\n",
- "D=C-E;\n",
- "B=D*931.5; #Binding energy of neutron(MeV)\n",
- "\n",
- "#Result\n",
- "print \"The binding energy of neutron is\",round(B,4),\"MeV\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The binding energy of neutron is 7.9336 MeV\n"
- ]
- }
- ],
- "prompt_number": 7
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example number 11.4, Page number 227"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#importing modules\n",
- "import math\n",
- "from __future__ import division\n",
- "\n",
- "#Variable declaration\n",
- "mBe=9.012182; #Atomic mass of beryllium(u)\n",
- "mHe=4.002603; #Atomic mass of helium\n",
- "mn=1.008665; #mass of neutron(kg)\n",
- "mC=12.000000; #Atomic mass of carbon\n",
- "\n",
- "#Calculation \n",
- "Q=(mBe+mHe-mn-mC)*931.5 #energy balance of the reaction(MeV)\n",
- "\n",
- "#Result\n",
- "print \"The Q-value is\",round(Q,1),\"MeV\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The Q-value is 5.7 MeV\n"
- ]
- }
- ],
- "prompt_number": 9
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example number 11.5, Page number 227"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#importing modules\n",
- "import math\n",
- "from __future__ import division\n",
- "\n",
- "#Variable declaration\n",
- "mLi=7.016004; #mass of Lithium(A=7)(u)\n",
- "mH=1.007825; #mass of Hydrogen(A=1)(u)\n",
- "mHe=4.002603; #mass of helium(A=4)(u)\n",
- "p=0.5; #energy of proton(MeV)\n",
- "\n",
- "#Calculation \n",
- "Q=(mLi+mH-2*(mHe))*931.5 #energy balance of the reaction(MeV)\n",
- "#The energy of 2 alpha particles is equal to the Q-value + energy of proton.\n",
- "Ealpha=(Q+p)/2; #energy of each alpha particle(MeV)\n",
- "\n",
- "#Result\n",
- "print \"The Q-value of the reaction is\",round(Q,2),\"MeV\"\n",
- "print \"The energy of each alpha particle is\",round(Ealpha,3),\"MeV\"\n",
- "print \"answer for energy in the book varies due to rounding off errors\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The Q-value of the reaction is 17.35 MeV\n",
- "The energy of each alpha particle is 8.924 MeV\n",
- "answer for energy in the book varies due to rounding off errors\n"
- ]
- }
- ],
- "prompt_number": 14
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example number 11.6, Page number 228"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#importing modules\n",
- "import math\n",
- "from __future__ import division\n",
- "\n",
- "#Variable declaration\n",
- "wt=1000; #weight(gm)\n",
- "A=235; #mass number of uranium\n",
- "N=(6.02*10**23/A)*wt; #no.of nuclei in 1kg of uranium\n",
- "Q=208; #energy-balance of the reaction\n",
- "\n",
- "#Calculation \n",
- "E=N*Q; #Energy released(MeV)\n",
- "#1MeV=4.45*10^-20kWh\n",
- "E=E*4.45*10**-20;\n",
- "\n",
- "#Result\n",
- "print \"The energy released is\",round(E/10**7,3),\"*10**7 kWh\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The energy released is 2.371 *10**7 kWh\n"
- ]
- }
- ],
- "prompt_number": 16
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example number 11.7, Page number 228"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#importing modules\n",
- "import math\n",
- "from __future__ import division\n",
- "\n",
- "#Variable declaration\n",
- "wt=5000; #weight(gm)\n",
- "A=235; #mass number of uranium\n",
- "Ef=208; #Energy released per fission(MeV)\n",
- "\n",
- "#Calculation \n",
- "N=(6.02*10**23/A)*wt; #number of nuclei in 5 Kg\n",
- "E=N*Ef; #Energy(MeV)\n",
- "E=E*1.6*10**-13; #Energy(J)\n",
- "T=24*60*60; #time\n",
- "P=E/T; #power(MW)\n",
- "\n",
- "#Result\n",
- "print \"The power output of a nuclear reactor is\",round(P/10**6),\"MW\"\n",
- "print \"answer given in the book is wrong\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The power output of a nuclear reactor is 4934.0 MW\n",
- "answer given in the book is wrong\n"
- ]
- }
- ],
- "prompt_number": 19
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example number 11.8, Page number 228"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#importing modules\n",
- "import math\n",
- "from __future__ import division\n",
- "\n",
- "#Variable declaration\n",
- "A=235; #mass number of uranium\n",
- "p=1000; #amount of electric power produced(MW)\n",
- "e=0.32; #energy conversion efficiency of the plant\n",
- "f=200; #fission energy per event(MeV)\n",
- "\n",
- "#Calculation \n",
- "I=p/e; #Input nuclear energy(MW)\n",
- "TE=I*(10**6)*3600*24*365; #total energy(J)\n",
- "EF=f*(10**6)*1.6*10**-19; #Energy released per fission event(J)\n",
- "N=TE/EF; #Number of nuclei required\n",
- "M=N*A/(6.02*10**23); #corresponding mass(g)\n",
- "\n",
- "#Result\n",
- "print \"The amount of uranium required is\",round(M*10**-3,1),\"kg\"\n",
- "print \"answer in the book varies due to rounding off errors\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The amount of uranium required is 1202.2 kg\n",
- "answer in the book varies due to rounding off errors\n"
- ]
- }
- ],
- "prompt_number": 22
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example number 11.9, Page number 229"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#importing modules\n",
- "import math\n",
- "from __future__ import division\n",
- "\n",
- "#Variable declaration\n",
- "q=1.6*10**-19; #charge of the particle(c)\n",
- "B=1; #magnetic field(T)\n",
- "m=1.67*10**-27; #mass of proton(kg)\n",
- "r=0.5; #radius(m)\n",
- "\n",
- "#Calculation \n",
- "omega=(q*B)/m; #angular frequency(radian/s)\n",
- "v=(omega/(2*math.pi))*10**-8; #frequency(MHz)\n",
- "s=omega*r; #speed of proton(m/s)\n",
- "K=(m*(s**2))*(1/2)*6.27*10**12; #kinetic energy of protons emerging from cyclotron(MeV)\n",
- "\n",
- "#Result\n",
- "print \"The frequency of oscillator to accelerate protons is\",round(omega/10**8,2),\"*10**8 radian/s\"\n",
- "print \"The speed of proton is\",round(s/10**7,1),\"*10**7 m/s\"\n",
- "print \"The kinetic energy of protons emerging from the cyclotron is\",int(K),\"MeV\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The frequency of oscillator to accelerate protons is 0.96 *10**8 radian/s\n",
- "The speed of proton is 4.8 *10**7 m/s\n",
- "The kinetic energy of protons emerging from the cyclotron is 12 MeV\n"
- ]
- }
- ],
- "prompt_number": 28
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example number 11.10, Page number 229"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#importing modules\n",
- "import math\n",
- "from __future__ import division\n",
- "\n",
- "#Variable declaration\n",
- "rho=1.83*10**17; #average density of carbon nucleus(kg/m^3)\n",
- "m=12; #mass(u)\n",
- "e=1.66*10**-27;\n",
- "\n",
- "#Calculation \n",
- "r=(m*e/((4/3)*math.pi*rho))**(1/3)*10**15; #radius(fm)\n",
- "\n",
- "#Result\n",
- "print \"The radius is\",round(r,2),\"fm\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The radius is 2.96 fm\n"
- ]
- }
- ],
- "prompt_number": 30
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example number 11.11, Page number 229"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#importing modules\n",
- "import math\n",
- "from __future__ import division\n",
- "\n",
- "#Variable declaration\n",
- "q=1.6*10**-19; #charge of the particle(c)\n",
- "B=5; #magnetic field(T)\n",
- "m=9.1*10**-31; #mass of electron(kg)\n",
- "\n",
- "#Calculation \n",
- "v=(q*B)/(2*math.pi*m); #cyclotron frequency(Hz)\n",
- "\n",
- "#Result\n",
- "print \"cyclotron frequency of electron is\",round(v/10**11,1),\"*10**11 Hz\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "cyclotron frequency of electron is 1.4 *10**11 Hz\n"
- ]
- }
- ],
- "prompt_number": 32
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example number 11.12, Page number 229"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#importing modules\n",
- "import math\n",
- "from __future__ import division\n",
- "\n",
- "#Variable declaration\n",
- "k=1.5; #maximum kinetic energy(MeV)\n",
- "m=1.67*10**-27; #mass of proton(kg)\n",
- "q=1.6*10**-19; #charge of particle(c)\n",
- "r=0.35; #radius(m)\n",
- "\n",
- "#Calculation \n",
- "B=math.sqrt(k*10**6*q*2*m)/(q*r); #magnetic field(T)\n",
- "\n",
- "#Result\n",
- "print \"The mgnetic field is\",round(B,1),\"T\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The mgnetic field is 0.5 T\n"
- ]
- }
- ],
- "prompt_number": 34
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example number 11.13, Page number 229"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#importing modules\n",
- "import math\n",
- "from __future__ import division\n",
- "\n",
- "#Variable declaration\n",
- "m=1.67*10**-27; #mass of proton(kg)\n",
- "q=1.6*10**-19; #charge of particle(q)\n",
- "v=25; #cyclotron frequency(MHz)\n",
- "\n",
- "#Calculation \n",
- "B=(v*10**6*2*math.pi*m)/q; #magnetic field(T)\n",
- "\n",
- "#Result\n",
- "print \"The required magnetic field is\",round(B,4),\"T\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The required magnetic field is 1.6395 T\n"
- ]
- }
- ],
- "prompt_number": 37
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example number 11.14, Page number 229"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#importing modules\n",
- "import math\n",
- "from __future__ import division\n",
- "\n",
- "#Variable declaration\n",
- "v=20; #cyclotron frequency(MHz)\n",
- "B=1.3; #magnetic field(T)\n",
- "\n",
- "#Calculation \n",
- "d=2*math.pi*v*10**6/B; #charge to mass ratio of proton(C/kg)\n",
- "\n",
- "#Result\n",
- "print \"charge to mass ratio of proton is\",round(d/10**6,2),\"*10**6 C/kg\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "charge to mass ratio of proton is 96.66 *10**6 C/kg\n"
- ]
- }
- ],
- "prompt_number": 39
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Engineering_Physics/Chapter13.ipynb b/Engineering_Physics/Chapter13.ipynb deleted file mode 100755 index 83323d13..00000000 --- a/Engineering_Physics/Chapter13.ipynb +++ /dev/null @@ -1,386 +0,0 @@ -{
- "metadata": {
- "name": "",
- "signature": "sha256:46a82c7fe1b65af7ee26b9fa38521b26c61cee31bc75dd5001fe45442416739c"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "heading",
- "level": 1,
- "metadata": {},
- "source": [
- "13: Optical Fibre"
- ]
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example number 13.1, Page number 250"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#importing modules\n",
- "import math\n",
- "from __future__ import division\n",
- "\n",
- "#Variable declaration\n",
- "n1=1.49; #refractive index of core\n",
- "n2=1.46; #refractive index of cladding\n",
- "\n",
- "#Calculation \n",
- "NA=math.sqrt((n1**2)-(n2**2)); #Numerical aperture\n",
- "\n",
- "#Result\n",
- "print \"The numerical aperture is\",round(NA,1)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The numerical aperture is 0.3\n"
- ]
- }
- ],
- "prompt_number": 2
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example number 13.2, Page number 250"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#importing modules\n",
- "import math\n",
- "from __future__ import division\n",
- "\n",
- "#Variable declaration\n",
- "NA=0.5; #numerical aperture of fibre \n",
- "n0=1; #refractive index of the medium(air)\n",
- "\n",
- "#Calculation \n",
- "i=math.asin(NA/n0); #acceptance angle(radian)\n",
- "i=i*180/math.pi; #angle(degrees)\n",
- "\n",
- "#Result\n",
- "print \"The acceptance angle is\",i,\"degrees\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The acceptance angle is 30.0 degrees\n"
- ]
- }
- ],
- "prompt_number": 3
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example number 13.3, Page number 250"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "\n",
- "#importing modules\n",
- "import math\n",
- "from __future__ import division\n",
- "\n",
- "#Variable declaration\n",
- "NA=0.25; #numerical apperture\n",
- "lamda=0.75; #wavelength(micro m)\n",
- "a=25; #core radius(micro m)\n",
- "\n",
- "#Calculation \n",
- "f=(2*math.pi*a*NA)/lamda; #normalised frequency\n",
- "Ng=(f**2)/2; #number of guided modes\n",
- "\n",
- "#Result\n",
- "print \"The number of guided modes is\",int(Ng)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The number of guided modes is 1370\n"
- ]
- }
- ],
- "prompt_number": 6
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example number 13.4, Page number 250"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#importing modules\n",
- "import math\n",
- "from __future__ import division\n",
- "\n",
- "#Variable declaration\n",
- "pi=100; #mean optical power launched(micro m)\n",
- "po=5; #mean optical power at fibre output(micro W)\n",
- "l=6; #length(km)\n",
- "\n",
- "#Calculation \n",
- "S=10*math.log10(pi/po); #signal attenuation(dB)\n",
- "Sk=S/l; #signal attenuation(dB/km)\n",
- "\n",
- "#Result\n",
- "print \"The signal attenuation is\",round(Sk,3),\"dB/km\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The signal attenuation is 2.168 dB/km\n"
- ]
- }
- ],
- "prompt_number": 8
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example number 13.5, Page number 250"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#importing modules\n",
- "import math\n",
- "from __future__ import division\n",
- "\n",
- "#Variable declaration\n",
- "ns=2.89; #sum of refractive indices of core & cladding\n",
- "nd=0.03; #difference of refractive indices of core & cladding\n",
- "\n",
- "#Calculation \n",
- "NA=math.sqrt(ns*nd); #numerical apperture\n",
- "\n",
- "#Result\n",
- "print \"The numerical apperture for the optical fibre is\",round(NA,2)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The numerical apperture for the optical fibre is 0.29\n"
- ]
- }
- ],
- "prompt_number": 10
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example number 13.6, Page number 250"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#importing modules\n",
- "import math\n",
- "from __future__ import division\n",
- "\n",
- "#Variable declaration\n",
- "NA=0.28; #numerical aperture\n",
- "a=30; #core radius(micro m)\n",
- "lamda=0.8; #wavelength(micro m)\n",
- "\n",
- "#Calculation \n",
- "f=(2*math.pi*a*NA)/lamda; #normalised frequency\n",
- "Ng=f**2/2; #number of guided modes\n",
- "\n",
- "#Result\n",
- "print \"The number of guided modes is\",int(Ng)\n",
- "print \"answer in the book varies due to rounding off errors\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The number of guided modes is 2176\n",
- "answer in the book varies due to rounding off errors\n"
- ]
- }
- ],
- "prompt_number": 14
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example number 13.7, Page number 250"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#importing modules\n",
- "import math\n",
- "from __future__ import division\n",
- "\n",
- "#Variable declaration\n",
- "S=2; #signal attenuation(dB/km)\n",
- "l=1; #length(km)\n",
- "p0=20; #mean optical power at fibre output(micro W)\n",
- "\n",
- "#Calculation \n",
- "pi=p0*10**(S/10); #mean optical power launched into fibre(micro W)\n",
- "\n",
- "#Result\n",
- "print \"The mean optical power launched into a fibre is\",round(pi,1),\"micro W\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The mean optical power launched into a fibre is 31.7 micro W\n"
- ]
- }
- ],
- "prompt_number": 16
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example number 13.8, Page number 251"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#importing modules\n",
- "import math\n",
- "from __future__ import division\n",
- "\n",
- "#Variable declaration\n",
- "S=2.3; #Signal attenuation(dB/km)\n",
- "l=4; #length(km)\n",
- "\n",
- "#Calculation \n",
- "S=S*l; #signal attenuation for 4km in dB\n",
- "P=10**(S/10); #ratio of mean optical power\n",
- "\n",
- "#Result\n",
- "print \"ratio of mean optical power is\",round(P,1)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "ratio of mean optical power is 8.3\n"
- ]
- }
- ],
- "prompt_number": 18
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example number 13.9, Page number 251"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#importing modules\n",
- "import math\n",
- "from __future__ import division\n",
- "\n",
- "#Variable declaration\n",
- "op=1/4; #ratio\n",
- "\n",
- "#Calculation \n",
- "#S=10*log(pi/po)\n",
- "S=10*math.log10(1/op); #signal attenuation(dB)\n",
- "\n",
- "#Result\n",
- "print \"Signal attenuation is\",int(S),\"dB\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Signal attenuation is 6 dB\n"
- ]
- }
- ],
- "prompt_number": 20
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Engineering_Physics/Chapter13_1.ipynb b/Engineering_Physics/Chapter13_1.ipynb deleted file mode 100755 index 83323d13..00000000 --- a/Engineering_Physics/Chapter13_1.ipynb +++ /dev/null @@ -1,386 +0,0 @@ -{
- "metadata": {
- "name": "",
- "signature": "sha256:46a82c7fe1b65af7ee26b9fa38521b26c61cee31bc75dd5001fe45442416739c"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "heading",
- "level": 1,
- "metadata": {},
- "source": [
- "13: Optical Fibre"
- ]
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example number 13.1, Page number 250"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#importing modules\n",
- "import math\n",
- "from __future__ import division\n",
- "\n",
- "#Variable declaration\n",
- "n1=1.49; #refractive index of core\n",
- "n2=1.46; #refractive index of cladding\n",
- "\n",
- "#Calculation \n",
- "NA=math.sqrt((n1**2)-(n2**2)); #Numerical aperture\n",
- "\n",
- "#Result\n",
- "print \"The numerical aperture is\",round(NA,1)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The numerical aperture is 0.3\n"
- ]
- }
- ],
- "prompt_number": 2
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example number 13.2, Page number 250"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#importing modules\n",
- "import math\n",
- "from __future__ import division\n",
- "\n",
- "#Variable declaration\n",
- "NA=0.5; #numerical aperture of fibre \n",
- "n0=1; #refractive index of the medium(air)\n",
- "\n",
- "#Calculation \n",
- "i=math.asin(NA/n0); #acceptance angle(radian)\n",
- "i=i*180/math.pi; #angle(degrees)\n",
- "\n",
- "#Result\n",
- "print \"The acceptance angle is\",i,\"degrees\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The acceptance angle is 30.0 degrees\n"
- ]
- }
- ],
- "prompt_number": 3
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example number 13.3, Page number 250"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "\n",
- "#importing modules\n",
- "import math\n",
- "from __future__ import division\n",
- "\n",
- "#Variable declaration\n",
- "NA=0.25; #numerical apperture\n",
- "lamda=0.75; #wavelength(micro m)\n",
- "a=25; #core radius(micro m)\n",
- "\n",
- "#Calculation \n",
- "f=(2*math.pi*a*NA)/lamda; #normalised frequency\n",
- "Ng=(f**2)/2; #number of guided modes\n",
- "\n",
- "#Result\n",
- "print \"The number of guided modes is\",int(Ng)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The number of guided modes is 1370\n"
- ]
- }
- ],
- "prompt_number": 6
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example number 13.4, Page number 250"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#importing modules\n",
- "import math\n",
- "from __future__ import division\n",
- "\n",
- "#Variable declaration\n",
- "pi=100; #mean optical power launched(micro m)\n",
- "po=5; #mean optical power at fibre output(micro W)\n",
- "l=6; #length(km)\n",
- "\n",
- "#Calculation \n",
- "S=10*math.log10(pi/po); #signal attenuation(dB)\n",
- "Sk=S/l; #signal attenuation(dB/km)\n",
- "\n",
- "#Result\n",
- "print \"The signal attenuation is\",round(Sk,3),\"dB/km\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The signal attenuation is 2.168 dB/km\n"
- ]
- }
- ],
- "prompt_number": 8
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example number 13.5, Page number 250"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#importing modules\n",
- "import math\n",
- "from __future__ import division\n",
- "\n",
- "#Variable declaration\n",
- "ns=2.89; #sum of refractive indices of core & cladding\n",
- "nd=0.03; #difference of refractive indices of core & cladding\n",
- "\n",
- "#Calculation \n",
- "NA=math.sqrt(ns*nd); #numerical apperture\n",
- "\n",
- "#Result\n",
- "print \"The numerical apperture for the optical fibre is\",round(NA,2)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The numerical apperture for the optical fibre is 0.29\n"
- ]
- }
- ],
- "prompt_number": 10
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example number 13.6, Page number 250"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#importing modules\n",
- "import math\n",
- "from __future__ import division\n",
- "\n",
- "#Variable declaration\n",
- "NA=0.28; #numerical aperture\n",
- "a=30; #core radius(micro m)\n",
- "lamda=0.8; #wavelength(micro m)\n",
- "\n",
- "#Calculation \n",
- "f=(2*math.pi*a*NA)/lamda; #normalised frequency\n",
- "Ng=f**2/2; #number of guided modes\n",
- "\n",
- "#Result\n",
- "print \"The number of guided modes is\",int(Ng)\n",
- "print \"answer in the book varies due to rounding off errors\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The number of guided modes is 2176\n",
- "answer in the book varies due to rounding off errors\n"
- ]
- }
- ],
- "prompt_number": 14
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example number 13.7, Page number 250"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#importing modules\n",
- "import math\n",
- "from __future__ import division\n",
- "\n",
- "#Variable declaration\n",
- "S=2; #signal attenuation(dB/km)\n",
- "l=1; #length(km)\n",
- "p0=20; #mean optical power at fibre output(micro W)\n",
- "\n",
- "#Calculation \n",
- "pi=p0*10**(S/10); #mean optical power launched into fibre(micro W)\n",
- "\n",
- "#Result\n",
- "print \"The mean optical power launched into a fibre is\",round(pi,1),\"micro W\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The mean optical power launched into a fibre is 31.7 micro W\n"
- ]
- }
- ],
- "prompt_number": 16
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example number 13.8, Page number 251"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#importing modules\n",
- "import math\n",
- "from __future__ import division\n",
- "\n",
- "#Variable declaration\n",
- "S=2.3; #Signal attenuation(dB/km)\n",
- "l=4; #length(km)\n",
- "\n",
- "#Calculation \n",
- "S=S*l; #signal attenuation for 4km in dB\n",
- "P=10**(S/10); #ratio of mean optical power\n",
- "\n",
- "#Result\n",
- "print \"ratio of mean optical power is\",round(P,1)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "ratio of mean optical power is 8.3\n"
- ]
- }
- ],
- "prompt_number": 18
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example number 13.9, Page number 251"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#importing modules\n",
- "import math\n",
- "from __future__ import division\n",
- "\n",
- "#Variable declaration\n",
- "op=1/4; #ratio\n",
- "\n",
- "#Calculation \n",
- "#S=10*log(pi/po)\n",
- "S=10*math.log10(1/op); #signal attenuation(dB)\n",
- "\n",
- "#Result\n",
- "print \"Signal attenuation is\",int(S),\"dB\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Signal attenuation is 6 dB\n"
- ]
- }
- ],
- "prompt_number": 20
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Engineering_Physics/Chapter1_1.ipynb b/Engineering_Physics/Chapter1_1.ipynb deleted file mode 100755 index e69de29b..00000000 --- a/Engineering_Physics/Chapter1_1.ipynb +++ /dev/null diff --git a/Engineering_Physics/Chapter2.ipynb b/Engineering_Physics/Chapter2.ipynb deleted file mode 100755 index b527497e..00000000 --- a/Engineering_Physics/Chapter2.ipynb +++ /dev/null @@ -1,431 +0,0 @@ -{
- "metadata": {
- "name": "",
- "signature": "sha256:ad2a8335720a8ceacd5db30334fb790c6a4c8fa5b69e23fcad6e232d80ed69c2"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "heading",
- "level": 1,
- "metadata": {},
- "source": [
- "2: Crystal Structure"
- ]
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example number 2.1, Page number 29"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#importing modules\n",
- "import math\n",
- "from __future__ import division\n",
- "\n",
- "#Variable declaration\n",
- "d=6.5*10**3; #density of silver bromide(Kg/m**3)\n",
- "m=187.8; #molecular weight of silver bromide\n",
- "\n",
- "#Calculation\n",
- "M=(4*m)/(6.022*10**26); #mass of ion in unit cell(kg)\n",
- "#d=mass of ions in unit cell/volume of unit cell\n",
- "a=((M/d)**(1/3))*10**10; #lattice parameter(Angstrom)\n",
- "\n",
- "#Result\n",
- "print \"The lattice parameter is\",round(a,2),\"Angstrom\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The lattice parameter is 5.77 Angstrom\n"
- ]
- }
- ],
- "prompt_number": 3
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example number 2.2, Page number 29"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#importing modules\n",
- "import math\n",
- "from __future__ import division\n",
- "\n",
- "#Variable declaration\n",
- "r=2.3; #atomic radius(Angstrom)\n",
- "\n",
- "#Calculation\n",
- "a=(4*r)/math.sqrt(3); \n",
- "fv=((a)**3-(2*(4/3)*math.pi*r**3))*10**-30; #free volume(m**3)\n",
- "\n",
- "#Result\n",
- "print \"The free volume per unit cell is\",round(fv*10**30,1),\"*10**-30 m**3\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The free volume per unit cell is 47.9 *10**-30 m**3\n"
- ]
- }
- ],
- "prompt_number": 6
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example number 2.3, Page number 29"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#importing modules\n",
- "import math\n",
- "from __future__ import division\n",
- "\n",
- "#Variable declaration\n",
- "k=8.625*10**-5; #Boltzmann constant(eV/K)\n",
- "#n1000/n500=ln[n1000/n500]=Ev/1000k\n",
- "Ev=1.08; #average energy required to create a vacancy(eV)\n",
- "\n",
- "#Calculation\n",
- "N=math.exp(Ev/(1000*k)); #ratio of vacancies\n",
- "\n",
- "#Result\n",
- "print \"The ratio of vacancies is\",round(N/10**5,1),\"*10**5\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The ratio of vacancies is 2.7 *10**5\n"
- ]
- }
- ],
- "prompt_number": 9
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example number 2.4, Page number 29"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#importing modules\n",
- "import math\n",
- "from __future__ import division\n",
- "\n",
- "#Variable declaration\n",
- "#n500=N*exp(-Ev/500k)\n",
- "k=8.625*10**-5; #Boltzmann constant(eV/K)\n",
- "Ev=0.95; #average energy required to create a vacancy\n",
- "\n",
- "#Calculation\n",
- "n=math.exp(-Ev/(500*k)); #n500/N\n",
- "\n",
- "#Result\n",
- "print \"The ratio of number of vacancies to the number of atoms is\",round(n*10**10,1),\"*10**-10\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The ratio of number of vacancies to the number of atoms is 2.7 *10**-10\n"
- ]
- }
- ],
- "prompt_number": 12
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example number 2.5, Page number 30"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#importing modules\n",
- "import math\n",
- "from __future__ import division\n",
- "\n",
- "#Variable declaration\n",
- "h=1;\n",
- "k=1;\n",
- "l=1;\n",
- "#d(hkl)=a/sqrt(h^2+k^2+l^2) where \"a\" is the lattice parameter\n",
- "r=0.18; #atomic radius(nm)\n",
- "\n",
- "#Calculation\n",
- "d111=(2*math.sqrt(2)*r)/math.sqrt((h**2)+(k**2)+(l**2)); #spacing(nm)\n",
- "\n",
- "#Result\n",
- "print \"The spacing of (111) planes in a monoatomic structure is\",round(d111,2),\"nm\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The spacing of (111) planes in a monoatomic structure is 0.29 nm\n"
- ]
- }
- ],
- "prompt_number": 14
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example number 2.6, Page number 30"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#importing modules\n",
- "import math\n",
- "from __future__ import division\n",
- "\n",
- "#Variable declaration\n",
- "M=200; #atomic weight \n",
- "a=5; #lattice parameter(angstrom)\n",
- "Na=6.022*(10**26);\n",
- "\n",
- "#Calculation\n",
- "rho=(2*M)/(Na*(a*10**-10)**3); #density of the structure(kg/m**3)\n",
- "\n",
- "#Result\n",
- "print \"The density of the bcc structure is\",round(rho/10**3,2),\"*10**3 kg/m**3\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The density of the bcc structure is 5.31 *10**3 kg/m**3\n"
- ]
- }
- ],
- "prompt_number": 17
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example number 2.7, Page number 30"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#importing modules\n",
- "import math\n",
- "from __future__ import division\n",
- "\n",
- "#Variable declaration\n",
- "#Free volume=a^3-[(4/3)*pi*r^3];for sc,a=2r\n",
- "#Therefore free volume =(2r)^3-[(4/3)*pi*r^3]\n",
- "fv=30.48*10**-30; #free volume per unit cell(m**3)\n",
- "\n",
- "#Calculation\n",
- "r=(fv/(8-(4/3)*math.pi))**(1/3)*(10**10); #atomic radius(angstrom) \n",
- "\n",
- "#Result\n",
- "print \"The atomic radius is\",round(r),\"Angstrom\" "
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The atomic radius is 2.0 Angstrom\n"
- ]
- }
- ],
- "prompt_number": 21
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example number 2.8, Page number 30"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#importing modules\n",
- "import math\n",
- "from __future__ import division\n",
- "\n",
- "#Variable declaration\n",
- "#free volume=a^3-[2*(4/3)*pi*r^3]\n",
- "#for bcc a=4r/3^(1/3)\n",
- "fv=61.72*(10**-30); #free volume(m**3)\n",
- "\n",
- "#Calculation\n",
- "a=-(fv/(1-math.pi*math.sqrt(3))/8)**1/3*10**31; #lattice parameter(angstrom)\n",
- "\n",
- "#Result\n",
- "print \"The lattice parameter is\",round(a,2),\"Angstrom\"\n",
- "print \"answer in the book varies due to rounding off errors\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The lattice parameter is 5.79 Angstrom\n",
- "answer in the book varies due to rounding off errors\n"
- ]
- }
- ],
- "prompt_number": 24
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example number 2.9, Page number 30"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#importing modules\n",
- "import math\n",
- "from __future__ import division\n",
- "\n",
- "#Variable declaration\n",
- "rho=9000; #density(kg/m**3)\n",
- "w=65; #atomic weight\n",
- "v=1; #volume(m**3)\n",
- "a=1.4; #average number of free electrons per atom\n",
- "\n",
- "#Calculation\n",
- "n=(rho*v)/(w/(6.022*10**26)); #number of atoms\n",
- "rhoe=n*a; #density of free electrons per atom(electrons/m**3)\n",
- "\n",
- "#Result\n",
- "print \"The density of free electrons is\",round(rhoe/10**29,3),\"*10**29 electrons/m**3\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The density of free electrons is 1.167 *10**29 electrons/m**3\n"
- ]
- }
- ],
- "prompt_number": 26
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example number 2.10, Page number 31"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#importing modules\n",
- "import math\n",
- "from __future__ import division\n",
- "\n",
- "#Variable declaration\n",
- "h=1;\n",
- "k=0;\n",
- "l=1;\n",
- "d101=0.5; #spacing of (101) plane\n",
- "\n",
- "#Calculation\n",
- "#d101=a/sqrt((h^2)+(k^2)+(l^2))\n",
- "a=d101*math.sqrt(2) #lattice parameter(Angstrom)\n",
- "\n",
- "#Result\n",
- "print \"The lattice parameter is\",round(a,1),\"Angstrom\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The lattice parameter is 0.7 Angstrom\n"
- ]
- }
- ],
- "prompt_number": 28
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Engineering_Physics/Chapter2_1.ipynb b/Engineering_Physics/Chapter2_1.ipynb deleted file mode 100755 index b527497e..00000000 --- a/Engineering_Physics/Chapter2_1.ipynb +++ /dev/null @@ -1,431 +0,0 @@ -{
- "metadata": {
- "name": "",
- "signature": "sha256:ad2a8335720a8ceacd5db30334fb790c6a4c8fa5b69e23fcad6e232d80ed69c2"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "heading",
- "level": 1,
- "metadata": {},
- "source": [
- "2: Crystal Structure"
- ]
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example number 2.1, Page number 29"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#importing modules\n",
- "import math\n",
- "from __future__ import division\n",
- "\n",
- "#Variable declaration\n",
- "d=6.5*10**3; #density of silver bromide(Kg/m**3)\n",
- "m=187.8; #molecular weight of silver bromide\n",
- "\n",
- "#Calculation\n",
- "M=(4*m)/(6.022*10**26); #mass of ion in unit cell(kg)\n",
- "#d=mass of ions in unit cell/volume of unit cell\n",
- "a=((M/d)**(1/3))*10**10; #lattice parameter(Angstrom)\n",
- "\n",
- "#Result\n",
- "print \"The lattice parameter is\",round(a,2),\"Angstrom\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The lattice parameter is 5.77 Angstrom\n"
- ]
- }
- ],
- "prompt_number": 3
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example number 2.2, Page number 29"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#importing modules\n",
- "import math\n",
- "from __future__ import division\n",
- "\n",
- "#Variable declaration\n",
- "r=2.3; #atomic radius(Angstrom)\n",
- "\n",
- "#Calculation\n",
- "a=(4*r)/math.sqrt(3); \n",
- "fv=((a)**3-(2*(4/3)*math.pi*r**3))*10**-30; #free volume(m**3)\n",
- "\n",
- "#Result\n",
- "print \"The free volume per unit cell is\",round(fv*10**30,1),\"*10**-30 m**3\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The free volume per unit cell is 47.9 *10**-30 m**3\n"
- ]
- }
- ],
- "prompt_number": 6
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example number 2.3, Page number 29"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#importing modules\n",
- "import math\n",
- "from __future__ import division\n",
- "\n",
- "#Variable declaration\n",
- "k=8.625*10**-5; #Boltzmann constant(eV/K)\n",
- "#n1000/n500=ln[n1000/n500]=Ev/1000k\n",
- "Ev=1.08; #average energy required to create a vacancy(eV)\n",
- "\n",
- "#Calculation\n",
- "N=math.exp(Ev/(1000*k)); #ratio of vacancies\n",
- "\n",
- "#Result\n",
- "print \"The ratio of vacancies is\",round(N/10**5,1),\"*10**5\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The ratio of vacancies is 2.7 *10**5\n"
- ]
- }
- ],
- "prompt_number": 9
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example number 2.4, Page number 29"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#importing modules\n",
- "import math\n",
- "from __future__ import division\n",
- "\n",
- "#Variable declaration\n",
- "#n500=N*exp(-Ev/500k)\n",
- "k=8.625*10**-5; #Boltzmann constant(eV/K)\n",
- "Ev=0.95; #average energy required to create a vacancy\n",
- "\n",
- "#Calculation\n",
- "n=math.exp(-Ev/(500*k)); #n500/N\n",
- "\n",
- "#Result\n",
- "print \"The ratio of number of vacancies to the number of atoms is\",round(n*10**10,1),\"*10**-10\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The ratio of number of vacancies to the number of atoms is 2.7 *10**-10\n"
- ]
- }
- ],
- "prompt_number": 12
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example number 2.5, Page number 30"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#importing modules\n",
- "import math\n",
- "from __future__ import division\n",
- "\n",
- "#Variable declaration\n",
- "h=1;\n",
- "k=1;\n",
- "l=1;\n",
- "#d(hkl)=a/sqrt(h^2+k^2+l^2) where \"a\" is the lattice parameter\n",
- "r=0.18; #atomic radius(nm)\n",
- "\n",
- "#Calculation\n",
- "d111=(2*math.sqrt(2)*r)/math.sqrt((h**2)+(k**2)+(l**2)); #spacing(nm)\n",
- "\n",
- "#Result\n",
- "print \"The spacing of (111) planes in a monoatomic structure is\",round(d111,2),\"nm\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The spacing of (111) planes in a monoatomic structure is 0.29 nm\n"
- ]
- }
- ],
- "prompt_number": 14
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example number 2.6, Page number 30"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#importing modules\n",
- "import math\n",
- "from __future__ import division\n",
- "\n",
- "#Variable declaration\n",
- "M=200; #atomic weight \n",
- "a=5; #lattice parameter(angstrom)\n",
- "Na=6.022*(10**26);\n",
- "\n",
- "#Calculation\n",
- "rho=(2*M)/(Na*(a*10**-10)**3); #density of the structure(kg/m**3)\n",
- "\n",
- "#Result\n",
- "print \"The density of the bcc structure is\",round(rho/10**3,2),\"*10**3 kg/m**3\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The density of the bcc structure is 5.31 *10**3 kg/m**3\n"
- ]
- }
- ],
- "prompt_number": 17
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example number 2.7, Page number 30"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#importing modules\n",
- "import math\n",
- "from __future__ import division\n",
- "\n",
- "#Variable declaration\n",
- "#Free volume=a^3-[(4/3)*pi*r^3];for sc,a=2r\n",
- "#Therefore free volume =(2r)^3-[(4/3)*pi*r^3]\n",
- "fv=30.48*10**-30; #free volume per unit cell(m**3)\n",
- "\n",
- "#Calculation\n",
- "r=(fv/(8-(4/3)*math.pi))**(1/3)*(10**10); #atomic radius(angstrom) \n",
- "\n",
- "#Result\n",
- "print \"The atomic radius is\",round(r),\"Angstrom\" "
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The atomic radius is 2.0 Angstrom\n"
- ]
- }
- ],
- "prompt_number": 21
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example number 2.8, Page number 30"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#importing modules\n",
- "import math\n",
- "from __future__ import division\n",
- "\n",
- "#Variable declaration\n",
- "#free volume=a^3-[2*(4/3)*pi*r^3]\n",
- "#for bcc a=4r/3^(1/3)\n",
- "fv=61.72*(10**-30); #free volume(m**3)\n",
- "\n",
- "#Calculation\n",
- "a=-(fv/(1-math.pi*math.sqrt(3))/8)**1/3*10**31; #lattice parameter(angstrom)\n",
- "\n",
- "#Result\n",
- "print \"The lattice parameter is\",round(a,2),\"Angstrom\"\n",
- "print \"answer in the book varies due to rounding off errors\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The lattice parameter is 5.79 Angstrom\n",
- "answer in the book varies due to rounding off errors\n"
- ]
- }
- ],
- "prompt_number": 24
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example number 2.9, Page number 30"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#importing modules\n",
- "import math\n",
- "from __future__ import division\n",
- "\n",
- "#Variable declaration\n",
- "rho=9000; #density(kg/m**3)\n",
- "w=65; #atomic weight\n",
- "v=1; #volume(m**3)\n",
- "a=1.4; #average number of free electrons per atom\n",
- "\n",
- "#Calculation\n",
- "n=(rho*v)/(w/(6.022*10**26)); #number of atoms\n",
- "rhoe=n*a; #density of free electrons per atom(electrons/m**3)\n",
- "\n",
- "#Result\n",
- "print \"The density of free electrons is\",round(rhoe/10**29,3),\"*10**29 electrons/m**3\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The density of free electrons is 1.167 *10**29 electrons/m**3\n"
- ]
- }
- ],
- "prompt_number": 26
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example number 2.10, Page number 31"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#importing modules\n",
- "import math\n",
- "from __future__ import division\n",
- "\n",
- "#Variable declaration\n",
- "h=1;\n",
- "k=0;\n",
- "l=1;\n",
- "d101=0.5; #spacing of (101) plane\n",
- "\n",
- "#Calculation\n",
- "#d101=a/sqrt((h^2)+(k^2)+(l^2))\n",
- "a=d101*math.sqrt(2) #lattice parameter(Angstrom)\n",
- "\n",
- "#Result\n",
- "print \"The lattice parameter is\",round(a,1),\"Angstrom\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The lattice parameter is 0.7 Angstrom\n"
- ]
- }
- ],
- "prompt_number": 28
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Engineering_Physics/Chapter4.ipynb b/Engineering_Physics/Chapter4.ipynb deleted file mode 100755 index f4145c55..00000000 --- a/Engineering_Physics/Chapter4.ipynb +++ /dev/null @@ -1,553 +0,0 @@ -{
- "metadata": {
- "name": "",
- "signature": "sha256:e9b50f0b4ca0520935774156fedb1fdaaf2b2fd5241b8184a650d42b25d657cd"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "heading",
- "level": 1,
- "metadata": {},
- "source": [
- "4: Interference"
- ]
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example number 4.1, Page number 69"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#importing modules\n",
- "import math\n",
- "from __future__ import division\n",
- "\n",
- "#Variable declaration\n",
- "i=40; #angle of incidence(degrees)\n",
- "mew=1.2; #refractive index\n",
- "t=0.23; #thickness of the film(micro m)\n",
- "\n",
- "#Calculation\n",
- "i=i*math.pi/180; #angle of incidence(radian)\n",
- "r=math.asin(math.sin(i)/mew); #angle of refraction(radian)\n",
- "lambda1=(2*mew*t*math.cos(r))*10**3; #wavelength absent(nm) \n",
- "lambda2=lambda1/2;\n",
- "\n",
- "#Result\n",
- "print \"The wavelength absent is\",round(lambda1,1),\"nm\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The wavelength absent is 466.1 nm\n"
- ]
- }
- ],
- "prompt_number": 2
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example number 4.2, Page number 69"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#importing modules\n",
- "import math\n",
- "from __future__ import division\n",
- "\n",
- "#Variable declaration\n",
- "lambda1=400*10**-9; #wavelength 1(m)\n",
- "lambda2=600*10**-9; #wavelength 2(m)\n",
- "#2*t=n*lambda\n",
- "n=150; \n",
- "\n",
- "#Calculation \n",
- "t=((n*lambda2)/2)*10**6; #thickness of the air film(micro meter)\n",
- "\n",
- "#Result\n",
- "print \"The thickness of the air film is\",t,\"micro m\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The thickness of the air film is 45.0 micro m\n"
- ]
- }
- ],
- "prompt_number": 4
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example number 4.3, Page number 70"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#importing modules\n",
- "import math\n",
- "from __future__ import division\n",
- "\n",
- "#Variable declaration\n",
- "lamda=600*10**-9; #wavelength(m)\n",
- "mew=2;\n",
- "theta=0.025; #wedge-angle(degrees)\n",
- "\n",
- "#Calculation \n",
- "theta=theta*math.pi/180; #wedge-angle(radian)\n",
- "x=(lamda/(2*mew*math.sin(theta)))*10**2; #bandwidth(cm)\n",
- "\n",
- "#Result\n",
- "print \"The bandwidth is\",round(x,3),\"cm\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The bandwidth is 0.034 cm\n"
- ]
- }
- ],
- "prompt_number": 10
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example number 4.4, Page number 70"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#importing modules\n",
- "import math\n",
- "from __future__ import division\n",
- "\n",
- "#Variable declaration\n",
- "xair=0.15; #bandwidth of air(cm)\n",
- "xliq=0.115; #bandwidth of liquid(cm)\n",
- "mewair=1; #refractive index of air\n",
- "\n",
- "#Calculation \n",
- "mewliq=(xair*mewair)/xliq; #refractive index of liquid\n",
- "\n",
- "#Result\n",
- "print \"The refractive index of liquid is\",round(mewliq,1)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The refractive index of liquid is 1.3\n"
- ]
- }
- ],
- "prompt_number": 12
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example number 4.5, Page number 70"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#importing modules\n",
- "import math\n",
- "from __future__ import division\n",
- "\n",
- "#Variable declaration\n",
- "n=9;\n",
- "lamda=589*10**-9; #wavelength of light used(m)\n",
- "R=0.95; #radius of curvature of lens(m)\n",
- "mew=1;\n",
- "\n",
- "#Calculation \n",
- "D=(math.sqrt((4*n*lamda*R)/mew))*10**2; #diameter of the ninth dark ring(m)\n",
- "\n",
- "#Result\n",
- "print \"The diameter of the ninth dark ring is\",round(D,2),\"cm\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The diameter of the ninth dark ring is 0.45 cm\n"
- ]
- }
- ],
- "prompt_number": 15
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example number 4.6, Page number 70"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#importing modules\n",
- "import math\n",
- "from __future__ import division\n",
- "\n",
- "#Variable declaration\n",
- "x=0.055; #distance in fringe shift(mm)\n",
- "n=200; #number of fringes\n",
- "\n",
- "#Calculation \n",
- "lamda=((2*x)/n)*10**6; #wavelength(nm)\n",
- "\n",
- "#Result\n",
- "print \"The wavelength of light used is\",lamda,\"nm\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The wavelength of light used is 550.0 nm\n"
- ]
- }
- ],
- "prompt_number": 17
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example number 4.7, Page number 70"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#importing modules\n",
- "import math\n",
- "from __future__ import division\n",
- "\n",
- "#Variable declaration\n",
- "n=50; #number of fringes\n",
- "lamda=500*10**-9; #wavelength of light used(m)\n",
- "mew=1.5; #refractive index of the plate\n",
- "\n",
- "#Calculation \n",
- "t=((n*lamda)/(2*(mew-1)))*10**6; #thickness of the plate(micro meter)\n",
- "\n",
- "#Result\n",
- "print \"The thickness of the plate is\",t,\"micro m\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The thickness of the plate is 25.0 micro m\n"
- ]
- }
- ],
- "prompt_number": 20
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example number 4.8, Page number 70"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#importing modules\n",
- "import math\n",
- "from __future__ import division\n",
- "\n",
- "#Variable declaration\n",
- "lamda=550*10**-9; #wavelength(m)\n",
- "mew=1.38; #refractive index\n",
- "\n",
- "#Calculation \n",
- "t=(lamda/(4*mew))*10**9; #thickness(nm)\n",
- "\n",
- "#Result\n",
- "print \"The minimum thickness of the plate for normal incidence of light is\",round(t,3),\"nm\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The minimum thickness of the plate for normal incidence of light is 99.638 nm\n"
- ]
- }
- ],
- "prompt_number": 23
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example number 4.9, Page number 70"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#importing modules\n",
- "import math\n",
- "from __future__ import division\n",
- "\n",
- "#Variable declaration\n",
- "i=35; #angle of incidence(degrees)\n",
- "mew=1.4; #refractive index\n",
- "n=50; \n",
- "lamda=459*10**-9; #wavelength(m)\n",
- "\n",
- "#Calculation \n",
- "i=i*math.pi/180; #angle of incidence(radian)\n",
- "r=math.asin(math.sin(i)/mew); #angle of refraction(radian)\n",
- "#2*mew*cos(r)=n*lambda\n",
- "#n(459)=(n+1)450\n",
- "t=(n*lamda/(2*mew*math.cos(r)))*10**6; #thickness of the film(micro meter)\n",
- "\n",
- "#Result\n",
- "print \"The thickness of the film is\",round(t,3),\"micro m\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The thickness of the film is 8.985 micro m\n"
- ]
- }
- ],
- "prompt_number": 26
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example number 4.10, Page number 71"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#importing modules\n",
- "import math\n",
- "from __future__ import division\n",
- "\n",
- "#Variable declaration\n",
- "lamda=500*10**-9; #wavelength(m)\n",
- "x=0.07; #observed band width(cm)\n",
- "mew=1; #refractive index\n",
- "\n",
- "#Calculation \n",
- "theta=(math.asin(lamda/(2*mew*x)))*10**2; #wedge angle(radian)\n",
- "theta=theta*180/math.pi; #wedge angle(degrees)\n",
- "\n",
- "#Result\n",
- "print \"The wedge angle is\",round(theta,2),\"degrees\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The wedge angle is 0.02 degrees\n"
- ]
- }
- ],
- "prompt_number": 31
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example number 4.11, Page number 71"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#importing modules\n",
- "import math\n",
- "from __future__ import division\n",
- "\n",
- "#Variable declaration\n",
- "dair=0.42; #diameter of certain rings(cm)\n",
- "dliq=0.38; #diameter of rings when liquid is introduced(cm)\n",
- "\n",
- "#Calculation \n",
- "mew=dair**2/dliq**2; #refractive index of liquid\n",
- "\n",
- "#Result\n",
- "print \"The refravtive index of liquid is\",round(mew,2)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The refravtive index of liquid is 1.22\n"
- ]
- }
- ],
- "prompt_number": 33
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example number 4.12, Page number 71"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#importing modules\n",
- "import math\n",
- "from __future__ import division\n",
- "\n",
- "#Variable declaration\n",
- "m=8; #eigth ring\n",
- "n=3; #third ring\n",
- "dm=0.4; #diameter of the eigth ring(cm)\n",
- "dn=0.2; #diameter of the third ring(cm)\n",
- "R=101; #Radius of curvature(cm)\n",
- "\n",
- "#Calculation \n",
- "lamda=(((dm**2)-(dn**2))/(4*R*(m-n))); #wavelength of light(cm) \n",
- "\n",
- "#Result\n",
- "print \"The wavelength of light used is\",round(lamda*10**5,4),\"*10**-5 cm\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The wavelength of light used is 5.9406 *10**-5 cm\n"
- ]
- }
- ],
- "prompt_number": 39
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example number 4.13, Page number 71"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#importing modules\n",
- "import math\n",
- "from __future__ import division\n",
- "\n",
- "#Variable declaration\n",
- "mew=1.38; #refractive index of magnesium floride\n",
- "t=175; #thickness of coating of magnesium fluoride(nm)\n",
- "\n",
- "#Calculation \n",
- "lamda=4*t*mew; #wavelength(nm)\n",
- "\n",
- "#Result\n",
- "print \"The wavelength which has high transmission is\",lamda,\"nm\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The wavelength which has high transmission is 966.0 nm\n"
- ]
- }
- ],
- "prompt_number": 41
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Engineering_Physics/Chapter4_1.ipynb b/Engineering_Physics/Chapter4_1.ipynb deleted file mode 100755 index f4145c55..00000000 --- a/Engineering_Physics/Chapter4_1.ipynb +++ /dev/null @@ -1,553 +0,0 @@ -{
- "metadata": {
- "name": "",
- "signature": "sha256:e9b50f0b4ca0520935774156fedb1fdaaf2b2fd5241b8184a650d42b25d657cd"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "heading",
- "level": 1,
- "metadata": {},
- "source": [
- "4: Interference"
- ]
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example number 4.1, Page number 69"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#importing modules\n",
- "import math\n",
- "from __future__ import division\n",
- "\n",
- "#Variable declaration\n",
- "i=40; #angle of incidence(degrees)\n",
- "mew=1.2; #refractive index\n",
- "t=0.23; #thickness of the film(micro m)\n",
- "\n",
- "#Calculation\n",
- "i=i*math.pi/180; #angle of incidence(radian)\n",
- "r=math.asin(math.sin(i)/mew); #angle of refraction(radian)\n",
- "lambda1=(2*mew*t*math.cos(r))*10**3; #wavelength absent(nm) \n",
- "lambda2=lambda1/2;\n",
- "\n",
- "#Result\n",
- "print \"The wavelength absent is\",round(lambda1,1),\"nm\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The wavelength absent is 466.1 nm\n"
- ]
- }
- ],
- "prompt_number": 2
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example number 4.2, Page number 69"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#importing modules\n",
- "import math\n",
- "from __future__ import division\n",
- "\n",
- "#Variable declaration\n",
- "lambda1=400*10**-9; #wavelength 1(m)\n",
- "lambda2=600*10**-9; #wavelength 2(m)\n",
- "#2*t=n*lambda\n",
- "n=150; \n",
- "\n",
- "#Calculation \n",
- "t=((n*lambda2)/2)*10**6; #thickness of the air film(micro meter)\n",
- "\n",
- "#Result\n",
- "print \"The thickness of the air film is\",t,\"micro m\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The thickness of the air film is 45.0 micro m\n"
- ]
- }
- ],
- "prompt_number": 4
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example number 4.3, Page number 70"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#importing modules\n",
- "import math\n",
- "from __future__ import division\n",
- "\n",
- "#Variable declaration\n",
- "lamda=600*10**-9; #wavelength(m)\n",
- "mew=2;\n",
- "theta=0.025; #wedge-angle(degrees)\n",
- "\n",
- "#Calculation \n",
- "theta=theta*math.pi/180; #wedge-angle(radian)\n",
- "x=(lamda/(2*mew*math.sin(theta)))*10**2; #bandwidth(cm)\n",
- "\n",
- "#Result\n",
- "print \"The bandwidth is\",round(x,3),\"cm\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The bandwidth is 0.034 cm\n"
- ]
- }
- ],
- "prompt_number": 10
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example number 4.4, Page number 70"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#importing modules\n",
- "import math\n",
- "from __future__ import division\n",
- "\n",
- "#Variable declaration\n",
- "xair=0.15; #bandwidth of air(cm)\n",
- "xliq=0.115; #bandwidth of liquid(cm)\n",
- "mewair=1; #refractive index of air\n",
- "\n",
- "#Calculation \n",
- "mewliq=(xair*mewair)/xliq; #refractive index of liquid\n",
- "\n",
- "#Result\n",
- "print \"The refractive index of liquid is\",round(mewliq,1)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The refractive index of liquid is 1.3\n"
- ]
- }
- ],
- "prompt_number": 12
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example number 4.5, Page number 70"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#importing modules\n",
- "import math\n",
- "from __future__ import division\n",
- "\n",
- "#Variable declaration\n",
- "n=9;\n",
- "lamda=589*10**-9; #wavelength of light used(m)\n",
- "R=0.95; #radius of curvature of lens(m)\n",
- "mew=1;\n",
- "\n",
- "#Calculation \n",
- "D=(math.sqrt((4*n*lamda*R)/mew))*10**2; #diameter of the ninth dark ring(m)\n",
- "\n",
- "#Result\n",
- "print \"The diameter of the ninth dark ring is\",round(D,2),\"cm\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The diameter of the ninth dark ring is 0.45 cm\n"
- ]
- }
- ],
- "prompt_number": 15
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example number 4.6, Page number 70"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#importing modules\n",
- "import math\n",
- "from __future__ import division\n",
- "\n",
- "#Variable declaration\n",
- "x=0.055; #distance in fringe shift(mm)\n",
- "n=200; #number of fringes\n",
- "\n",
- "#Calculation \n",
- "lamda=((2*x)/n)*10**6; #wavelength(nm)\n",
- "\n",
- "#Result\n",
- "print \"The wavelength of light used is\",lamda,\"nm\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The wavelength of light used is 550.0 nm\n"
- ]
- }
- ],
- "prompt_number": 17
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example number 4.7, Page number 70"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#importing modules\n",
- "import math\n",
- "from __future__ import division\n",
- "\n",
- "#Variable declaration\n",
- "n=50; #number of fringes\n",
- "lamda=500*10**-9; #wavelength of light used(m)\n",
- "mew=1.5; #refractive index of the plate\n",
- "\n",
- "#Calculation \n",
- "t=((n*lamda)/(2*(mew-1)))*10**6; #thickness of the plate(micro meter)\n",
- "\n",
- "#Result\n",
- "print \"The thickness of the plate is\",t,\"micro m\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The thickness of the plate is 25.0 micro m\n"
- ]
- }
- ],
- "prompt_number": 20
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example number 4.8, Page number 70"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#importing modules\n",
- "import math\n",
- "from __future__ import division\n",
- "\n",
- "#Variable declaration\n",
- "lamda=550*10**-9; #wavelength(m)\n",
- "mew=1.38; #refractive index\n",
- "\n",
- "#Calculation \n",
- "t=(lamda/(4*mew))*10**9; #thickness(nm)\n",
- "\n",
- "#Result\n",
- "print \"The minimum thickness of the plate for normal incidence of light is\",round(t,3),\"nm\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The minimum thickness of the plate for normal incidence of light is 99.638 nm\n"
- ]
- }
- ],
- "prompt_number": 23
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example number 4.9, Page number 70"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#importing modules\n",
- "import math\n",
- "from __future__ import division\n",
- "\n",
- "#Variable declaration\n",
- "i=35; #angle of incidence(degrees)\n",
- "mew=1.4; #refractive index\n",
- "n=50; \n",
- "lamda=459*10**-9; #wavelength(m)\n",
- "\n",
- "#Calculation \n",
- "i=i*math.pi/180; #angle of incidence(radian)\n",
- "r=math.asin(math.sin(i)/mew); #angle of refraction(radian)\n",
- "#2*mew*cos(r)=n*lambda\n",
- "#n(459)=(n+1)450\n",
- "t=(n*lamda/(2*mew*math.cos(r)))*10**6; #thickness of the film(micro meter)\n",
- "\n",
- "#Result\n",
- "print \"The thickness of the film is\",round(t,3),\"micro m\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The thickness of the film is 8.985 micro m\n"
- ]
- }
- ],
- "prompt_number": 26
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example number 4.10, Page number 71"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#importing modules\n",
- "import math\n",
- "from __future__ import division\n",
- "\n",
- "#Variable declaration\n",
- "lamda=500*10**-9; #wavelength(m)\n",
- "x=0.07; #observed band width(cm)\n",
- "mew=1; #refractive index\n",
- "\n",
- "#Calculation \n",
- "theta=(math.asin(lamda/(2*mew*x)))*10**2; #wedge angle(radian)\n",
- "theta=theta*180/math.pi; #wedge angle(degrees)\n",
- "\n",
- "#Result\n",
- "print \"The wedge angle is\",round(theta,2),\"degrees\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The wedge angle is 0.02 degrees\n"
- ]
- }
- ],
- "prompt_number": 31
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example number 4.11, Page number 71"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#importing modules\n",
- "import math\n",
- "from __future__ import division\n",
- "\n",
- "#Variable declaration\n",
- "dair=0.42; #diameter of certain rings(cm)\n",
- "dliq=0.38; #diameter of rings when liquid is introduced(cm)\n",
- "\n",
- "#Calculation \n",
- "mew=dair**2/dliq**2; #refractive index of liquid\n",
- "\n",
- "#Result\n",
- "print \"The refravtive index of liquid is\",round(mew,2)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The refravtive index of liquid is 1.22\n"
- ]
- }
- ],
- "prompt_number": 33
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example number 4.12, Page number 71"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#importing modules\n",
- "import math\n",
- "from __future__ import division\n",
- "\n",
- "#Variable declaration\n",
- "m=8; #eigth ring\n",
- "n=3; #third ring\n",
- "dm=0.4; #diameter of the eigth ring(cm)\n",
- "dn=0.2; #diameter of the third ring(cm)\n",
- "R=101; #Radius of curvature(cm)\n",
- "\n",
- "#Calculation \n",
- "lamda=(((dm**2)-(dn**2))/(4*R*(m-n))); #wavelength of light(cm) \n",
- "\n",
- "#Result\n",
- "print \"The wavelength of light used is\",round(lamda*10**5,4),\"*10**-5 cm\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The wavelength of light used is 5.9406 *10**-5 cm\n"
- ]
- }
- ],
- "prompt_number": 39
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example number 4.13, Page number 71"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#importing modules\n",
- "import math\n",
- "from __future__ import division\n",
- "\n",
- "#Variable declaration\n",
- "mew=1.38; #refractive index of magnesium floride\n",
- "t=175; #thickness of coating of magnesium fluoride(nm)\n",
- "\n",
- "#Calculation \n",
- "lamda=4*t*mew; #wavelength(nm)\n",
- "\n",
- "#Result\n",
- "print \"The wavelength which has high transmission is\",lamda,\"nm\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The wavelength which has high transmission is 966.0 nm\n"
- ]
- }
- ],
- "prompt_number": 41
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Engineering_Physics/Chapter5.ipynb b/Engineering_Physics/Chapter5.ipynb deleted file mode 100755 index d125b365..00000000 --- a/Engineering_Physics/Chapter5.ipynb +++ /dev/null @@ -1,469 +0,0 @@ -{
- "metadata": {
- "name": "",
- "signature": "sha256:f4c05fd79d4d56cbd4b08f847aeb0bba767b388c9bbe1bea8066d97e3ac78212"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "heading",
- "level": 1,
- "metadata": {},
- "source": [
- "5: Diffraction"
- ]
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example number 5.1, Page number 86"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#importing modules\n",
- "import math\n",
- "from __future__ import division\n",
- "\n",
- "#Variable declaration\n",
- "n=1;\n",
- "lamda=600*10**-9; #wavelength(m)\n",
- "theta=35; #angle at which first minimum falls(degrees)\n",
- "\n",
- "#Calculation \n",
- "theta=theta*math.pi/180; #angle at which first minimum falls(radian)\n",
- "d=((n*lamda)/math.sin(theta))*10**6; #width of the slit(micro m)\n",
- "\n",
- "#Result\n",
- "print \"The width of the slit is\",round(d,2),\"micro m\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The width of the slit is 1.05 micro m\n"
- ]
- }
- ],
- "prompt_number": 3
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example number 5.2, Page number 86"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#importing modules\n",
- "import math\n",
- "from __future__ import division\n",
- "\n",
- "#Variable declaration\n",
- "D=0.95; #distance of the screen from the slit(m)\n",
- "lamda=589*10**-9; #wavelength(m)\n",
- "d=0.5*10**-3; #width of the slit(m)\n",
- "\n",
- "#Calculation \n",
- "y=((2*D*lamda)/d)*10**3; #width of a central band(mm)\n",
- "\n",
- "#Result\n",
- "print \"The width of the central band is\",round(y,2),\"mm\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The width of the central band is 2.24 mm\n"
- ]
- }
- ],
- "prompt_number": 5
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example number 5.3, Page number 86"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#importing modules\n",
- "import math\n",
- "from __future__ import division\n",
- "\n",
- "#Variable declaration\n",
- "D=1.1; #distance of the screen from the slit(m)\n",
- "lamda=589*10**-9; #wavelength(m)\n",
- "y=4.5*10**-3; #distance of first minimum on either side of central maximum(m)\n",
- "\n",
- "#Calculation \n",
- "d=((D*lamda)/y)*10**3 #slit width(mm)\n",
- "\n",
- "#Result\n",
- "print \"The slit width is\",round(d,3),\"mm\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The slit width is 0.144 mm\n"
- ]
- }
- ],
- "prompt_number": 8
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example number 5.4, Page number 86"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#importing modules\n",
- "import math\n",
- "from __future__ import division\n",
- "\n",
- "#Variable declaration\n",
- "n=4;\n",
- "lamda=589.6*10**-9; #wavelength(m)\n",
- "D=0.95; #distance of the screen from the slit(m)\n",
- "w=0.28*10**-3; #width of the slit(m)\n",
- "\n",
- "#Calculation \n",
- "d=((n*lamda*D)/w)*10**3; #distance between centres(mm)\n",
- "\n",
- "#Result\n",
- "print \"The distance between centres of central maximum and the fourth dark fringe is\",int(d),\"mm\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The distance between centres of central maximum and the fourth dark fringe is 8 mm\n"
- ]
- }
- ],
- "prompt_number": 11
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example number 5.5, Page number 86"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#importing modules\n",
- "import math\n",
- "from __future__ import division\n",
- "\n",
- "#Variable declaration\n",
- "s=5*math.pi/2; #secondary maximum\n",
- "\n",
- "#Calculation \n",
- "I=(math.sin(s)/s)**2; #I2/I0\n",
- "\n",
- "#Result\n",
- "print \"Ratio of intensities of central & second secondary maximum is\",round(I,3)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Ratio of intensities of central & second secondary maximum is 0.016\n"
- ]
- }
- ],
- "prompt_number": 13
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example number 5.6, Page number 86"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#importing modules\n",
- "import math\n",
- "from __future__ import division\n",
- "\n",
- "#Variable declaration\n",
- "lamda=450*10**-9; #wavelength(m)\n",
- "n=2;\n",
- "dlambda=1*10**-9; #difference in wavelength(m)\n",
- "\n",
- "#Calculation \n",
- "N=lamda/(n*dlambda); #minimum number of lines per cm \n",
- "\n",
- "#Result\n",
- "print \"The minimum number of lines per cm is\",N/2"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The minimum number of lines per cm is 112.5\n"
- ]
- }
- ],
- "prompt_number": 16
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example number 5.7, Page number 86"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#importing modules\n",
- "import math\n",
- "from __future__ import division\n",
- "\n",
- "#Variable declaration\n",
- "n=1;\n",
- "lamda=650*10**-9; #wavelength(m)\n",
- "d=2*10**-6; #width of the slit(m)\n",
- "\n",
- "#Calculation \n",
- "theta=math.asin((n*lamda)/d); #angle at which first minimum will be observed(radian)\n",
- "theta=theta*180/math.pi; #angle at which first minimum will be observed(degrees)\n",
- "\n",
- "#Result\n",
- "print \"The angle at which first minimum will be observed is\",round(theta,3),\"degrees\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The angle at which first minimum will be observed is 18.966 degrees\n"
- ]
- }
- ],
- "prompt_number": 22
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example number 5.8, Page number 87"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#importing modules\n",
- "import math\n",
- "from __future__ import division\n",
- "\n",
- "#Variable declaration\n",
- "lamda=600*10**-9; #wavelength(m)\n",
- "y=2*10**-3; #width of the central band(m)\n",
- "D=1; #distance of the screen from the slit(m)\n",
- "\n",
- "#Calculation \n",
- "d=((2*D*lamda)/y)*10**3; #slit width(mm)\n",
- "\n",
- "#Result\n",
- "print \"The slit width is\",d,\"mm\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The slit width is 0.6 mm\n"
- ]
- }
- ],
- "prompt_number": 24
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example number 5.9, Page number 87"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#importing modules\n",
- "import math\n",
- "from __future__ import division\n",
- "\n",
- "#Variable declaration\n",
- "y=6*10**-3; #first minimum is observed(m)\n",
- "d=90*10**-6; #slit width(m)\n",
- "D=0.98; #distance of the screen from the slit(m)\n",
- "\n",
- "#Calculation \n",
- "lamda=((y*d)/D)*10**9; #wavelength(nm)\n",
- "\n",
- "#Result\n",
- "print \"The wavelength of light used is\",int(lamda),\"nm\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The wavelength of light used is 551 nm\n"
- ]
- }
- ],
- "prompt_number": 27
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example number 5.10, Page number 87"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#importing modules\n",
- "import math\n",
- "from __future__ import division\n",
- "\n",
- "#Variable declaration\n",
- "n=1;\n",
- "lambda1=450*10**-9; #wavelength of first spectral line(m)\n",
- "d=1/5000; #number of lines\n",
- "\n",
- "#Calculation \n",
- "theta1=math.asin((n*lambda1)/d); \n",
- "theta1=round(theta1*10**2*180/math.pi);\n",
- "theta2=theta1+2.97;\n",
- "theta2=theta2*math.pi/180;\n",
- "lambda2=d*math.sin(theta2)/n; #wavelength of second spectral line(nm)\n",
- "\n",
- "#Result\n",
- "print \"The wavelength of second spectral line is\",int(lambda2*10**7),\"nm\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The wavelength of second spectral line is 550 nm\n"
- ]
- }
- ],
- "prompt_number": 41
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example number 5.11, Page number 87"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#importing modules\n",
- "import math\n",
- "from __future__ import division\n",
- "\n",
- "#Variable declaration\n",
- "n=3;\n",
- "lamda=700*10**-9; #wavelength(m)\n",
- "theta=90; #angle(degrees)\n",
- "\n",
- "#Calculation \n",
- "theta=theta*math.pi/180; #angle(radian)\n",
- "d=n*lamda/math.sin(theta); #grating element(m)\n",
- "\n",
- "#Result\n",
- "print \"The minimum grating element required to observe the entire third order spectrum is\",d*10**6,\"*10**-6 m\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The minimum grating element required to observe the entire third order spectrum is 2.1 *10**-6 m\n"
- ]
- }
- ],
- "prompt_number": 45
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Engineering_Physics/Chapter5_1.ipynb b/Engineering_Physics/Chapter5_1.ipynb deleted file mode 100755 index d125b365..00000000 --- a/Engineering_Physics/Chapter5_1.ipynb +++ /dev/null @@ -1,469 +0,0 @@ -{
- "metadata": {
- "name": "",
- "signature": "sha256:f4c05fd79d4d56cbd4b08f847aeb0bba767b388c9bbe1bea8066d97e3ac78212"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "heading",
- "level": 1,
- "metadata": {},
- "source": [
- "5: Diffraction"
- ]
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example number 5.1, Page number 86"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#importing modules\n",
- "import math\n",
- "from __future__ import division\n",
- "\n",
- "#Variable declaration\n",
- "n=1;\n",
- "lamda=600*10**-9; #wavelength(m)\n",
- "theta=35; #angle at which first minimum falls(degrees)\n",
- "\n",
- "#Calculation \n",
- "theta=theta*math.pi/180; #angle at which first minimum falls(radian)\n",
- "d=((n*lamda)/math.sin(theta))*10**6; #width of the slit(micro m)\n",
- "\n",
- "#Result\n",
- "print \"The width of the slit is\",round(d,2),\"micro m\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The width of the slit is 1.05 micro m\n"
- ]
- }
- ],
- "prompt_number": 3
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example number 5.2, Page number 86"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#importing modules\n",
- "import math\n",
- "from __future__ import division\n",
- "\n",
- "#Variable declaration\n",
- "D=0.95; #distance of the screen from the slit(m)\n",
- "lamda=589*10**-9; #wavelength(m)\n",
- "d=0.5*10**-3; #width of the slit(m)\n",
- "\n",
- "#Calculation \n",
- "y=((2*D*lamda)/d)*10**3; #width of a central band(mm)\n",
- "\n",
- "#Result\n",
- "print \"The width of the central band is\",round(y,2),\"mm\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The width of the central band is 2.24 mm\n"
- ]
- }
- ],
- "prompt_number": 5
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example number 5.3, Page number 86"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#importing modules\n",
- "import math\n",
- "from __future__ import division\n",
- "\n",
- "#Variable declaration\n",
- "D=1.1; #distance of the screen from the slit(m)\n",
- "lamda=589*10**-9; #wavelength(m)\n",
- "y=4.5*10**-3; #distance of first minimum on either side of central maximum(m)\n",
- "\n",
- "#Calculation \n",
- "d=((D*lamda)/y)*10**3 #slit width(mm)\n",
- "\n",
- "#Result\n",
- "print \"The slit width is\",round(d,3),\"mm\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The slit width is 0.144 mm\n"
- ]
- }
- ],
- "prompt_number": 8
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example number 5.4, Page number 86"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#importing modules\n",
- "import math\n",
- "from __future__ import division\n",
- "\n",
- "#Variable declaration\n",
- "n=4;\n",
- "lamda=589.6*10**-9; #wavelength(m)\n",
- "D=0.95; #distance of the screen from the slit(m)\n",
- "w=0.28*10**-3; #width of the slit(m)\n",
- "\n",
- "#Calculation \n",
- "d=((n*lamda*D)/w)*10**3; #distance between centres(mm)\n",
- "\n",
- "#Result\n",
- "print \"The distance between centres of central maximum and the fourth dark fringe is\",int(d),\"mm\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The distance between centres of central maximum and the fourth dark fringe is 8 mm\n"
- ]
- }
- ],
- "prompt_number": 11
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example number 5.5, Page number 86"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#importing modules\n",
- "import math\n",
- "from __future__ import division\n",
- "\n",
- "#Variable declaration\n",
- "s=5*math.pi/2; #secondary maximum\n",
- "\n",
- "#Calculation \n",
- "I=(math.sin(s)/s)**2; #I2/I0\n",
- "\n",
- "#Result\n",
- "print \"Ratio of intensities of central & second secondary maximum is\",round(I,3)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Ratio of intensities of central & second secondary maximum is 0.016\n"
- ]
- }
- ],
- "prompt_number": 13
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example number 5.6, Page number 86"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#importing modules\n",
- "import math\n",
- "from __future__ import division\n",
- "\n",
- "#Variable declaration\n",
- "lamda=450*10**-9; #wavelength(m)\n",
- "n=2;\n",
- "dlambda=1*10**-9; #difference in wavelength(m)\n",
- "\n",
- "#Calculation \n",
- "N=lamda/(n*dlambda); #minimum number of lines per cm \n",
- "\n",
- "#Result\n",
- "print \"The minimum number of lines per cm is\",N/2"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The minimum number of lines per cm is 112.5\n"
- ]
- }
- ],
- "prompt_number": 16
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example number 5.7, Page number 86"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#importing modules\n",
- "import math\n",
- "from __future__ import division\n",
- "\n",
- "#Variable declaration\n",
- "n=1;\n",
- "lamda=650*10**-9; #wavelength(m)\n",
- "d=2*10**-6; #width of the slit(m)\n",
- "\n",
- "#Calculation \n",
- "theta=math.asin((n*lamda)/d); #angle at which first minimum will be observed(radian)\n",
- "theta=theta*180/math.pi; #angle at which first minimum will be observed(degrees)\n",
- "\n",
- "#Result\n",
- "print \"The angle at which first minimum will be observed is\",round(theta,3),\"degrees\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The angle at which first minimum will be observed is 18.966 degrees\n"
- ]
- }
- ],
- "prompt_number": 22
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example number 5.8, Page number 87"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#importing modules\n",
- "import math\n",
- "from __future__ import division\n",
- "\n",
- "#Variable declaration\n",
- "lamda=600*10**-9; #wavelength(m)\n",
- "y=2*10**-3; #width of the central band(m)\n",
- "D=1; #distance of the screen from the slit(m)\n",
- "\n",
- "#Calculation \n",
- "d=((2*D*lamda)/y)*10**3; #slit width(mm)\n",
- "\n",
- "#Result\n",
- "print \"The slit width is\",d,\"mm\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The slit width is 0.6 mm\n"
- ]
- }
- ],
- "prompt_number": 24
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example number 5.9, Page number 87"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#importing modules\n",
- "import math\n",
- "from __future__ import division\n",
- "\n",
- "#Variable declaration\n",
- "y=6*10**-3; #first minimum is observed(m)\n",
- "d=90*10**-6; #slit width(m)\n",
- "D=0.98; #distance of the screen from the slit(m)\n",
- "\n",
- "#Calculation \n",
- "lamda=((y*d)/D)*10**9; #wavelength(nm)\n",
- "\n",
- "#Result\n",
- "print \"The wavelength of light used is\",int(lamda),\"nm\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The wavelength of light used is 551 nm\n"
- ]
- }
- ],
- "prompt_number": 27
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example number 5.10, Page number 87"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#importing modules\n",
- "import math\n",
- "from __future__ import division\n",
- "\n",
- "#Variable declaration\n",
- "n=1;\n",
- "lambda1=450*10**-9; #wavelength of first spectral line(m)\n",
- "d=1/5000; #number of lines\n",
- "\n",
- "#Calculation \n",
- "theta1=math.asin((n*lambda1)/d); \n",
- "theta1=round(theta1*10**2*180/math.pi);\n",
- "theta2=theta1+2.97;\n",
- "theta2=theta2*math.pi/180;\n",
- "lambda2=d*math.sin(theta2)/n; #wavelength of second spectral line(nm)\n",
- "\n",
- "#Result\n",
- "print \"The wavelength of second spectral line is\",int(lambda2*10**7),\"nm\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The wavelength of second spectral line is 550 nm\n"
- ]
- }
- ],
- "prompt_number": 41
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example number 5.11, Page number 87"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#importing modules\n",
- "import math\n",
- "from __future__ import division\n",
- "\n",
- "#Variable declaration\n",
- "n=3;\n",
- "lamda=700*10**-9; #wavelength(m)\n",
- "theta=90; #angle(degrees)\n",
- "\n",
- "#Calculation \n",
- "theta=theta*math.pi/180; #angle(radian)\n",
- "d=n*lamda/math.sin(theta); #grating element(m)\n",
- "\n",
- "#Result\n",
- "print \"The minimum grating element required to observe the entire third order spectrum is\",d*10**6,\"*10**-6 m\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The minimum grating element required to observe the entire third order spectrum is 2.1 *10**-6 m\n"
- ]
- }
- ],
- "prompt_number": 45
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Engineering_Physics/Chapter6.ipynb b/Engineering_Physics/Chapter6.ipynb deleted file mode 100755 index 271f7718..00000000 --- a/Engineering_Physics/Chapter6.ipynb +++ /dev/null @@ -1,437 +0,0 @@ -{
- "metadata": {
- "name": "",
- "signature": "sha256:015049a6d28a54143e382d872ce51260f52be159a8159c04fe93d876c0cea685"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "heading",
- "level": 1,
- "metadata": {},
- "source": [
- "6: Polarisation"
- ]
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example number 6.1, Page number 108"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#importing modules\n",
- "import math\n",
- "from __future__ import division\n",
- "\n",
- "#Variable declaration\n",
- "mew=1.63; #refractive index of the glass plate\n",
- "\n",
- "#Calculation \n",
- "#tan ip=mew\n",
- "ip=math.atan(mew); #ip=polarising angle(radian)\n",
- "ip=ip*180/math.pi; #ip=polarising angle(degrees)\n",
- "#ip+r=90\n",
- "r=90-ip; #angle of refraction(degrees)\n",
- "rd=int(r); #angle(degrees)\n",
- "rm=round(60*(r-rd)); #angle(minutes)\n",
- "\n",
- "#Result\n",
- "print \"The angle of refraction is\",rd,\"degrees\",rm,\"minutes\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The angle of refraction is 31 degrees 32.0 minutes\n"
- ]
- }
- ],
- "prompt_number": 4
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example number 6.2, Page number 108"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#importing modules\n",
- "import math\n",
- "from __future__ import division\n",
- "\n",
- "#Variable declaration\n",
- "#I=I0(cos^2(teta))\n",
- "theta=50; #angle made between two principle planes(degrees)\n",
- "\n",
- "#Calculation \n",
- "theta=theta*math.pi/180; #angle(radian)\n",
- "I=(math.cos(theta))**2; #incident unpolarized light\n",
- "#percentage of incident unpolarised light is (I/I0)*100 where I0 is incident polarised light\n",
- "p=I*100; #percentage of incident unpolarized light(%)\n",
- "\n",
- "#Result\n",
- "print \"The percentage of incident unpolarized light is\",int(p),\"%\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The percentage of incident unpolarized light is 41 %\n"
- ]
- }
- ],
- "prompt_number": 6
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example number 6.3, Page number 108"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#importing modules\n",
- "import math\n",
- "from __future__ import division\n",
- "\n",
- "#Variable declaration\n",
- "#I=I0*cos^2(teta)\n",
- "#cos^2(teta)=I/I0\n",
- "a=0.08; #a=I/I0;where I=incident unpolarized light & I0=incident polarized light\n",
- "\n",
- "#Calculation \n",
- "theta=math.acos(math.sqrt(a)); #angle between planes of transmission of analyser and polariser(radian)\n",
- "theta=theta*180/math.pi; #angle(degrees)\n",
- "thetad=int(theta); #angle(degrees)\n",
- "thetam=round(60*(theta-thetad)); #angle(minutes)\n",
- "\n",
- "#Result\n",
- "print \"The angle between the planes of transmission of analyser & polariser is +(or)- \",thetad,\"degrees\",thetam,\"minutes\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The angle between the planes of transmission of analyser & polariser is +(or)- 73 degrees 34.0 minutes\n"
- ]
- }
- ],
- "prompt_number": 9
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example number 6.4, Page number 108"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#importing modules\n",
- "import math\n",
- "from __future__ import division\n",
- "\n",
- "#Variable declaration\n",
- "#IE=A^2(cos^2(teta));IO=A^2(sin^2(teta))\n",
- "#I0/IE=tan^2(teta)\n",
- "theta=40; #angle made between incident beam & optic axis(degrees)\n",
- "\n",
- "#Calculation \n",
- "theta=theta*math.pi/180; #angle(radian)\n",
- "a=math.tan(theta)**2; #I0/IE\n",
- "\n",
- "#Result\n",
- "print \"I0/IE=\",round(a,1)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "I0/IE= 0.7\n"
- ]
- }
- ],
- "prompt_number": 11
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example number 6.5, Page number 108"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#importing modules\n",
- "import math\n",
- "from __future__ import division\n",
- "\n",
- "#Variable declaration\n",
- "lamda=589; #wavelength of light(nm)\n",
- "mew0=1.54; #refractive index for ordinary wave\n",
- "mewE=1.55; #refractive index for extraordinary wave\n",
- "\n",
- "#Calculation \n",
- "t=lamda/(4*(mewE-mew0))*10**-3; #thickness(micro m)\n",
- "\n",
- "#Result\n",
- "print \"The thickness of a quarter-wave plate is\",t,\"micro m\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The thickness of a quarter-wave plate is 14.725 micro m\n"
- ]
- }
- ],
- "prompt_number": 13
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example number 6.6, Page number 109"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#importing modules\n",
- "import math\n",
- "from __future__ import division\n",
- "\n",
- "#Variable declaration\n",
- "ip=52; #angle of polarization(degrees)\n",
- "\n",
- "#Calculation \n",
- "ip=ip*math.pi/180; #angle(radian)\n",
- "mew=math.tan(ip); #refractive index of the material surface\n",
- "\n",
- "#Result\n",
- "print \"The refractive index of the material surface is\",round(mew,2)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The refractive index of the material surface is 1.28\n"
- ]
- }
- ],
- "prompt_number": 15
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example number 6.7, Page number 109"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#importing modules\n",
- "import math\n",
- "from __future__ import division\n",
- "\n",
- "#Variable declaration\n",
- "r=33; #angle of refraction(degrees)\n",
- "\n",
- "#Calculation \n",
- "ip=90-r; #polarising angle(degrees)\n",
- "ip=ip*math.pi/180; #angle(radian)\n",
- "mew=math.tan(ip); #refractive index of quartz\n",
- "\n",
- "#Result\n",
- "print \"The refractive index of quartz is\",round(mew,2)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The refractive index of quartz is 1.54\n"
- ]
- }
- ],
- "prompt_number": 18
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example number 6.8, Page number 109"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#importing modules\n",
- "import math\n",
- "from __future__ import division\n",
- "\n",
- "#Variable declaration\n",
- "#IE=A^2*cos^2(teta);IO=A^2*sin^2(teta)\n",
- "#I0/IE=tan^2(teta)=0.65\n",
- "a=0.65; #ratio of intensities of ordinary & extraordinary light\n",
- "\n",
- "#Calculation \n",
- "theta=math.atan(math.sqrt(a)); #angle made by plane of vibration of the incident light with optic axis(radian)\n",
- "theta=theta*180/math.pi; #angle(degrees)\n",
- "thetad=int(theta); #angle(degrees)\n",
- "thetam=int(60*(theta-thetad));\n",
- "\n",
- "#Result\n",
- "print \"The angle made by the plane of vibration of incident light with the optic axis is\",thetad,\"degrees\",thetam,\"minutes\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The angle made by the plane of vibration of incident light with the optic axis is 38 degrees 52 minutes\n"
- ]
- }
- ],
- "prompt_number": 20
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example number 6.9, Page number 109"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#importing modules\n",
- "import math\n",
- "from __future__ import division\n",
- "\n",
- "#Variable declaration\n",
- "mew0=1.544; #refractive index of ordinary waves\n",
- "mewE=1.553; #refractive index of extraordinary waves\n",
- "lamda=550; #wavelength(nm) \n",
- "t=9;\n",
- "\n",
- "#Calculation \n",
- "delta=((2*180)/(lamda*(10**-9)))*(mewE-mew0)*t*(10**-6); #phase difference(degrees)\n",
- "\n",
- "#Result\n",
- "print \"The phase difference between O and E rays is\",int(delta),\"degrees\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The phase difference between O and E rays is 53 degrees\n"
- ]
- }
- ],
- "prompt_number": 23
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example number 6.10, Page number 109"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#importing modules\n",
- "import math\n",
- "from __future__ import division\n",
- "\n",
- "#Variable declaration\n",
- "delta=50; #phase difference(degrees)\n",
- "mewE=1.544; #refractive index of extraordinary waves\n",
- "mew0=1.553; #refractive index of ordinary waves\n",
- "t=8; #thickness(nm)\n",
- "\n",
- "#Calculation \n",
- "lamda=((2*180)/delta)*(mew0-mewE)*t*10**-6*10**9; #wavelength of light incident(nm)\n",
- "\n",
- "#Result\n",
- "print \"The wavelength of light incident is\",lamda,\"nm\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The wavelength of light incident is 518.4 nm\n"
- ]
- }
- ],
- "prompt_number": 25
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Engineering_Physics/Chapter6_1.ipynb b/Engineering_Physics/Chapter6_1.ipynb deleted file mode 100755 index 271f7718..00000000 --- a/Engineering_Physics/Chapter6_1.ipynb +++ /dev/null @@ -1,437 +0,0 @@ -{
- "metadata": {
- "name": "",
- "signature": "sha256:015049a6d28a54143e382d872ce51260f52be159a8159c04fe93d876c0cea685"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "heading",
- "level": 1,
- "metadata": {},
- "source": [
- "6: Polarisation"
- ]
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example number 6.1, Page number 108"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#importing modules\n",
- "import math\n",
- "from __future__ import division\n",
- "\n",
- "#Variable declaration\n",
- "mew=1.63; #refractive index of the glass plate\n",
- "\n",
- "#Calculation \n",
- "#tan ip=mew\n",
- "ip=math.atan(mew); #ip=polarising angle(radian)\n",
- "ip=ip*180/math.pi; #ip=polarising angle(degrees)\n",
- "#ip+r=90\n",
- "r=90-ip; #angle of refraction(degrees)\n",
- "rd=int(r); #angle(degrees)\n",
- "rm=round(60*(r-rd)); #angle(minutes)\n",
- "\n",
- "#Result\n",
- "print \"The angle of refraction is\",rd,\"degrees\",rm,\"minutes\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The angle of refraction is 31 degrees 32.0 minutes\n"
- ]
- }
- ],
- "prompt_number": 4
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example number 6.2, Page number 108"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#importing modules\n",
- "import math\n",
- "from __future__ import division\n",
- "\n",
- "#Variable declaration\n",
- "#I=I0(cos^2(teta))\n",
- "theta=50; #angle made between two principle planes(degrees)\n",
- "\n",
- "#Calculation \n",
- "theta=theta*math.pi/180; #angle(radian)\n",
- "I=(math.cos(theta))**2; #incident unpolarized light\n",
- "#percentage of incident unpolarised light is (I/I0)*100 where I0 is incident polarised light\n",
- "p=I*100; #percentage of incident unpolarized light(%)\n",
- "\n",
- "#Result\n",
- "print \"The percentage of incident unpolarized light is\",int(p),\"%\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The percentage of incident unpolarized light is 41 %\n"
- ]
- }
- ],
- "prompt_number": 6
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example number 6.3, Page number 108"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#importing modules\n",
- "import math\n",
- "from __future__ import division\n",
- "\n",
- "#Variable declaration\n",
- "#I=I0*cos^2(teta)\n",
- "#cos^2(teta)=I/I0\n",
- "a=0.08; #a=I/I0;where I=incident unpolarized light & I0=incident polarized light\n",
- "\n",
- "#Calculation \n",
- "theta=math.acos(math.sqrt(a)); #angle between planes of transmission of analyser and polariser(radian)\n",
- "theta=theta*180/math.pi; #angle(degrees)\n",
- "thetad=int(theta); #angle(degrees)\n",
- "thetam=round(60*(theta-thetad)); #angle(minutes)\n",
- "\n",
- "#Result\n",
- "print \"The angle between the planes of transmission of analyser & polariser is +(or)- \",thetad,\"degrees\",thetam,\"minutes\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The angle between the planes of transmission of analyser & polariser is +(or)- 73 degrees 34.0 minutes\n"
- ]
- }
- ],
- "prompt_number": 9
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example number 6.4, Page number 108"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#importing modules\n",
- "import math\n",
- "from __future__ import division\n",
- "\n",
- "#Variable declaration\n",
- "#IE=A^2(cos^2(teta));IO=A^2(sin^2(teta))\n",
- "#I0/IE=tan^2(teta)\n",
- "theta=40; #angle made between incident beam & optic axis(degrees)\n",
- "\n",
- "#Calculation \n",
- "theta=theta*math.pi/180; #angle(radian)\n",
- "a=math.tan(theta)**2; #I0/IE\n",
- "\n",
- "#Result\n",
- "print \"I0/IE=\",round(a,1)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "I0/IE= 0.7\n"
- ]
- }
- ],
- "prompt_number": 11
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example number 6.5, Page number 108"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#importing modules\n",
- "import math\n",
- "from __future__ import division\n",
- "\n",
- "#Variable declaration\n",
- "lamda=589; #wavelength of light(nm)\n",
- "mew0=1.54; #refractive index for ordinary wave\n",
- "mewE=1.55; #refractive index for extraordinary wave\n",
- "\n",
- "#Calculation \n",
- "t=lamda/(4*(mewE-mew0))*10**-3; #thickness(micro m)\n",
- "\n",
- "#Result\n",
- "print \"The thickness of a quarter-wave plate is\",t,\"micro m\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The thickness of a quarter-wave plate is 14.725 micro m\n"
- ]
- }
- ],
- "prompt_number": 13
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example number 6.6, Page number 109"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#importing modules\n",
- "import math\n",
- "from __future__ import division\n",
- "\n",
- "#Variable declaration\n",
- "ip=52; #angle of polarization(degrees)\n",
- "\n",
- "#Calculation \n",
- "ip=ip*math.pi/180; #angle(radian)\n",
- "mew=math.tan(ip); #refractive index of the material surface\n",
- "\n",
- "#Result\n",
- "print \"The refractive index of the material surface is\",round(mew,2)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The refractive index of the material surface is 1.28\n"
- ]
- }
- ],
- "prompt_number": 15
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example number 6.7, Page number 109"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#importing modules\n",
- "import math\n",
- "from __future__ import division\n",
- "\n",
- "#Variable declaration\n",
- "r=33; #angle of refraction(degrees)\n",
- "\n",
- "#Calculation \n",
- "ip=90-r; #polarising angle(degrees)\n",
- "ip=ip*math.pi/180; #angle(radian)\n",
- "mew=math.tan(ip); #refractive index of quartz\n",
- "\n",
- "#Result\n",
- "print \"The refractive index of quartz is\",round(mew,2)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The refractive index of quartz is 1.54\n"
- ]
- }
- ],
- "prompt_number": 18
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example number 6.8, Page number 109"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#importing modules\n",
- "import math\n",
- "from __future__ import division\n",
- "\n",
- "#Variable declaration\n",
- "#IE=A^2*cos^2(teta);IO=A^2*sin^2(teta)\n",
- "#I0/IE=tan^2(teta)=0.65\n",
- "a=0.65; #ratio of intensities of ordinary & extraordinary light\n",
- "\n",
- "#Calculation \n",
- "theta=math.atan(math.sqrt(a)); #angle made by plane of vibration of the incident light with optic axis(radian)\n",
- "theta=theta*180/math.pi; #angle(degrees)\n",
- "thetad=int(theta); #angle(degrees)\n",
- "thetam=int(60*(theta-thetad));\n",
- "\n",
- "#Result\n",
- "print \"The angle made by the plane of vibration of incident light with the optic axis is\",thetad,\"degrees\",thetam,\"minutes\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The angle made by the plane of vibration of incident light with the optic axis is 38 degrees 52 minutes\n"
- ]
- }
- ],
- "prompt_number": 20
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example number 6.9, Page number 109"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#importing modules\n",
- "import math\n",
- "from __future__ import division\n",
- "\n",
- "#Variable declaration\n",
- "mew0=1.544; #refractive index of ordinary waves\n",
- "mewE=1.553; #refractive index of extraordinary waves\n",
- "lamda=550; #wavelength(nm) \n",
- "t=9;\n",
- "\n",
- "#Calculation \n",
- "delta=((2*180)/(lamda*(10**-9)))*(mewE-mew0)*t*(10**-6); #phase difference(degrees)\n",
- "\n",
- "#Result\n",
- "print \"The phase difference between O and E rays is\",int(delta),\"degrees\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The phase difference between O and E rays is 53 degrees\n"
- ]
- }
- ],
- "prompt_number": 23
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example number 6.10, Page number 109"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#importing modules\n",
- "import math\n",
- "from __future__ import division\n",
- "\n",
- "#Variable declaration\n",
- "delta=50; #phase difference(degrees)\n",
- "mewE=1.544; #refractive index of extraordinary waves\n",
- "mew0=1.553; #refractive index of ordinary waves\n",
- "t=8; #thickness(nm)\n",
- "\n",
- "#Calculation \n",
- "lamda=((2*180)/delta)*(mew0-mewE)*t*10**-6*10**9; #wavelength of light incident(nm)\n",
- "\n",
- "#Result\n",
- "print \"The wavelength of light incident is\",lamda,\"nm\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The wavelength of light incident is 518.4 nm\n"
- ]
- }
- ],
- "prompt_number": 25
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Engineering_Physics/Chapter7.ipynb b/Engineering_Physics/Chapter7.ipynb deleted file mode 100755 index 36caeba7..00000000 --- a/Engineering_Physics/Chapter7.ipynb +++ /dev/null @@ -1,582 +0,0 @@ -{
- "metadata": {
- "name": "",
- "signature": "sha256:604dfd31225e3c2fe12afc104ed18461bd1bcabd558389f5cc69df386ba9091d"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "heading",
- "level": 1,
- "metadata": {},
- "source": [
- "7: Motion of a charged particle"
- ]
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example number 7.1, Page number 132"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#importing modules\n",
- "import math\n",
- "from __future__ import division\n",
- "\n",
- "#Variable declaration\n",
- "e=1.6*10**-19; #charge of the electron(c)\n",
- "V=18; #potential difference(kV)\n",
- "m=9.1*10**-31; #mass of the electron(kg)\n",
- "\n",
- "#Calculation \n",
- "K=e*V*10**3; #Kinetic energy(J)\n",
- "v=math.sqrt((2*e*V*10**3)/m); #speed of electron(m/s)\n",
- "\n",
- "#Result\n",
- "print \"The kinetic energy of electron is\",K*10**16,\"*10**-16 J\"\n",
- "print \"Speed of the electron is\",round(v/10**7,3),\"*10**7 m/s\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The kinetic energy of electron is 28.8 *10**-16 J\n",
- "Speed of the electron is 7.956 *10**7 m/s\n"
- ]
- }
- ],
- "prompt_number": 5
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example number 7.2, Page number 133"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#importing modules\n",
- "import math\n",
- "from __future__ import division\n",
- "\n",
- "#Variable declaration\n",
- "m=9.1*10**-31; #mass of electron(kg)\n",
- "vx=4*10**6; #velocity along x-axis(m/s)\n",
- "E=1500; #electric field strength(N/C)\n",
- "l=0.07; #length in y-axis(m)\n",
- "q=1.6*10**-19; #charge of electron(c)\n",
- "\n",
- "#Calculation \n",
- "y=(-q*E*(l**2))/(2*m*(vx**2))*10**2; #vertical displacement of electron(cm)\n",
- "\n",
- "#Result\n",
- "print \"The vertical displacement of electron when it leaves the electric field is\",round(y,3),\"cm\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The vertical displacement of electron when it leaves the electric field is -4.038 cm\n"
- ]
- }
- ],
- "prompt_number": 7
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example number 7.3, Page number 133"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#importing modules\n",
- "import math\n",
- "from __future__ import division\n",
- "\n",
- "#Variable declaration\n",
- "u=5*10**5; #velocity(m/s)\n",
- "m=1.67*10**-27; #mass of proton(kg)\n",
- "q=1.6*10**-19;\n",
- "E=500; #electric field(N/C)\n",
- "theta=42; #angle(degrees)\n",
- "\n",
- "#Calculation \n",
- "theta=theta*math.pi/180; #angle(radian)\n",
- "t=((u*m*math.sin(theta))/(q*E))*10**6; #time required for the proton(micro s)\n",
- "\n",
- "#Result\n",
- "print \"The time required for the proton is\",round(t,2),\"micro s\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The time required for the proton is 6.98 micro s\n"
- ]
- }
- ],
- "prompt_number": 9
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example number 7.4, Page number 133"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#importing modules\n",
- "import math\n",
- "from __future__ import division\n",
- "\n",
- "#Variable declaration\n",
- "m=1.67*10**-27; #mass of proton(kg)\n",
- "q=1.6*10**-19;\n",
- "B=0.36; #magnetic field(T)\n",
- "R=0.2; #radius(m)\n",
- "\n",
- "#Calculation \n",
- "v=(q*B*R)/m; #orbital speed of proton(m/s)\n",
- "\n",
- "#Result\n",
- "print \"The orbital speed of proton is\",round(v/10**6,1),\"*10**6 m/s\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The orbital speed of proton is 6.9 *10**6 m/s\n"
- ]
- }
- ],
- "prompt_number": 12
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example number 7.5, Page number 133"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#importing modules\n",
- "import math\n",
- "from __future__ import division\n",
- "\n",
- "#Variable declaration\n",
- "v=2*10**6; #speed(m/s)\n",
- "theta=30; #angle at which proton enters at the origin of coordinate system(degrees)\n",
- "B=0.3; #magnetic field(T)\n",
- "m=1.67*10**-27; #mass of proton(kg)\n",
- "q=1.6*10**-19;\n",
- "\n",
- "#Calculation \n",
- "theta=theta*math.pi/180; #angle(radian)\n",
- "vp=v*math.sin(theta); #v(perpendicular component)\n",
- "vpa=v*math.cos(theta); #v(parallel component)\n",
- "p=(vpa*2*math.pi*m)/(q*B); #pitch of the helix described by the proton\n",
- "R=((m*vp)/(q*B))*10**2; #radius of the trajectory\n",
- "\n",
- "#Result\n",
- "print \"the pitch of the helix is\",round(p,2),\"m\"\n",
- "print \"the radius of trajectory is\",round(R,2),\"cm\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "the pitch of the helix is 0.38 m\n",
- "the radius of trajectory is 3.48 cm\n"
- ]
- }
- ],
- "prompt_number": 14
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example number 7.6, Page number 133"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#importing modules\n",
- "import math\n",
- "from __future__ import division\n",
- "\n",
- "#Variable declaration\n",
- "V=25; #deflecting voltage(V)\n",
- "l=0.03; #length of deflecting planes(m)\n",
- "d=0.75; #distance between 2 deflecting plates(cm)\n",
- "Va=800; #final anode voltage(V)\n",
- "D=0.2; #distance between the screen and the plates(m)\n",
- "e=1.6*10**-19;\n",
- "m=9.1*10**-31; #mass of electron(kg)\n",
- "\n",
- "#Calculation \n",
- "y=(((V*l)/(2*d*Va))*(D+(l/2)))*10**4; #displacement produced(cm)\n",
- "a=((V*l)/(2*d*Va))*10**2;\n",
- "alpha=math.atan(a); #angle made by the beam with the axis(radian)\n",
- "alpha1=alpha*180/math.pi; #angle(degrees)\n",
- "v=((math.sqrt((2*e*Va)/m))/math.cos(alpha)); #velocity of electron(v)\n",
- "\n",
- "#Result\n",
- "print \"the displacement produced is\",round(y,2),\"cm\"\n",
- "print \"the angle made by the beam with the axis is\",round(alpha1,2),\"degrees\"\n",
- "print \"velocity of electrons is\",round(v/10**7,2),\"*10**7 m/s\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "the displacement produced is 1.34 cm\n",
- "the angle made by the beam with the axis is 3.58 degrees\n",
- "velocity of electrons is 1.68 *10**7 m/s\n"
- ]
- }
- ],
- "prompt_number": 18
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example number 7.7, Page number 134"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#importing modules\n",
- "import math\n",
- "from __future__ import division\n",
- "\n",
- "#Variable declaration\n",
- "e=1.6*10**-19;\n",
- "B=5*10**-5; #magnetic field(Wb/m**2)\n",
- "l=0.04; #length of magnetic field along the axis(m)\n",
- "m=9.1*10**-31; #mass of electron(kg)\n",
- "D=0.25; #distance of the screen from the field(m)\n",
- "Va=600; #final anode voltage(V)\n",
- "\n",
- "#Calculation \n",
- "y=(((e*B*l)/m)*math.sqrt(m/(2*e*Va))*(D+(l/2)))*10**2; #displacement of the electron(cm)\n",
- "\n",
- "#Result\n",
- "print \"the displacement of the electron beam spot on the screen is\",round(y,2),\"cm\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "the displacement of the electron beam spot on the screen is 0.65 cm\n"
- ]
- }
- ],
- "prompt_number": 20
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example number 7.8, Page number 134"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#importing modules\n",
- "import math\n",
- "from __future__ import division\n",
- "\n",
- "#Variable declaration\n",
- "E=2.5*10**4; #electric field(V/m)\n",
- "B=0.18; #magnetic field(T)\n",
- "B1=0.22; #magnetic field in the main chamber(T)\n",
- "m2=13; #mass number of carbon(kg)\n",
- "m1=12; #mass number of carbon(kg)\n",
- "e=1.6*10**-9;\n",
- "q=1.67*10**-27;\n",
- "\n",
- "#Calculation \n",
- "v=E/B; #velocity of particles(m/s)\n",
- "s=((2*v*(m2-m1)*q)/(e*B1))*10**12; #seperation on photographic plate(cm)\n",
- "\n",
- "#Result\n",
- "print \"the seperation on photographic plate is\",round(s,3),\"cm\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "the seperation on photographic plate is 1.318 cm\n"
- ]
- }
- ],
- "prompt_number": 22
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example number 7.9, Page number 134"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#importing modules\n",
- "import math\n",
- "from __future__ import division\n",
- "\n",
- "#Variable declaration\n",
- "v=5.6*10**6; #speed of the electron(m/s)\n",
- "m=9.1*10**-31; #mass of electron(kg)\n",
- "e=1.6*10**-19;\n",
- "s=0.03; #distance travelled(m)\n",
- "\n",
- "#Calculation \n",
- "E=(m*(v)**2)/(2*e*s); #intensity of electric field(N/C)\n",
- "\n",
- "#Result\n",
- "print \"The intensity of electric field is\",round(E),\"N/C\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The intensity of electric field is 2973.0 N/C\n"
- ]
- }
- ],
- "prompt_number": 24
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example number 7.10, Page number 134"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#importing modules\n",
- "import math\n",
- "from __future__ import division\n",
- "\n",
- "#Variable declaration\n",
- "v=5*10**7;\n",
- "B=0.4; #magnetic field(T)\n",
- "r=0.711*10**-3; #radius of the circle(m)\n",
- "\n",
- "#Calculation \n",
- "Q=v/(B*r); #charge to mass ratio(C/kg)\n",
- "\n",
- "#Result\n",
- "print \"The charge to mass ratio is\",round(Q/10**10,2),\"*10**10 C/kg\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The charge to mass ratio is 17.58 *10**10 C/kg\n"
- ]
- }
- ],
- "prompt_number": 27
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example number 7.11, Page number 135"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#importing modules\n",
- "import math\n",
- "from __future__ import division\n",
- "\n",
- "#Variable declaration\n",
- "m=9.1*10**-31; #mass of electron(kg)\n",
- "v=3*10**7; #speed of electron(m/s)\n",
- "R=0.05; #radius of the circle(m)\n",
- "q=1.6*10**-31;\n",
- "\n",
- "#Calculation \n",
- "B=((m*v)/(q*R))*10**-9; #magnetic field(mT)\n",
- "\n",
- "#Result\n",
- "print \"The magnetic field to bend a beam is\",round(B,1),\"mT\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The magnetic field to bend a beam is 3.4 mT\n"
- ]
- }
- ],
- "prompt_number": 29
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example number 7.12, Page number 135"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#importing modules\n",
- "import math\n",
- "from __future__ import division\n",
- "\n",
- "#Variable declaration\n",
- "m=9.1*10**-31; #mass of electron(kg)\n",
- "q=1.6*10**-19;\n",
- "t=8*10**-9; #time(ns)\n",
- "\n",
- "#Calculation \n",
- "B=(2*math.pi*m*500)/(q*t); #magnetic field(T)\n",
- "\n",
- "#Result\n",
- "print \"The magnetic field is\",round(B,2),\"T\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The magnetic field is 2.23 T\n"
- ]
- }
- ],
- "prompt_number": 31
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example number 7.13, Page number 135"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#importing modules\n",
- "import math\n",
- "from __future__ import division\n",
- "\n",
- "#Variable declaration\n",
- "v=9.15*10**7; #cyclotron frequency of proton(Hz)\n",
- "m=1.67*10**-27; #mass of proton(kg)\n",
- "q=1.6*10**-19;\n",
- "\n",
- "#Calculation \n",
- "B=(2*math.pi*v*m)/q; #magnetic field(T)\n",
- "\n",
- "#Result\n",
- "print \"The magnetic field is\",int(B),\"T\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The magnetic field is 6 T\n"
- ]
- }
- ],
- "prompt_number": 33
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Engineering_Physics/Chapter7_1.ipynb b/Engineering_Physics/Chapter7_1.ipynb deleted file mode 100755 index 36caeba7..00000000 --- a/Engineering_Physics/Chapter7_1.ipynb +++ /dev/null @@ -1,582 +0,0 @@ -{
- "metadata": {
- "name": "",
- "signature": "sha256:604dfd31225e3c2fe12afc104ed18461bd1bcabd558389f5cc69df386ba9091d"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "heading",
- "level": 1,
- "metadata": {},
- "source": [
- "7: Motion of a charged particle"
- ]
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example number 7.1, Page number 132"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#importing modules\n",
- "import math\n",
- "from __future__ import division\n",
- "\n",
- "#Variable declaration\n",
- "e=1.6*10**-19; #charge of the electron(c)\n",
- "V=18; #potential difference(kV)\n",
- "m=9.1*10**-31; #mass of the electron(kg)\n",
- "\n",
- "#Calculation \n",
- "K=e*V*10**3; #Kinetic energy(J)\n",
- "v=math.sqrt((2*e*V*10**3)/m); #speed of electron(m/s)\n",
- "\n",
- "#Result\n",
- "print \"The kinetic energy of electron is\",K*10**16,\"*10**-16 J\"\n",
- "print \"Speed of the electron is\",round(v/10**7,3),\"*10**7 m/s\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The kinetic energy of electron is 28.8 *10**-16 J\n",
- "Speed of the electron is 7.956 *10**7 m/s\n"
- ]
- }
- ],
- "prompt_number": 5
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example number 7.2, Page number 133"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#importing modules\n",
- "import math\n",
- "from __future__ import division\n",
- "\n",
- "#Variable declaration\n",
- "m=9.1*10**-31; #mass of electron(kg)\n",
- "vx=4*10**6; #velocity along x-axis(m/s)\n",
- "E=1500; #electric field strength(N/C)\n",
- "l=0.07; #length in y-axis(m)\n",
- "q=1.6*10**-19; #charge of electron(c)\n",
- "\n",
- "#Calculation \n",
- "y=(-q*E*(l**2))/(2*m*(vx**2))*10**2; #vertical displacement of electron(cm)\n",
- "\n",
- "#Result\n",
- "print \"The vertical displacement of electron when it leaves the electric field is\",round(y,3),\"cm\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The vertical displacement of electron when it leaves the electric field is -4.038 cm\n"
- ]
- }
- ],
- "prompt_number": 7
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example number 7.3, Page number 133"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#importing modules\n",
- "import math\n",
- "from __future__ import division\n",
- "\n",
- "#Variable declaration\n",
- "u=5*10**5; #velocity(m/s)\n",
- "m=1.67*10**-27; #mass of proton(kg)\n",
- "q=1.6*10**-19;\n",
- "E=500; #electric field(N/C)\n",
- "theta=42; #angle(degrees)\n",
- "\n",
- "#Calculation \n",
- "theta=theta*math.pi/180; #angle(radian)\n",
- "t=((u*m*math.sin(theta))/(q*E))*10**6; #time required for the proton(micro s)\n",
- "\n",
- "#Result\n",
- "print \"The time required for the proton is\",round(t,2),\"micro s\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The time required for the proton is 6.98 micro s\n"
- ]
- }
- ],
- "prompt_number": 9
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example number 7.4, Page number 133"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#importing modules\n",
- "import math\n",
- "from __future__ import division\n",
- "\n",
- "#Variable declaration\n",
- "m=1.67*10**-27; #mass of proton(kg)\n",
- "q=1.6*10**-19;\n",
- "B=0.36; #magnetic field(T)\n",
- "R=0.2; #radius(m)\n",
- "\n",
- "#Calculation \n",
- "v=(q*B*R)/m; #orbital speed of proton(m/s)\n",
- "\n",
- "#Result\n",
- "print \"The orbital speed of proton is\",round(v/10**6,1),\"*10**6 m/s\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The orbital speed of proton is 6.9 *10**6 m/s\n"
- ]
- }
- ],
- "prompt_number": 12
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example number 7.5, Page number 133"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#importing modules\n",
- "import math\n",
- "from __future__ import division\n",
- "\n",
- "#Variable declaration\n",
- "v=2*10**6; #speed(m/s)\n",
- "theta=30; #angle at which proton enters at the origin of coordinate system(degrees)\n",
- "B=0.3; #magnetic field(T)\n",
- "m=1.67*10**-27; #mass of proton(kg)\n",
- "q=1.6*10**-19;\n",
- "\n",
- "#Calculation \n",
- "theta=theta*math.pi/180; #angle(radian)\n",
- "vp=v*math.sin(theta); #v(perpendicular component)\n",
- "vpa=v*math.cos(theta); #v(parallel component)\n",
- "p=(vpa*2*math.pi*m)/(q*B); #pitch of the helix described by the proton\n",
- "R=((m*vp)/(q*B))*10**2; #radius of the trajectory\n",
- "\n",
- "#Result\n",
- "print \"the pitch of the helix is\",round(p,2),\"m\"\n",
- "print \"the radius of trajectory is\",round(R,2),\"cm\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "the pitch of the helix is 0.38 m\n",
- "the radius of trajectory is 3.48 cm\n"
- ]
- }
- ],
- "prompt_number": 14
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example number 7.6, Page number 133"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#importing modules\n",
- "import math\n",
- "from __future__ import division\n",
- "\n",
- "#Variable declaration\n",
- "V=25; #deflecting voltage(V)\n",
- "l=0.03; #length of deflecting planes(m)\n",
- "d=0.75; #distance between 2 deflecting plates(cm)\n",
- "Va=800; #final anode voltage(V)\n",
- "D=0.2; #distance between the screen and the plates(m)\n",
- "e=1.6*10**-19;\n",
- "m=9.1*10**-31; #mass of electron(kg)\n",
- "\n",
- "#Calculation \n",
- "y=(((V*l)/(2*d*Va))*(D+(l/2)))*10**4; #displacement produced(cm)\n",
- "a=((V*l)/(2*d*Va))*10**2;\n",
- "alpha=math.atan(a); #angle made by the beam with the axis(radian)\n",
- "alpha1=alpha*180/math.pi; #angle(degrees)\n",
- "v=((math.sqrt((2*e*Va)/m))/math.cos(alpha)); #velocity of electron(v)\n",
- "\n",
- "#Result\n",
- "print \"the displacement produced is\",round(y,2),\"cm\"\n",
- "print \"the angle made by the beam with the axis is\",round(alpha1,2),\"degrees\"\n",
- "print \"velocity of electrons is\",round(v/10**7,2),\"*10**7 m/s\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "the displacement produced is 1.34 cm\n",
- "the angle made by the beam with the axis is 3.58 degrees\n",
- "velocity of electrons is 1.68 *10**7 m/s\n"
- ]
- }
- ],
- "prompt_number": 18
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example number 7.7, Page number 134"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#importing modules\n",
- "import math\n",
- "from __future__ import division\n",
- "\n",
- "#Variable declaration\n",
- "e=1.6*10**-19;\n",
- "B=5*10**-5; #magnetic field(Wb/m**2)\n",
- "l=0.04; #length of magnetic field along the axis(m)\n",
- "m=9.1*10**-31; #mass of electron(kg)\n",
- "D=0.25; #distance of the screen from the field(m)\n",
- "Va=600; #final anode voltage(V)\n",
- "\n",
- "#Calculation \n",
- "y=(((e*B*l)/m)*math.sqrt(m/(2*e*Va))*(D+(l/2)))*10**2; #displacement of the electron(cm)\n",
- "\n",
- "#Result\n",
- "print \"the displacement of the electron beam spot on the screen is\",round(y,2),\"cm\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "the displacement of the electron beam spot on the screen is 0.65 cm\n"
- ]
- }
- ],
- "prompt_number": 20
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example number 7.8, Page number 134"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#importing modules\n",
- "import math\n",
- "from __future__ import division\n",
- "\n",
- "#Variable declaration\n",
- "E=2.5*10**4; #electric field(V/m)\n",
- "B=0.18; #magnetic field(T)\n",
- "B1=0.22; #magnetic field in the main chamber(T)\n",
- "m2=13; #mass number of carbon(kg)\n",
- "m1=12; #mass number of carbon(kg)\n",
- "e=1.6*10**-9;\n",
- "q=1.67*10**-27;\n",
- "\n",
- "#Calculation \n",
- "v=E/B; #velocity of particles(m/s)\n",
- "s=((2*v*(m2-m1)*q)/(e*B1))*10**12; #seperation on photographic plate(cm)\n",
- "\n",
- "#Result\n",
- "print \"the seperation on photographic plate is\",round(s,3),\"cm\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "the seperation on photographic plate is 1.318 cm\n"
- ]
- }
- ],
- "prompt_number": 22
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example number 7.9, Page number 134"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#importing modules\n",
- "import math\n",
- "from __future__ import division\n",
- "\n",
- "#Variable declaration\n",
- "v=5.6*10**6; #speed of the electron(m/s)\n",
- "m=9.1*10**-31; #mass of electron(kg)\n",
- "e=1.6*10**-19;\n",
- "s=0.03; #distance travelled(m)\n",
- "\n",
- "#Calculation \n",
- "E=(m*(v)**2)/(2*e*s); #intensity of electric field(N/C)\n",
- "\n",
- "#Result\n",
- "print \"The intensity of electric field is\",round(E),\"N/C\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The intensity of electric field is 2973.0 N/C\n"
- ]
- }
- ],
- "prompt_number": 24
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example number 7.10, Page number 134"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#importing modules\n",
- "import math\n",
- "from __future__ import division\n",
- "\n",
- "#Variable declaration\n",
- "v=5*10**7;\n",
- "B=0.4; #magnetic field(T)\n",
- "r=0.711*10**-3; #radius of the circle(m)\n",
- "\n",
- "#Calculation \n",
- "Q=v/(B*r); #charge to mass ratio(C/kg)\n",
- "\n",
- "#Result\n",
- "print \"The charge to mass ratio is\",round(Q/10**10,2),\"*10**10 C/kg\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The charge to mass ratio is 17.58 *10**10 C/kg\n"
- ]
- }
- ],
- "prompt_number": 27
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example number 7.11, Page number 135"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#importing modules\n",
- "import math\n",
- "from __future__ import division\n",
- "\n",
- "#Variable declaration\n",
- "m=9.1*10**-31; #mass of electron(kg)\n",
- "v=3*10**7; #speed of electron(m/s)\n",
- "R=0.05; #radius of the circle(m)\n",
- "q=1.6*10**-31;\n",
- "\n",
- "#Calculation \n",
- "B=((m*v)/(q*R))*10**-9; #magnetic field(mT)\n",
- "\n",
- "#Result\n",
- "print \"The magnetic field to bend a beam is\",round(B,1),\"mT\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The magnetic field to bend a beam is 3.4 mT\n"
- ]
- }
- ],
- "prompt_number": 29
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example number 7.12, Page number 135"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#importing modules\n",
- "import math\n",
- "from __future__ import division\n",
- "\n",
- "#Variable declaration\n",
- "m=9.1*10**-31; #mass of electron(kg)\n",
- "q=1.6*10**-19;\n",
- "t=8*10**-9; #time(ns)\n",
- "\n",
- "#Calculation \n",
- "B=(2*math.pi*m*500)/(q*t); #magnetic field(T)\n",
- "\n",
- "#Result\n",
- "print \"The magnetic field is\",round(B,2),\"T\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The magnetic field is 2.23 T\n"
- ]
- }
- ],
- "prompt_number": 31
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example number 7.13, Page number 135"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#importing modules\n",
- "import math\n",
- "from __future__ import division\n",
- "\n",
- "#Variable declaration\n",
- "v=9.15*10**7; #cyclotron frequency of proton(Hz)\n",
- "m=1.67*10**-27; #mass of proton(kg)\n",
- "q=1.6*10**-19;\n",
- "\n",
- "#Calculation \n",
- "B=(2*math.pi*v*m)/q; #magnetic field(T)\n",
- "\n",
- "#Result\n",
- "print \"The magnetic field is\",int(B),\"T\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The magnetic field is 6 T\n"
- ]
- }
- ],
- "prompt_number": 33
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Engineering_Physics/Chapter8.ipynb b/Engineering_Physics/Chapter8.ipynb deleted file mode 100755 index b3ec194e..00000000 --- a/Engineering_Physics/Chapter8.ipynb +++ /dev/null @@ -1,360 +0,0 @@ -{
- "metadata": {
- "name": "",
- "signature": "sha256:a95b407b682939fdad30498a4e63981a88538f3262f7c6d2067bc16aa9ba5b35"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "heading",
- "level": 1,
- "metadata": {},
- "source": [
- "8: Magnetic materials and Spectroscopy"
- ]
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example number 8.1, Page number 153"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#importing modules\n",
- "import math\n",
- "from __future__ import division\n",
- "\n",
- "#Variable declaration\n",
- "mew=0.9*10**-23; #magnetic dipole moment(J/T)\n",
- "B=0.72; #magnetic field applied(T)\n",
- "k=1.38*10**-23; #boltzmann constant\n",
- "\n",
- "#Calculation \n",
- "T=(2*mew*B)/(3*k); #temperature(K)\n",
- "\n",
- "#Result\n",
- "print \"The temperature is\",round(T,2),\"K\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The temperature is 0.31 K\n"
- ]
- }
- ],
- "prompt_number": 2
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example number 8.2, Page number 153"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#importing modules\n",
- "import math\n",
- "from __future__ import division\n",
- "\n",
- "#Variable declaration\n",
- "#(C=mew0*M*T)/B.\n",
- "#Therefore M=(C*B)/(mew0*T)\n",
- "C=2*10**-3; #C is curies constant(K)\n",
- "B=0.4; #applied magnetic field(T)\n",
- "mew0=4*math.pi*10**-7;\n",
- "T=300; #temperature(K)\n",
- "\n",
- "#Calculation \n",
- "M=(C*B)/(mew0*T); #magnetisation(A/m)\n",
- "\n",
- "#Result\n",
- "print \"magnetisation is\",round(M,2),\"A/m\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "magnetisation is 2.12 A/m\n"
- ]
- }
- ],
- "prompt_number": 4
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example number 8.3, Page number 153"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#importing modules\n",
- "import math\n",
- "from __future__ import division\n",
- "\n",
- "#Variable declaration\n",
- "e=1.6*10**-19;\n",
- "B=0.35; #magnetic field(T)\n",
- "lamda=500*10**-9; #wavelength(m)\n",
- "m=9.1*10**-31;\n",
- "c=3*10**8; #speed of light \n",
- "\n",
- "#Calculation \n",
- "deltalambda=(e*B*(lamda)**2)/(4*(math.pi)*m*c*10**-9); #Zeeman shift in wave length(nm)\n",
- "\n",
- "#Result\n",
- "print \"Zeeman shift in wave length is\",round(deltalambda,5),\"nm\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Zeeman shift in wave length is 0.00408 nm\n"
- ]
- }
- ],
- "prompt_number": 7
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example number 8.4, Page number 154"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#importing modules\n",
- "import math\n",
- "from __future__ import division\n",
- "\n",
- "#Variable declaration\n",
- "#T=(C*B)/(mew0*B)\n",
- "C=2.1*10**-3; #C is curie's constant(K)\n",
- "B=0.38; #magnetic field(T)\n",
- "mew0=4*math.pi*10**-7; #molecular magnetic moment\n",
- "M=2.15; #magnetisation(A/m)\n",
- "\n",
- "#Calculation \n",
- "T=(C*B)/(mew0*M); #temperature(K)\n",
- "\n",
- "#Result\n",
- "print \"Temperature is\",round(T,1),\"K\"\n",
- "print \"answer in the book varies due to rounding off errors\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Temperature is 295.4 K\n",
- "answer in the book varies due to rounding off errors\n"
- ]
- }
- ],
- "prompt_number": 9
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example number 8.5, Page number 154"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#importing modules\n",
- "import math\n",
- "from __future__ import division\n",
- "\n",
- "#Variable declaration\n",
- "#(M1*T1)=(M2*T2).Therefore M2=(M1*T1)/T2\n",
- "M1=2; #Initial magnetisation(A/m)\n",
- "T1=305; #Initial temperature(K)\n",
- "T2=321;\t\t #final temperature(K)\t\n",
- "\n",
- "#Calculation \n",
- "M2=(M1*T1)/T2; #magnetisation at 321K(A/m)\n",
- "\n",
- "#Result\n",
- "print \"Magnetisation at 321 K is\",round(M2,1),\"A/m\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Magnetisation at 321 K is 1.9 A/m\n"
- ]
- }
- ],
- "prompt_number": 11
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example number 8.6, Page number 154"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#importing modules\n",
- "import math\n",
- "from __future__ import division\n",
- "\n",
- "#Variable declaration\n",
- "mew0=4*math.pi*10**-7; #molecular magnetic moment\n",
- "M=4; #magnetisation(A/m)\n",
- "T=310; #temperature(K)\n",
- "C=1.9*10**-3; #Curie's constant(K)\n",
- "\n",
- "#Calculation \n",
- "B=(mew0*M*T)/C; #magnetic field(T)\n",
- "\n",
- "#Result\n",
- "print \"Magnetic field is\",round(B,2),\"T\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Magnetic field is 0.82 T\n"
- ]
- }
- ],
- "prompt_number": 13
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example number 8.7, Page number 154"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#importing modules\n",
- "import math\n",
- "from __future__ import division\n",
- "\n",
- "#Variable declaration\n",
- "#e/m is gyromagnetic ratio.\n",
- "deltalambda=0.01*10**-9; #Zeeman shift(m)\n",
- "c=3*10**8; #speed of light in vacuum(m/s)\n",
- "lamda=600*10**-9; #wavelength(m)\n",
- "e=1.6*10**-19;\n",
- "m=9.1*10**-31;\n",
- "\n",
- "#Calculation \n",
- "B=(deltalambda*4*math.pi*m*c)/(e*(lamda)**2); #uniform magnetic field(T)\n",
- "\n",
- "#Result\n",
- "print \"Magnetic field is\",round(B,4),\"T\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Magnetic field is 0.5956 T\n"
- ]
- }
- ],
- "prompt_number": 16
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example number 8.8, Page number 154"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#importing modules\n",
- "import math\n",
- "from __future__ import division\n",
- "\n",
- "#Variable declaration\n",
- "deltalambda=0.01*10**-9; #Zeeman shift(m)\n",
- "c=3*10**8; #speed of light in vacuum(m/s)\n",
- "B=0.78; #magnetic field(T)\n",
- "lamda=550*10**-9; #wavelength(m)\n",
- "\n",
- "#Calculation \n",
- "Y=(deltalambda*4*math.pi*3*10**8)/(B*(lamda)**2); #e/m ratio(C/kg)\n",
- "\n",
- "#Result\n",
- "print \"e/m ratio is\",round(Y/10**11,1),\"*10**11 C/kg\"\n",
- "print \"answer in the book varies due to rounding off errors\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "e/m ratio is 1.6 *10**11 C/kg\n",
- "answer in the book varies due to rounding off errors\n"
- ]
- }
- ],
- "prompt_number": 21
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Engineering_Physics/Chapter8_1.ipynb b/Engineering_Physics/Chapter8_1.ipynb deleted file mode 100755 index b3ec194e..00000000 --- a/Engineering_Physics/Chapter8_1.ipynb +++ /dev/null @@ -1,360 +0,0 @@ -{
- "metadata": {
- "name": "",
- "signature": "sha256:a95b407b682939fdad30498a4e63981a88538f3262f7c6d2067bc16aa9ba5b35"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "heading",
- "level": 1,
- "metadata": {},
- "source": [
- "8: Magnetic materials and Spectroscopy"
- ]
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example number 8.1, Page number 153"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#importing modules\n",
- "import math\n",
- "from __future__ import division\n",
- "\n",
- "#Variable declaration\n",
- "mew=0.9*10**-23; #magnetic dipole moment(J/T)\n",
- "B=0.72; #magnetic field applied(T)\n",
- "k=1.38*10**-23; #boltzmann constant\n",
- "\n",
- "#Calculation \n",
- "T=(2*mew*B)/(3*k); #temperature(K)\n",
- "\n",
- "#Result\n",
- "print \"The temperature is\",round(T,2),\"K\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The temperature is 0.31 K\n"
- ]
- }
- ],
- "prompt_number": 2
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example number 8.2, Page number 153"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#importing modules\n",
- "import math\n",
- "from __future__ import division\n",
- "\n",
- "#Variable declaration\n",
- "#(C=mew0*M*T)/B.\n",
- "#Therefore M=(C*B)/(mew0*T)\n",
- "C=2*10**-3; #C is curies constant(K)\n",
- "B=0.4; #applied magnetic field(T)\n",
- "mew0=4*math.pi*10**-7;\n",
- "T=300; #temperature(K)\n",
- "\n",
- "#Calculation \n",
- "M=(C*B)/(mew0*T); #magnetisation(A/m)\n",
- "\n",
- "#Result\n",
- "print \"magnetisation is\",round(M,2),\"A/m\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "magnetisation is 2.12 A/m\n"
- ]
- }
- ],
- "prompt_number": 4
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example number 8.3, Page number 153"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#importing modules\n",
- "import math\n",
- "from __future__ import division\n",
- "\n",
- "#Variable declaration\n",
- "e=1.6*10**-19;\n",
- "B=0.35; #magnetic field(T)\n",
- "lamda=500*10**-9; #wavelength(m)\n",
- "m=9.1*10**-31;\n",
- "c=3*10**8; #speed of light \n",
- "\n",
- "#Calculation \n",
- "deltalambda=(e*B*(lamda)**2)/(4*(math.pi)*m*c*10**-9); #Zeeman shift in wave length(nm)\n",
- "\n",
- "#Result\n",
- "print \"Zeeman shift in wave length is\",round(deltalambda,5),\"nm\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Zeeman shift in wave length is 0.00408 nm\n"
- ]
- }
- ],
- "prompt_number": 7
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example number 8.4, Page number 154"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#importing modules\n",
- "import math\n",
- "from __future__ import division\n",
- "\n",
- "#Variable declaration\n",
- "#T=(C*B)/(mew0*B)\n",
- "C=2.1*10**-3; #C is curie's constant(K)\n",
- "B=0.38; #magnetic field(T)\n",
- "mew0=4*math.pi*10**-7; #molecular magnetic moment\n",
- "M=2.15; #magnetisation(A/m)\n",
- "\n",
- "#Calculation \n",
- "T=(C*B)/(mew0*M); #temperature(K)\n",
- "\n",
- "#Result\n",
- "print \"Temperature is\",round(T,1),\"K\"\n",
- "print \"answer in the book varies due to rounding off errors\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Temperature is 295.4 K\n",
- "answer in the book varies due to rounding off errors\n"
- ]
- }
- ],
- "prompt_number": 9
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example number 8.5, Page number 154"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#importing modules\n",
- "import math\n",
- "from __future__ import division\n",
- "\n",
- "#Variable declaration\n",
- "#(M1*T1)=(M2*T2).Therefore M2=(M1*T1)/T2\n",
- "M1=2; #Initial magnetisation(A/m)\n",
- "T1=305; #Initial temperature(K)\n",
- "T2=321;\t\t #final temperature(K)\t\n",
- "\n",
- "#Calculation \n",
- "M2=(M1*T1)/T2; #magnetisation at 321K(A/m)\n",
- "\n",
- "#Result\n",
- "print \"Magnetisation at 321 K is\",round(M2,1),\"A/m\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Magnetisation at 321 K is 1.9 A/m\n"
- ]
- }
- ],
- "prompt_number": 11
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example number 8.6, Page number 154"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#importing modules\n",
- "import math\n",
- "from __future__ import division\n",
- "\n",
- "#Variable declaration\n",
- "mew0=4*math.pi*10**-7; #molecular magnetic moment\n",
- "M=4; #magnetisation(A/m)\n",
- "T=310; #temperature(K)\n",
- "C=1.9*10**-3; #Curie's constant(K)\n",
- "\n",
- "#Calculation \n",
- "B=(mew0*M*T)/C; #magnetic field(T)\n",
- "\n",
- "#Result\n",
- "print \"Magnetic field is\",round(B,2),\"T\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Magnetic field is 0.82 T\n"
- ]
- }
- ],
- "prompt_number": 13
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example number 8.7, Page number 154"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#importing modules\n",
- "import math\n",
- "from __future__ import division\n",
- "\n",
- "#Variable declaration\n",
- "#e/m is gyromagnetic ratio.\n",
- "deltalambda=0.01*10**-9; #Zeeman shift(m)\n",
- "c=3*10**8; #speed of light in vacuum(m/s)\n",
- "lamda=600*10**-9; #wavelength(m)\n",
- "e=1.6*10**-19;\n",
- "m=9.1*10**-31;\n",
- "\n",
- "#Calculation \n",
- "B=(deltalambda*4*math.pi*m*c)/(e*(lamda)**2); #uniform magnetic field(T)\n",
- "\n",
- "#Result\n",
- "print \"Magnetic field is\",round(B,4),\"T\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Magnetic field is 0.5956 T\n"
- ]
- }
- ],
- "prompt_number": 16
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example number 8.8, Page number 154"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#importing modules\n",
- "import math\n",
- "from __future__ import division\n",
- "\n",
- "#Variable declaration\n",
- "deltalambda=0.01*10**-9; #Zeeman shift(m)\n",
- "c=3*10**8; #speed of light in vacuum(m/s)\n",
- "B=0.78; #magnetic field(T)\n",
- "lamda=550*10**-9; #wavelength(m)\n",
- "\n",
- "#Calculation \n",
- "Y=(deltalambda*4*math.pi*3*10**8)/(B*(lamda)**2); #e/m ratio(C/kg)\n",
- "\n",
- "#Result\n",
- "print \"e/m ratio is\",round(Y/10**11,1),\"*10**11 C/kg\"\n",
- "print \"answer in the book varies due to rounding off errors\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "e/m ratio is 1.6 *10**11 C/kg\n",
- "answer in the book varies due to rounding off errors\n"
- ]
- }
- ],
- "prompt_number": 21
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Engineering_Physics/Chapter9.ipynb b/Engineering_Physics/Chapter9.ipynb deleted file mode 100755 index 29c8edee..00000000 --- a/Engineering_Physics/Chapter9.ipynb +++ /dev/null @@ -1,576 +0,0 @@ -{
- "metadata": {
- "name": "",
- "signature": "sha256:495ac96015f20ad80c50d2c1722e924721169f0ca7b7ca56739c5ef92f3f2a43"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "heading",
- "level": 1,
- "metadata": {},
- "source": [
- "9: Quantum Theory"
- ]
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example number 9.1, Page number 171"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#importing modules\n",
- "import math\n",
- "from __future__ import division\n",
- "\n",
- "#Variable declaration\n",
- "r=0.05; #radius of the wire(mm)\n",
- "l=4; #length of the wire(cm)\n",
- "e=1;\n",
- "T=3000; #temperature(K)\n",
- "s=5.6703*10**-8; #stefan's constant \n",
- "\n",
- "#Calculation \n",
- "A=2*math.pi*r*l*10**-5; #area(m**2)\n",
- "p=s*T**4*A*e; #power radiated by the filament(W)\n",
- "\n",
- "#Result\n",
- "print \"The power radiated by the filament is\",round(p,2),\"W\"\n",
- "print \"answer given in the book is wrong\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The power radiated by the filament is 57.72 W\n",
- "answer given in the book is wrong\n"
- ]
- }
- ],
- "prompt_number": 7
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example number 9.2, Page number 171"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#importing modules\n",
- "import math\n",
- "from __future__ import division\n",
- "\n",
- "#Variable declaration\n",
- "h=6.626*10**-34; #plancks constant\n",
- "c=3*10**8; #speed of light(m/s)\n",
- "lamda=550; #wavelength(nm)\n",
- "\n",
- "#Calculation \n",
- "E=(h*c)/(lamda*10**-9); #energy of photon(J)\n",
- "Es=0.1/E; #number of photons(per square cm per second)\n",
- "\n",
- "#Result\n",
- "print \"The number of photons are\",round(Es/10**17,2),\"*10**17 per square cm per second\"\n",
- "print \"answer in the book varies due to rounding off errors\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The number of photons are 2.77 *10**17 per square cm per second\n",
- "answer in the book varies due to rounding off errors\n"
- ]
- }
- ],
- "prompt_number": 11
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example number 9.3, Page number 171"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#importing modules\n",
- "import math\n",
- "from __future__ import division\n",
- "\n",
- "#Variable declaration\n",
- "h=6.626*10**-34; #plancks constant\n",
- "c=3*10**8; #speed of light(m/s)\n",
- "lamda=300*10**-9; #wavelength(m)\n",
- "e=1.6*10**-19;\n",
- "phi=2.2; #work function(eV)\n",
- "\n",
- "#Calculation \n",
- "E=(h*c)/lamda; #energy of photon(J)\n",
- "Kmax=(E-(phi*e))/e; #maximum kinetic energy(eV)\n",
- "\n",
- "#Result\n",
- "print \"The maximum kinetic energy is\",round(Kmax,2),\"eV\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The maximum kinetic energy is 1.94 eV\n"
- ]
- }
- ],
- "prompt_number": 17
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example number 9.4, Page number 172"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#importing modules\n",
- "import math\n",
- "from __future__ import division\n",
- "\n",
- "#Variable declaration\n",
- "h=6.626*10**-34; #plancks constant\n",
- "c=3*10**8; #speed of light(m/s)\n",
- "lamda=175*10**-9; #wavelength of light(m)\n",
- "w=5; #work function of nickel(eV)\n",
- "\n",
- "#Calculation \n",
- "E=(h*c)/(lamda*1.6*10**-19); #Energy of 200 nm photon(eV)\n",
- "#From photoelectric equation E-w is the potential difference\n",
- "p=E-w; #potential difference required to stop the fastest electron(eV)\n",
- "\n",
- "#Result\n",
- "print \"The potential difference that should be applied is\",round(p,1),\"V\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The potential difference that should be applied is 2.1 V\n"
- ]
- }
- ],
- "prompt_number": 21
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example number 9.5, Page number 172"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#importing modules\n",
- "import math\n",
- "from __future__ import division\n",
- "\n",
- "#Variable declaration\n",
- "h=6.626*10**-34; #plancks constant\n",
- "c=3*10**8; #speed of light(m/s)\n",
- "e=1.6*10**-19;\n",
- "V=50; #accelerating voltage(kV)\n",
- "\n",
- "#Calculation \n",
- "lambdamin=((h*c)/(e*V*10**3))*10**9; #shortest wavelength of X-rays(nm)\n",
- "\n",
- "#Result\n",
- "print \"The shortest wavelength of X-rays is\",round(lambdamin,4),\"nm\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The shortest wavelength of X-rays is 0.0248 nm\n"
- ]
- }
- ],
- "prompt_number": 23
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example number 9.6, Page number 172"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#importing modules\n",
- "import math\n",
- "from __future__ import division\n",
- "\n",
- "#Variable declaration\n",
- "lambda1=0.708; #wavelength of a certain line in an X-ray spectrum(angstrom)\n",
- "Z1=42; #atomic number\n",
- "Z2=24;\n",
- "a=1; #screening constant\n",
- "\n",
- "#Calculation \n",
- "lambda2=(lambda1*(Z1-a)**2)/((Z2-a)**2); #wavelength of same line(angstrom)\n",
- "\n",
- "#Result\n",
- "print \"The wavelength of same line is\",round(lambda2,2),\"angstrom\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The wavelength of same line is 2.25 angstrom\n"
- ]
- }
- ],
- "prompt_number": 25
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example number 9.7, Page number 172"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#importing modules\n",
- "import math\n",
- "from __future__ import division\n",
- "\n",
- "#Variable declaration\n",
- "#From Bragg's law 2*d*sin(teta)=n*lambda\n",
- "n=1;\n",
- "lamda=0.32; #wavelength(nm)\n",
- "theta=28; #angle at which first order Bragg's reflection is observed(degrees)\n",
- "\n",
- "#Calculation \n",
- "theta=theta*math.pi/180; #angle(radian)\n",
- "d=lamda/(2*math.sin(theta)); #distance between atomic planes(nm)\n",
- "\n",
- "#Result\n",
- "print \"The distance between atomic planes is\",round(d,2),\"nm\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The distance between atomic planes is 0.34 nm\n"
- ]
- }
- ],
- "prompt_number": 28
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example number 9.8, Page number 172"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#importing modules\n",
- "import math\n",
- "from __future__ import division\n",
- "\n",
- "#Variable declaration\n",
- "h=6.626*10**-34; #plancks constant\n",
- "theta=50; #angle(degrees)\n",
- "m=9.1*10**-31; #mass of electron(kg)\n",
- "c=3*10**8; #speed of light(m/s)\n",
- "\n",
- "#Calculation \n",
- "theta=theta*math.pi/180; #angle(radian)\n",
- "deltalambda=(h/(m*c))*(1-math.cos(theta))*10**12; \n",
- "lambdafin=2.5; #wavelength of scattered X-rays\n",
- "lambdainit=lambdafin-deltalambda; #wavelength of X-rays in the incident beam(pm)\n",
- "\n",
- "#Result\n",
- "print \"The wavelength of X-rays in the incident beam is\",round(lambdainit,2),\"pm\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The wavelength of X-rays in the incident beam is 1.63 pm\n"
- ]
- }
- ],
- "prompt_number": 30
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example number 9.9, Page number 172"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#importing modules\n",
- "import math\n",
- "from __future__ import division\n",
- "\n",
- "#Variable declaration\n",
- "h=6.626*10**-34; #plancks constant\n",
- "c=3*10**8; #speed of light(m/s)\n",
- "lamda=500*10**-9; #wavelength of laser(m)\n",
- "t=20*10**-3; #time(s)\n",
- "N=2.52*10**16; #number of photons in a 20ms pulse\n",
- "\n",
- "#Calculation \n",
- "E=(h*c)/lamda; #Energy of 500 nm photon(J)\n",
- "p=E*N/t; #power of the laser(W)\n",
- "\n",
- "#Result\n",
- "print \"The power of the laser is\",round(p,1),\"W\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The power of the laser is 0.5 W\n"
- ]
- }
- ],
- "prompt_number": 34
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example number 9.10, Page number 173"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#importing modules\n",
- "import math\n",
- "from __future__ import division\n",
- "\n",
- "#Variable declaration\n",
- "h=6.626*10**-34; #plancks constant\n",
- "c=3*10**8; #speed of light(m/s)\n",
- "lamda=350*10**-9; #threshold wavelength(m)\n",
- "e=1.6*10**-19;\n",
- "\n",
- "#Calculation \n",
- "W=h*c/(lamda*e); #work function of the surface(eV)\n",
- "\n",
- "#Result\n",
- "print \"The work function of the surface is\",round(W,2),\"eV\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The work function of the surface is 3.55 eV\n"
- ]
- }
- ],
- "prompt_number": 40
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example number 9.11, Page number 173"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#importing modules\n",
- "import math\n",
- "from __future__ import division\n",
- "\n",
- "#Variable declaration\n",
- "h=6.626*10**-34; #plancks constant\n",
- "c=3*10**8; #speed of light(m/s)\n",
- "e=1.6*10**-19;\n",
- "lambdamin=0.02*10**-9; #minimum wavelength(m)\n",
- "\n",
- "#Calculation \n",
- "V=(h*c/(lambdamin*e))*10**-3; #accelerating voltage(kV)\n",
- "\n",
- "#Result\n",
- "print \"The accelerating voltage needed to produce minimum wavelength is\",round(V,4),\"kV\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The accelerating voltage needed to produce minimum wavelength is 62.1187 kV\n"
- ]
- }
- ],
- "prompt_number": 42
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example number 9.12, Page number 173"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#importing modules\n",
- "import math\n",
- "from __future__ import division\n",
- "\n",
- "#Variable declaration\n",
- "#According to Bragg's eq.2*d*sin(teta)=n*lambda\n",
- "n=2; #since second order Bragg's eq.\n",
- "d=5; #since d=5(lambda)\n",
- "lamda=1;\n",
- "\n",
- "#Calculation \n",
- "a=(n*lamda)/(2*5*lamda);\n",
- "theta=math.asin(a); #angle of second order Braggs reflection(radian)\n",
- "theta=theta*180/math.pi; #angle(degrees)\n",
- "\n",
- "#Result\n",
- "print \"The angle of second order Braggs reflection is\",round(theta,2),\"degrees\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The angle of second order Braggs reflection is 11.54 degrees\n"
- ]
- }
- ],
- "prompt_number": 45
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example number 9.13, Page number 173"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#importing modules\n",
- "import math\n",
- "from __future__ import division\n",
- "\n",
- "#Variable declaration\n",
- "h=6.626*10**-34; #plancks constant\n",
- "c=3*10**8; #speed of light(m/s)\n",
- "lamda=0.03; #wavelength(nm)\n",
- "p=80/100;\n",
- "\n",
- "#Calculation \n",
- "E=(h*c)/(lamda*10**-9); #energy of photon(J) \n",
- "TE=E/p; #Total energy.E=80% of TE(J)\n",
- "TE=TE*(10**-3)/e; #Total energy(keV)\n",
- "\n",
- "#Result\n",
- "print \"The electron must have been accelerated through a potential difference of\",round(TE,3),\"kV\" "
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The electron must have been accelerated through a potential difference of 51.766 kV\n"
- ]
- }
- ],
- "prompt_number": 49
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Engineering_Physics/Chapter9_1.ipynb b/Engineering_Physics/Chapter9_1.ipynb deleted file mode 100755 index 29c8edee..00000000 --- a/Engineering_Physics/Chapter9_1.ipynb +++ /dev/null @@ -1,576 +0,0 @@ -{
- "metadata": {
- "name": "",
- "signature": "sha256:495ac96015f20ad80c50d2c1722e924721169f0ca7b7ca56739c5ef92f3f2a43"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "heading",
- "level": 1,
- "metadata": {},
- "source": [
- "9: Quantum Theory"
- ]
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example number 9.1, Page number 171"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#importing modules\n",
- "import math\n",
- "from __future__ import division\n",
- "\n",
- "#Variable declaration\n",
- "r=0.05; #radius of the wire(mm)\n",
- "l=4; #length of the wire(cm)\n",
- "e=1;\n",
- "T=3000; #temperature(K)\n",
- "s=5.6703*10**-8; #stefan's constant \n",
- "\n",
- "#Calculation \n",
- "A=2*math.pi*r*l*10**-5; #area(m**2)\n",
- "p=s*T**4*A*e; #power radiated by the filament(W)\n",
- "\n",
- "#Result\n",
- "print \"The power radiated by the filament is\",round(p,2),\"W\"\n",
- "print \"answer given in the book is wrong\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The power radiated by the filament is 57.72 W\n",
- "answer given in the book is wrong\n"
- ]
- }
- ],
- "prompt_number": 7
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example number 9.2, Page number 171"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#importing modules\n",
- "import math\n",
- "from __future__ import division\n",
- "\n",
- "#Variable declaration\n",
- "h=6.626*10**-34; #plancks constant\n",
- "c=3*10**8; #speed of light(m/s)\n",
- "lamda=550; #wavelength(nm)\n",
- "\n",
- "#Calculation \n",
- "E=(h*c)/(lamda*10**-9); #energy of photon(J)\n",
- "Es=0.1/E; #number of photons(per square cm per second)\n",
- "\n",
- "#Result\n",
- "print \"The number of photons are\",round(Es/10**17,2),\"*10**17 per square cm per second\"\n",
- "print \"answer in the book varies due to rounding off errors\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The number of photons are 2.77 *10**17 per square cm per second\n",
- "answer in the book varies due to rounding off errors\n"
- ]
- }
- ],
- "prompt_number": 11
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example number 9.3, Page number 171"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#importing modules\n",
- "import math\n",
- "from __future__ import division\n",
- "\n",
- "#Variable declaration\n",
- "h=6.626*10**-34; #plancks constant\n",
- "c=3*10**8; #speed of light(m/s)\n",
- "lamda=300*10**-9; #wavelength(m)\n",
- "e=1.6*10**-19;\n",
- "phi=2.2; #work function(eV)\n",
- "\n",
- "#Calculation \n",
- "E=(h*c)/lamda; #energy of photon(J)\n",
- "Kmax=(E-(phi*e))/e; #maximum kinetic energy(eV)\n",
- "\n",
- "#Result\n",
- "print \"The maximum kinetic energy is\",round(Kmax,2),\"eV\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The maximum kinetic energy is 1.94 eV\n"
- ]
- }
- ],
- "prompt_number": 17
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example number 9.4, Page number 172"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#importing modules\n",
- "import math\n",
- "from __future__ import division\n",
- "\n",
- "#Variable declaration\n",
- "h=6.626*10**-34; #plancks constant\n",
- "c=3*10**8; #speed of light(m/s)\n",
- "lamda=175*10**-9; #wavelength of light(m)\n",
- "w=5; #work function of nickel(eV)\n",
- "\n",
- "#Calculation \n",
- "E=(h*c)/(lamda*1.6*10**-19); #Energy of 200 nm photon(eV)\n",
- "#From photoelectric equation E-w is the potential difference\n",
- "p=E-w; #potential difference required to stop the fastest electron(eV)\n",
- "\n",
- "#Result\n",
- "print \"The potential difference that should be applied is\",round(p,1),\"V\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The potential difference that should be applied is 2.1 V\n"
- ]
- }
- ],
- "prompt_number": 21
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example number 9.5, Page number 172"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#importing modules\n",
- "import math\n",
- "from __future__ import division\n",
- "\n",
- "#Variable declaration\n",
- "h=6.626*10**-34; #plancks constant\n",
- "c=3*10**8; #speed of light(m/s)\n",
- "e=1.6*10**-19;\n",
- "V=50; #accelerating voltage(kV)\n",
- "\n",
- "#Calculation \n",
- "lambdamin=((h*c)/(e*V*10**3))*10**9; #shortest wavelength of X-rays(nm)\n",
- "\n",
- "#Result\n",
- "print \"The shortest wavelength of X-rays is\",round(lambdamin,4),\"nm\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The shortest wavelength of X-rays is 0.0248 nm\n"
- ]
- }
- ],
- "prompt_number": 23
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example number 9.6, Page number 172"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#importing modules\n",
- "import math\n",
- "from __future__ import division\n",
- "\n",
- "#Variable declaration\n",
- "lambda1=0.708; #wavelength of a certain line in an X-ray spectrum(angstrom)\n",
- "Z1=42; #atomic number\n",
- "Z2=24;\n",
- "a=1; #screening constant\n",
- "\n",
- "#Calculation \n",
- "lambda2=(lambda1*(Z1-a)**2)/((Z2-a)**2); #wavelength of same line(angstrom)\n",
- "\n",
- "#Result\n",
- "print \"The wavelength of same line is\",round(lambda2,2),\"angstrom\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The wavelength of same line is 2.25 angstrom\n"
- ]
- }
- ],
- "prompt_number": 25
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example number 9.7, Page number 172"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#importing modules\n",
- "import math\n",
- "from __future__ import division\n",
- "\n",
- "#Variable declaration\n",
- "#From Bragg's law 2*d*sin(teta)=n*lambda\n",
- "n=1;\n",
- "lamda=0.32; #wavelength(nm)\n",
- "theta=28; #angle at which first order Bragg's reflection is observed(degrees)\n",
- "\n",
- "#Calculation \n",
- "theta=theta*math.pi/180; #angle(radian)\n",
- "d=lamda/(2*math.sin(theta)); #distance between atomic planes(nm)\n",
- "\n",
- "#Result\n",
- "print \"The distance between atomic planes is\",round(d,2),\"nm\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The distance between atomic planes is 0.34 nm\n"
- ]
- }
- ],
- "prompt_number": 28
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example number 9.8, Page number 172"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#importing modules\n",
- "import math\n",
- "from __future__ import division\n",
- "\n",
- "#Variable declaration\n",
- "h=6.626*10**-34; #plancks constant\n",
- "theta=50; #angle(degrees)\n",
- "m=9.1*10**-31; #mass of electron(kg)\n",
- "c=3*10**8; #speed of light(m/s)\n",
- "\n",
- "#Calculation \n",
- "theta=theta*math.pi/180; #angle(radian)\n",
- "deltalambda=(h/(m*c))*(1-math.cos(theta))*10**12; \n",
- "lambdafin=2.5; #wavelength of scattered X-rays\n",
- "lambdainit=lambdafin-deltalambda; #wavelength of X-rays in the incident beam(pm)\n",
- "\n",
- "#Result\n",
- "print \"The wavelength of X-rays in the incident beam is\",round(lambdainit,2),\"pm\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The wavelength of X-rays in the incident beam is 1.63 pm\n"
- ]
- }
- ],
- "prompt_number": 30
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example number 9.9, Page number 172"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#importing modules\n",
- "import math\n",
- "from __future__ import division\n",
- "\n",
- "#Variable declaration\n",
- "h=6.626*10**-34; #plancks constant\n",
- "c=3*10**8; #speed of light(m/s)\n",
- "lamda=500*10**-9; #wavelength of laser(m)\n",
- "t=20*10**-3; #time(s)\n",
- "N=2.52*10**16; #number of photons in a 20ms pulse\n",
- "\n",
- "#Calculation \n",
- "E=(h*c)/lamda; #Energy of 500 nm photon(J)\n",
- "p=E*N/t; #power of the laser(W)\n",
- "\n",
- "#Result\n",
- "print \"The power of the laser is\",round(p,1),\"W\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The power of the laser is 0.5 W\n"
- ]
- }
- ],
- "prompt_number": 34
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example number 9.10, Page number 173"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#importing modules\n",
- "import math\n",
- "from __future__ import division\n",
- "\n",
- "#Variable declaration\n",
- "h=6.626*10**-34; #plancks constant\n",
- "c=3*10**8; #speed of light(m/s)\n",
- "lamda=350*10**-9; #threshold wavelength(m)\n",
- "e=1.6*10**-19;\n",
- "\n",
- "#Calculation \n",
- "W=h*c/(lamda*e); #work function of the surface(eV)\n",
- "\n",
- "#Result\n",
- "print \"The work function of the surface is\",round(W,2),\"eV\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The work function of the surface is 3.55 eV\n"
- ]
- }
- ],
- "prompt_number": 40
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example number 9.11, Page number 173"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#importing modules\n",
- "import math\n",
- "from __future__ import division\n",
- "\n",
- "#Variable declaration\n",
- "h=6.626*10**-34; #plancks constant\n",
- "c=3*10**8; #speed of light(m/s)\n",
- "e=1.6*10**-19;\n",
- "lambdamin=0.02*10**-9; #minimum wavelength(m)\n",
- "\n",
- "#Calculation \n",
- "V=(h*c/(lambdamin*e))*10**-3; #accelerating voltage(kV)\n",
- "\n",
- "#Result\n",
- "print \"The accelerating voltage needed to produce minimum wavelength is\",round(V,4),\"kV\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The accelerating voltage needed to produce minimum wavelength is 62.1187 kV\n"
- ]
- }
- ],
- "prompt_number": 42
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example number 9.12, Page number 173"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#importing modules\n",
- "import math\n",
- "from __future__ import division\n",
- "\n",
- "#Variable declaration\n",
- "#According to Bragg's eq.2*d*sin(teta)=n*lambda\n",
- "n=2; #since second order Bragg's eq.\n",
- "d=5; #since d=5(lambda)\n",
- "lamda=1;\n",
- "\n",
- "#Calculation \n",
- "a=(n*lamda)/(2*5*lamda);\n",
- "theta=math.asin(a); #angle of second order Braggs reflection(radian)\n",
- "theta=theta*180/math.pi; #angle(degrees)\n",
- "\n",
- "#Result\n",
- "print \"The angle of second order Braggs reflection is\",round(theta,2),\"degrees\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The angle of second order Braggs reflection is 11.54 degrees\n"
- ]
- }
- ],
- "prompt_number": 45
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example number 9.13, Page number 173"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#importing modules\n",
- "import math\n",
- "from __future__ import division\n",
- "\n",
- "#Variable declaration\n",
- "h=6.626*10**-34; #plancks constant\n",
- "c=3*10**8; #speed of light(m/s)\n",
- "lamda=0.03; #wavelength(nm)\n",
- "p=80/100;\n",
- "\n",
- "#Calculation \n",
- "E=(h*c)/(lamda*10**-9); #energy of photon(J) \n",
- "TE=E/p; #Total energy.E=80% of TE(J)\n",
- "TE=TE*(10**-3)/e; #Total energy(keV)\n",
- "\n",
- "#Result\n",
- "print \"The electron must have been accelerated through a potential difference of\",round(TE,3),\"kV\" "
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The electron must have been accelerated through a potential difference of 51.766 kV\n"
- ]
- }
- ],
- "prompt_number": 49
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Engineering_Physics/README.txt b/Engineering_Physics/README.txt deleted file mode 100755 index 98363f45..00000000 --- a/Engineering_Physics/README.txt +++ /dev/null @@ -1,10 +0,0 @@ -Contributed By: SINDHU ARROJU -Course: btech -College/Institute/Organization: JNTUH -Department/Designation: Computer Science -Book Title: Engineering Physics -Author: P. V. Naik -Publisher: Pearson Education(Singapore) -Year of publication: 2004 -Isbn: 8129705206 -Edition: 1
\ No newline at end of file diff --git a/Engineering_Physics/chapter1.ipynb b/Engineering_Physics/chapter1.ipynb deleted file mode 100755 index 69791eee..00000000 --- a/Engineering_Physics/chapter1.ipynb +++ /dev/null @@ -1,862 +0,0 @@ -{
- "metadata": {
- "name": "",
- "signature": "sha256:0e6e83846c51774a2df15d43a35bd12afc7130920f3c7a5b4cb2d5a70d8dae4a"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "heading",
- "level": 1,
- "metadata": {},
- "source": [
- "Chapter1:RELATIVISTIC MECHANICS"
- ]
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Ex1.1:pg-12"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "# example 1\n",
- "import math\n",
- "#to calculate length of the bar measured by the ststionary observer\n",
- "lo =1 #length in metre\n",
- "v=0.75*3*10**8 #speed (m/s)\n",
- "c=3*10**8 #light speed(m/s)\n",
- "l=lo*math.sqrt(1-(v**2/c**2))\n",
- "print \"\\n the length of bar in is\",round(l,2),\"m\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- " the length of bar in is 0.66 m\n"
- ]
- }
- ],
- "prompt_number": 11
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Ex1.2:pg-12"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "# example 2\n",
- "import math\n",
- "#to calculate velocity of rocket \n",
- "#lo be the length at rest \n",
- "l=99.0/100 #length is 99 per cent of its length at rest is l=(99/100)lo\n",
- "c=3*10**8 #light speed(m/s)\n",
- "v=c*math.sqrt(1-l**2) #formula is v=c math.sqrt(1-(l/lo)**2)\n",
- "print\" the velocity of rocket is v=\",\"{:.2e}\".format(v),\"m/s\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- " the velocity of rocket is v= 4.23e+07 m/s\n"
- ]
- }
- ],
- "prompt_number": 21
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Ex1.4:pg-13"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "#to percentage contraction of a rod\n",
- "c=3.0*10**8 #light speed(m/s)\n",
- "v=0.8*c #velocity(m/s)\n",
- "#let lo be the length of the rod in the frame in which it is at rest\n",
- "#s' is the frame which is moving with a speed 0.8c in a direction making an angle 60 with x-axis\n",
- "#components of lo along perpendicular to the direction of motion are lo cos60 and lo sin60 respectively\n",
- "l1=math.cos(math.pi/3)*math.sqrt(1-(v/c)**2) #length of the rod alond the direction of motion =lo math.cos(pi/3)math.sqrt(1-(v/c)**2)\n",
- "l2=math.sin(math.pi/3) #length of the rod perpendicular to the direction of motion =lo sin60\n",
- "l=math.sqrt(l1**2+l2**2) # length of the moving rod\n",
- "per=(1-l)*100/1\n",
- "print \"percentage contraction of a rod is\",round(per,2),\"%\"\n",
- "\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "percentage contraction of a rod is 8.35 %\n"
- ]
- }
- ],
- "prompt_number": 23
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Ex1.7:pg-14"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "# example 7\n",
- "import math\n",
- "#to calculate velocity of the circular lamina \n",
- "c=3*10**8 #light speed (m/s)\n",
- "#R'=R/2 (radius)\n",
- "#R'=R math.sqrt(1-(v/c)**2)\n",
- "v=(math.sqrt(3)/2)*c \n",
- "print \"velocity of the circular lamina relative to frame s is v=\",v,\"m/s\"\n",
- "#answer is given in terms of c in the textbook\n",
- "\n",
- "\n",
- "\n",
- "\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "velocity of the circular lamina relative to frame s is v= 259807621.135 m/s\n"
- ]
- }
- ],
- "prompt_number": 4
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Ex1.8:pg-17"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "# example 8\n",
- "import math\n",
- "#to calculate speed of the clock\n",
- "#clock should record l=59 minutes for each hour recorded by clocks stationary with respect to the observer\n",
- "l=59.0 \n",
- "lo=60\n",
- "c=3*10**8 #light speed (m/s)\n",
- "v=math.sqrt(c**2*(1-l**2/lo**2))\n",
- "print \"speed of the clock is v =\",\"{:.2e}\".format(v),\"m/s\"\n",
- "\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "speed of the clock is v = 5.45e+07 m/s\n"
- ]
- }
- ],
- "prompt_number": 31
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Ex1.9:pg-17"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "#to calculate distance travelled by the beam \n",
- "deltat0=2.5*10**-8 #proper half life of pi mesons in (s)\n",
- "c=3*10**8 #light speed (m/s)\n",
- "v=0.8*c #mesons velocity (m/s)\n",
- "deltat=deltat0/math.sqrt(1-(v/c)**2) #half life (s)\n",
- "#No=initial flux ,N=flux after time t\n",
- "#N=N0 e**(-t/T)\n",
- "#N=N0/e**2 (given)=No e(-t/T)\n",
- "#t=2 deltat\n",
- "d=2*deltat*v #d=vt\n",
- "print \"distance travelled by the beam is d=\",d,\"m\" \n",
- "#answer is given in the textbook=19.96 m\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "distance travelled by the beam is d= 20.0 m\n"
- ]
- }
- ],
- "prompt_number": 6
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Ex1.10:pg-17"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "#to calculate distance travelled by the particle \n",
- "deltat0=2*10**-8 #proper half life to of the particle in (s) \n",
- "c=3*10**8 #light speed (m/s)\n",
- "v=0.96*c #speed of the particle (m/s)\n",
- "deltat=(deltat0)/(math.sqrt(1-(v/c)**2)) #half life in the laboratory frame t in (s) \n",
- "#t=deltat (flux of the beam falls to (1/2) times initial flux)\n",
- "d=v*deltat #d=vt\n",
- "print \"distance travelled by the particle in this time is d=\",d,\"m\"\n",
- "#answer is given wrong in the textbook =20.45 m\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "distance travelled by the particle in this time is d= 20.5714285714 m\n"
- ]
- }
- ],
- "prompt_number": 8
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Ex1.11:pg-18"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "#to calculate speed\n",
- "deltat0=1440 #proper time interval measured by an observer moving with the clock (min)\n",
- "deltat=1444 #time interval measured by a stationary observer (min)\n",
- "c=3*10**8 #light speed (m/s)\n",
- "v=c*math.sqrt(1-(deltat0/deltat)**2)\n",
- "print \" moving clock appears to lose 4min in 24 hours from the stationary observer is v=\",\"{:.1e}\".format(v),\"m/s\"\n",
- "#answer is given wrong in the book =2.32*10**7 m/s\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- " moving clock appears to lose 4min in 24 hours from the stationary observer is v= 3.0e+08 m/s\n"
- ]
- }
- ],
- "prompt_number": 33
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Ex1.12:pg-18"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "#to calculate velocity of beta particle\n",
- "c=3*10**8 #light velocity(m/s)\n",
- "u1=0.9*c #velocity of the beta particle relative to the atom in the direction of motion\n",
- "v=0.25*c #velocity of the radioactive atom relative to an experimenter\n",
- "u=(u1+v)/(1+u1*v/c**2) \n",
- "print \" velocity of the beta particle as observed by the experimenter is u=\",\"{:.3e}\".format(u),\"m/s\"\n",
- "#answer is given in terms of c in the book =0.94c\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- " velocity of the beta particle as observed by the experimenter is u= 2.816e+08 m/s\n"
- ]
- }
- ],
- "prompt_number": 34
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Ex1.13:pg-18"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "#to calculate velocity\n",
- "c=3*10**8 # light velocity\n",
- "v=0.75*c #speed of A\n",
- "ux=-0.85*c #speed of B\n",
- "ux1=(ux-v)/(1-ux*v/c**2)\n",
- "print\"\\n the velocity of B with respect to A (m/s) is\",\"{:.3e}\".format(ux1),\"m/s\"\n",
- "#answer is given in terms of c in the book=-0.9771c\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- " the velocity of B with respect to A (m/s) is -2.931e+08 m/s\n"
- ]
- }
- ],
- "prompt_number": 36
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Ex1.14:pg-19"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "#to calculate velocity in the laboratory frame\n",
- "c=3*10**8 #light speed (m/s)\n",
- "v=0.8*c #velocity relative to laboratory along positive direction of x-axis\n",
- "#given that u'=3 i+4 j+12 k (m/s)\n",
- "ux1=3 #in (m/s)\n",
- "uy1=4 #in (m/s)\n",
- "uz1=12 #in (m/s)\n",
- "ux=(ux1+v)/(1+v*ux1/c**2)\n",
- "uy=(uy1*math.sqrt(1-(v/c)**2))/(1+v*ux1/c**2)\n",
- "uz=(uz1*math.sqrt(1-(v/c)**2))/(1+v*ux1/c**2)\n",
- "print \"u=ux i+uy j+uz k\"\n",
- "print \"where\"\n",
- "print \"ux=\", \"{:.1e}\".format(ux),\"m/s\"\n",
- "print \"uy=\",round(uy,1),\"m/s\"\n",
- "print \"uz=\",round(uz,1),\"m/s\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "u=ux i+uy j+uz k\n",
- "where\n",
- "ux= 2.4e+08 m/s\n",
- "uy= 2.4 m/s\n",
- "uz= 7.2 m/s\n"
- ]
- }
- ],
- "prompt_number": 41
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Ex1.17:pg-21"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "# to calculate velocity of the particle\n",
- "c=3.0*10**8 #light speed (m/s)\n",
- "v=0.4*c #velocity of frame s' relative to s along axis x\n",
- "ux=0.8*c*(1.0/2) #component of velocity u(=0.8 c) of the particle along x axis ux=0.8 c cos60\n",
- "uy=0.8*c*sin (math.pi/3) #component of the velocity u of the particle along y axis \n",
- "ux1=(ux-v)/(1-ux*v/c**2)\n",
- "uy1=uy*math.sqrt(1-(v/c)**2)/(1-(ux*v/c**2))\n",
- "print \"resultant velocity as observed by a person in frame s1 is u1=ux1 i+uy1 j\"\n",
- "print \"where\"\n",
- "print \"ux1=\",ux1,\"m/s\"\n",
- "print \"uy1=\",\"{:.1e}\".format(uy1),\"m/s\"\n",
- "#answer is given in terms of c in the book i.e. uy1=0.756c m/s\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "resultant velocity as observed by a person in frame s1 is u1=ux1 i+uy1 j\n",
- "where\n",
- "ux1= 0.0 m/s\n",
- "uy1= 2.3e+08 m/s\n"
- ]
- }
- ],
- "prompt_number": 48
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Ex1.18:pg-25"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "#to calculate mass, momentum,total energy,kinetic energy\n",
- "c=3*10**8 #light speed (m/s)\n",
- "v=c/sqrt (2) #velocity (m/s)\n",
- "#let mo be the rest mass of the particle \n",
- "#relativistic mass m of the particle is m=mo/math.sqrt(1-(v/c)**2)\n",
- "m=1/sqrt (1-v**2/c**2) #in kg\n",
- "print \"mass m=\",round(m,2),\" mo\"\n",
- "#momentum p of the particle is p=mv\n",
- "p=m*v #in kg-m/s\n",
- "print \"momentum p=\",\"{:.1e}\".format(p),\" mo\"\n",
- "#total energy E of the particle\n",
- "E=m*c**2 #in J\n",
- "print \"energy E=\",\"{:.3e}\".format(E,3),\" mo\"\n",
- "#kinetic energy K=E-mo c**2\n",
- "K=E-c**2 #in J\n",
- "print \"kinetic energy K=\",\"{:.3e}\".format(K),\" mo\"\n",
- "#answer is given in terms of m0 and c in the book\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "mass m= 1.41 mo\n",
- "momentum p= 3.0e+08 mo\n",
- "energy E= 1.273e+17 mo\n",
- "kinetic energy K= 3.728e+16 mo\n"
- ]
- }
- ],
- "prompt_number": 56
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Ex1.19:pg-26"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "#to calculate velocity of the parcticle \n",
- "c=3*10**8 #light speed(m/s)\n",
- "# we know that E(energy)=mc**2\n",
- "# mo=rest mass\n",
- "#E=3 moc**2=mc**2 or m=3 mo (given that total energy of the particle is thrice its rest energy) \n",
- "m=3.0 # relativistic mass \n",
- "#formula is v=c math.sqrt(1-(mo/m)**2)\n",
- "v=math.sqrt(c**2*(1-(1/m)**2))\n",
- "print \"velocity of the particle is v=\",\"{:.3e}\".format(v),\"m/s\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "velocity of the particle is v= 2.828e+08 m/s\n"
- ]
- }
- ],
- "prompt_number": 57
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Ex1.20:pg-26"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "#to calculate mass(m),speed(v) of an electron\n",
- "K=1.5*10**6*1.6*10**-19 #kinetic energy(J)\n",
- "m0=9.11*10**-31 #rest mass of an electron(kg)\n",
- "c=3*10**8 # velocity of light in vacuum(m/s)\n",
- "m=(K/c**2)+m0 #relativistic kinetic energy(k=(m-mo)c**2)\n",
- "print \"mass is m=\",\"{:.2e}\".format(m),\"kg \"\n",
- "v=c*math.sqrt(1-m0**2/m**2)\n",
- "print \"speed of an electron is v=\",\"{:.1e}\".format(v),\"m/s\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "mass is m= 3.58e-30 kg \n",
- "speed of an electron is v= 2.9e+08 m/s\n"
- ]
- }
- ],
- "prompt_number": 59
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Ex1.21:pg-27"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "#to calculate work to be done \n",
- "E=0.5*10**6 #rest energy of electron (MeV) E=m0*c**2\n",
- "v1=0.6*3*10**8 #speed of electron in (m/s)\n",
- "v2=0.8*3*10**8\n",
- "c=3.0*10**8 #speed of light in (m/s)\n",
- "K1=E*((1.0/math.sqrt(1-v1**2/c**2))-1) #kinetic energy in (eV)\n",
- "K2=E*((1.0/math.sqrt(1-v2**2/c**2))-1)\n",
- "w=(K2-K1)*1.6*10**-19\n",
- "print \"amount of work to be done is w=\",\"{:.1e}\".format(w),\"J\"\n",
- "\n",
- "\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "amount of work to be done is w= 3.3e-14 J\n"
- ]
- }
- ],
- "prompt_number": 64
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Ex1.22:pg-27"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "#to calculate speed\n",
- "c=3*10**8 #light speed (m/s)\n",
- "m=2.25 #mass m of a body be 2.25 times its rest mass mo i.e. m=2.25m0\n",
- "#formula is v=c math.sqrt(1-(m0/m)**2)\n",
- "v=c*math.sqrt(1-(1/m)**2)\n",
- "print \" speed is v=\",\"{:.3e}\".format(v),\"m/s\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- " speed is v= 2.687e+08 m/s\n"
- ]
- }
- ],
- "prompt_number": 67
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Ex1.23:pg-27"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "#to calculate speed of the rocket\n",
- "m0=50.0 #weight of man on the earth(kg)\n",
- "m=50.5 #weight of man in rocket ship (kg)\n",
- "c=3*10**8 #speed of light(m/s)\n",
- "v=c*math.sqrt(1-m0**2/m**2)\n",
- "print \"speed of the rocket is v=\",\"{:.2e}\".format(v),\"m/s\" \n",
- "#to calculate speed of electron\n",
- "m0=9.11*10**-31 #mass of electron =rest mass of proton\n",
- "m=1.67*10**-27\n",
- "v=c*math.sqrt(1-m0**2/m**2)\n",
- "print \"speed of an electron is v=\",\"{:.2e}\".format(v),\"m/s\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "speed of the rocket is v= 4.21e+07 m/s\n",
- "speed of an electron is v= 3.00e+08 m/s\n"
- ]
- }
- ],
- "prompt_number": 71
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Ex1.24:pg-28"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "#to calculate velocity\n",
- "c=3*10**8 #light speed (m/s)\n",
- "#K(kinetic energy)=(m-mo(rest mass))c**2\n",
- "#it can also be written as mc**2=K+m0c**2\n",
- "#given that K=2m0c**2(rest mass energy)\n",
- "#m=3m0\n",
- "m=3.0 #relativistic mass\n",
- "#formula is v=c math.sqrt(1-(m0/m)**2)\n",
- "v=c*math.sqrt(1-(1/m)**2)\n",
- "print \"velocity of a body is v=\",\"{:.3e}\".format(v),\"m/s\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "velocity of a body is v= 2.828e+08 m/s\n"
- ]
- }
- ],
- "prompt_number": 73
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Ex1.25:pg-28"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "#to calculate kinetic energy ,momentum of electron\n",
- "m0=9.11*10**-31 #its rest mass (kg)\n",
- "c=3*10**8 #light velocity in (m/s)\n",
- "m=11*m0 #mass of moving electron is 11 times its rest mass\n",
- "K=(m-m0)*c**2/(1.6*10**-19) # kinetic energy\n",
- "print \"kinetic energy is K=\",\"{:.2e}\".format(K),\"eV\" \n",
- "v=c*math.sqrt(1-(m0/m)**2) #velocity(m/s)\n",
- "p=m*v #momentum\n",
- "print \"momentum is p=\",\"{:.2e}\".format(p),\"kg m/s\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "kinetic energy is K= 5.12e+06 eV\n",
- "momentum is p= 2.99e-21 kg m/s\n"
- ]
- }
- ],
- "prompt_number": 74
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Ex1.26:pg-29\n",
- " "
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "#to calculate proton gain in mass\n",
- "c=3*10**8 #light speed(m/s)\n",
- "K=500*10**6*1.6*10**-19 #kinetic energy (J)\n",
- "deltam=K/c**2\n",
- "print \"proton gain in mass is delm=\",\"{:.2e}\".format(deltam),\"kg\"\n",
- "#answer is given wrong in the book=8.89*10**28 kg\n",
- "\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- " proton gain in mass is delm= 8.89e-28 kg\n"
- ]
- }
- ],
- "prompt_number": 76
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Ex1.27:pg-29"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "#to calculate speed of 0.1MeV electron\n",
- "E=0.512*10**6 #rest mass energy E=m0*c**2\n",
- "c=3*10**8 #velocity of light (m/s)\n",
- "K=0.1*10**6 #kinetic energy (MeV)\n",
- "v=c*math.sqrt(1-(E/(K+E))**2) \n",
- "print \"speed of electron is v=\",\"{:.3e}\".format(v),\"m/s\" \n",
- "#to calculate mass and speed of 2MeV electron\n",
- "E=2*10**6*1.6*10**-19 #in (J)\n",
- "m=E/c**2 \n",
- "print \"mass is m=\",\"{:.2e}\".format(m),\"kg\"\n",
- "m0=9.11*10**-31 #electron mass (kg)\n",
- "v=c*math.sqrt(1-m0**2/m**2)\n",
- "print \"speed is v=\",\"{:.2e}\".format(v),\"m/s\"\n",
- "\n",
- "\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "speed of electron is v= 1.643e+08 m/s\n",
- "mass is m= 3.56e-30 kg\n",
- "speed is v= 2.90e+08 m/s\n"
- ]
- }
- ],
- "prompt_number": 78
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [],
- "language": "python",
- "metadata": {},
- "outputs": []
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Engineering_Physics/chapter10.ipynb b/Engineering_Physics/chapter10.ipynb deleted file mode 100755 index 2fc444d1..00000000 --- a/Engineering_Physics/chapter10.ipynb +++ /dev/null @@ -1,293 +0,0 @@ -{
- "metadata": {
- "name": "",
- "signature": "sha256:a79be9375f475037255b00d45fe0971e7606cfc4e70e2a086d7c77b20c6b87d7"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "heading",
- "level": 1,
- "metadata": {},
- "source": [
- "Chapter10:MAGNETIC PROPERTIES OF MATERIALS"
- ]
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Ex10.1:pg-288"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "#to calculate permeability and susceptibility of the bar\n",
- "phi=2.4*10**-5 #magnetic flux in weber\n",
- "A=0.2*10**-4 #cross sectional area in m**2\n",
- "B=phi/A #magnetic induction in N/Am\n",
- "H=1200 #magnetising field in A/m\n",
- "mu=B/H\n",
- "print \"permeability is mu=\",\"{:.2e}\".format(mu),\"N/A**2\"\n",
- "muo=4*math.pi*10**-7 \n",
- "chim=(mu/muo)-1\n",
- "print \"susceptibility is chim=\",round(chim,2),\"unitless\"\n",
- "#the answer is given wrong in the book (round off error) chim=737\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "permeability is mu= 1.00e-03 N/A**2\n",
- "susceptibility is chim= 794.77 unitless\n"
- ]
- }
- ],
- "prompt_number": 2
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Ex10.2:pg-289"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "#to calculate current should be sent through the solenoid\n",
- "l=.10 #length in m\n",
- "N=50 #number of turns\n",
- "H=5*10**3 #magnetising field in A/m\n",
- "i=H*l/N \n",
- "print \"current is i=\",i,\"A\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "current is i= 10.0 A\n"
- ]
- }
- ],
- "prompt_number": 2
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Ex10.3:pg-289"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "#to calculate magnetic moment of the rod\n",
- "#formula is B=muo*(H+I)\n",
- "#where H=ni\n",
- "n=500.0 #number of turns in turns/m\n",
- "i=0.5 #current passed through the solenoid in A\n",
- "mur=1200.0 #relative permeability\n",
- "I=(mur-1)*n*i #intensity of magnetisation in A/m\n",
- "V=10**-3 #volume in m**3\n",
- "M=I*V\n",
- "print \"the magnetic moment of the rod is M=\",round(M),\"A-m**2\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "the magnetic moment of the rod is M= 300.0 A-m**2\n"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Ex10.4:pg-290"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "#to calculate flux density,magnetic intensity,permeability of iron\n",
- "phi=2.0*10**-6 #flux in the ring in weber\n",
- "A=10**-4 #cross-sectional area in m**2\n",
- "B=phi/A \n",
- "print \"flux density is B=\",\"{:.2e}\".format(B),\"weber/m**2\"\n",
- "N=200.0 #number of turns\n",
- "i=0.30 #current flows in the windings in A\n",
- "l=0.2 #length in m\n",
- "H=N*i/l\n",
- "print \"magnetic intensity is H=\",H,\"A-turn/m\"\n",
- "mu=B/H\n",
- "print \"permeability is mu=\",round(mu,6),\"weber/A-m\"\n",
- "muo=4*math.pi*10**-7 \n",
- "mur=mu/muo\n",
- "print \"relative permeability is mur=\",round(mur,2),\"unitless\"\n",
- "\n",
- "#the answer for H in the book is wrong\n",
- "#the answer for mu in book is wrong due to wrong H\n",
- "#the answer in book for mur is wrong due to \n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "flux density is B= 2.00e-02 weber/m**2\n",
- "magnetic intensity is H= 300.0 A-turn/m\n",
- "permeability is mu= 6.7e-05 weber/A-m\n",
- "relative permeability is mur= 53.05 unitless\n"
- ]
- }
- ],
- "prompt_number": 3
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Ex10.5:pg-294"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "#to calculate number of ampere turns\n",
- "l=0.5 #length in m\n",
- "mu=6.5*10**-3 #permeability of iron in henry/m\n",
- "A=2*10**-4 #area of cross-section in m**-4\n",
- "R=l/(mu*A) #reluctance in A-turns/weber\n",
- "flux=4*10**-4 #in weber\n",
- "mmf=flux*R\n",
- "print \"the number of ampere turns is mmf=\",round(mmf,1),\"ampere-turns\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "the number of ampere turns is mmf= 153.8 ampere-turns\n"
- ]
- }
- ],
- "prompt_number": 9
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Ex10.6:pg-294"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "#to calculate relative permeability of the medium\n",
- "phi=1.5*10**-3 #magnetic flux in weber\n",
- "l=math.pi*50*10**-2 #length in m\n",
- "A=10*10**-4 #area of cross-section\n",
- "N=1000 #number of turns\n",
- "i=5 #current in A\n",
- "muo=4*math.pi*10**-7\n",
- "#phi(magnetic flux)=m.m.f/reluctance\n",
- "#phi=N*i*muo*mur*A/l\n",
- "#we get,\n",
- "mur=phi*l/(N*i*A*muo)\n",
- "print \"relative permeability of the medium is mur=\",mur,\"unitless\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "relative permeability of the medium is mur= 375.0 unitless\n"
- ]
- }
- ],
- "prompt_number": 9
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Ex10.7:pg-295"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "#to calculate magnetising current\n",
- "#formula is phi(magnetic flux)=m.m.f/reluctance\n",
- "#phi=N*i*mu*A/l--------eq(1)\n",
- "#phi=BA------------eq(2)\n",
- "B=0.20 #magnetic flux density in weber/m**2\n",
- "l=1 #average length of the circuit in m \n",
- "N=100 #number of turns\n",
- "mu=7.3*10**-3 #in h.m\n",
- "#from eq(1)and eq(2),we get\n",
- "i=B*l/(N*mu)\n",
- "print \"magnetising current is i=\",round(i,3),\"A\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "magnetising current is i= 0.274 A\n"
- ]
- }
- ],
- "prompt_number": 10
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Engineering_Physics/chapter11.ipynb b/Engineering_Physics/chapter11.ipynb deleted file mode 100755 index ffad9cdd..00000000 --- a/Engineering_Physics/chapter11.ipynb +++ /dev/null @@ -1,60 +0,0 @@ -{
- "metadata": {
- "name": "",
- "signature": "sha256:63e794f26755ea447faa48915f48af2a1824429b5991c82dc73f1b261af07c7e"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "heading",
- "level": 1,
- "metadata": {},
- "source": [
- "Chapter11:ULTRASONICS"
- ]
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Ex11.1:pg-303"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "#to calculate fundemental frequency\n",
- "Y=7.9*10**10 #Young modulus for quartz in Nm**-2\n",
- "rho=2.65*10**3 #density of quartz in kg/m**3\n",
- "#the velocity of longitudinal wave is given by\n",
- "v=math.sqrt(Y/rho) #in m/s\n",
- "#for fundamental mode of vibration ,thickness is given by lamda/2\n",
- "lamda=2*0.001 #wavelength in m\n",
- "nu=v/lamda\n",
- "print \"the fundamental frequency is nu=\",round(nu,2),\"Hz\"\n",
- "#answer is given wrong in the book ,nu=2730 Hz\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "the fundamental frequency is nu= 2729987.21 Hz\n"
- ]
- }
- ],
- "prompt_number": 2
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Engineering_Physics/chapter12.ipynb b/Engineering_Physics/chapter12.ipynb deleted file mode 100755 index 35c30635..00000000 --- a/Engineering_Physics/chapter12.ipynb +++ /dev/null @@ -1,550 +0,0 @@ -{
- "metadata": {
- "name": "",
- "signature": "sha256:024efb73c06fed2399cd64a4d2a669b874446e3757757251d146f1996aec28e9"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "heading",
- "level": 1,
- "metadata": {},
- "source": [
- "Chapter12:ELECTROMAGNETICS"
- ]
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Ex12.1:pg-322"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "#to calculate electric flux\n",
- "#electric flux through a surface is phi=vector(E)*vector(s)\n",
- "#where vector E=2i+4j+7k,vector s=10j\n",
- "E=4 #E=4j\n",
- "s=10 #s=10j\n",
- "phi=E*s\n",
- "print \"electric flux is phi=\",phi,\"units\"\n",
- "#to calculate flux coming out of any face of the cube\n",
- "q=1 #charge in coulomb\n",
- "epsilon0=8.85*10**-12 #permittivity in free space in coul**2/N-m**2\n",
- "phi1=q/(6*epsilon0)\n",
- "print \"flux coming out of any face of the cube is phi1=\",\"{:.2e}\".format(phi1),\"N-m**2/coul**2\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "electric flux is phi= 40 units\n",
- "flux coming out of any face of the cube is phi1= 1.88e+10 N-m**2/coul**2\n"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Ex12.2:pg-322"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "#to calculate electric field at a point from centre of the shell\n",
- "q=0.2*10**-6 #charge\n",
- "r=3 #radius\n",
- "epsilon0=8.85*10**-12\n",
- "E=q/(4*math.pi*epsilon0*r**2)\n",
- "print \"electric field at a point from centre of the shell is E=\",round(E),\"N/coulomb\"\n",
- "#to calculate electric field at a point just outside the shell\n",
- "R=0.25 #radius\n",
- "E=q/(4*math.pi*epsilon0*R**2)\n",
- "print \"electric field at a point just outside the shell is E=\",\"{:.2e}\".format(E),\"N/coulomb\"\n",
- "#to calculate the electric field at a point inside the shell\n",
- "#when the point is situated inside the spherical shell,the electric field is zero\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "electric field at a point from centre of the shell is E= 200.0 N/coulomb\n",
- "electric field at a point just outside the shell is E= 2.88e+04 N/coulomb\n"
- ]
- }
- ],
- "prompt_number": 2
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Ex12.3:pg-323"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "#to calculate electric field at a point on earth vertically below the wire\n",
- "lamda=10**-4 #wavelength in coulomb/m\n",
- "r=4 #radius in m\n",
- "epsilon0=8.854*10**-12\n",
- "E=2*lamda/(4*math.pi*epsilon0*r)\n",
- "print \"electric field at a point on earth vertically below the wire is E=\",\"{:.1e}\".format(E),\"N/coulomb\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "electric field at a point on earth vertically below the wire is E= 4.5e+05 N/coulomb\n"
- ]
- }
- ],
- "prompt_number": 4
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Ex12.4:pg-323"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "#to calculate separation between those equipotential surfaces \n",
- "V=5 #potential difference\n",
- "epsilon0=8.85*10**-12 #permittivity of free space\n",
- "sigma=1*10**-7 #in c/m**2\n",
- "#electric field due to an infinite sheet of surface charge density is given by E=sigma/(2*epsilon0) eq(1)\n",
- "#E=V/d eq(2)\n",
- "#from eq(1) and eq(2),we get \n",
- "d=(2*epsilon0*V)/sigma\n",
- "print \"separation between those equipotential surfaces is d=\",\"{:.2e}\".format(d),\"m\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "separation between those equipotential surfaces is d= 8.85e-04 m\n"
- ]
- }
- ],
- "prompt_number": 5
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Ex12.5:pg-324"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "#to calculate force per unit area\n",
- "#force of attraction per unit area is given by F=(epsilon0*E**2)/2 eq(1)\n",
- "#E=V/d eq(2)\n",
- "epsilon0=8.85*10**-12 #permittivity of free space\n",
- "d=1*10**-3 #distance\n",
- "V=100 #potential difference in volts\n",
- "#from eq(1) and eq(2),we get\n",
- "F=(epsilon0*V**2)/(2*d**2)\n",
- "print \"force per unit area is F=\",\"{:.2e}\".format(F),\"N/m**2\"\n",
- "#answer is given incorrect in the book ,F=4.425*10**-12\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "force per unit area is F= 4.42e-02 N/m**2\n"
- ]
- }
- ],
- "prompt_number": 6
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Ex12.6:pg-324"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "#to calculate charge\n",
- "#let charge be q coulomb ,then the surface density of charge i.e. sigma=q/(4*math.pi*r**2)..............eq(1)\n",
- "#outward pull per unit area =sigma**2/(2*epsilon0)............eq(2)\n",
- "#put eq(1) in eq(2),we get q**2/(4*math.pi*r**2)**2*(2*epsilon0)..............eq(3)\n",
- "#pressure due to surface tension =4*T/r............eq(4)\n",
- "T=27\n",
- "r=1.5*10**-2\n",
- "epsilon0=8.85*10**-12\n",
- "#equate eq(3) and eq(4),we get\n",
- "q=math.sqrt(4*T*((4*math.pi*r**2)**2)*2*epsilon0/r)\n",
- "print \"charge is q=\",\"{:.3e}\".format(q),\"coulomb\"\n",
- "#answer is given wrong in the book,square of 4*math.pi*r**2 is not taken in the solution.\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "charge is q= 1.009e-06 coulomb\n"
- ]
- }
- ],
- "prompt_number": 8
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Ex12.7:pg-325"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "#to calculate increase in radius \n",
- "q=4.8*10**-8 #charge in coulomb\n",
- "r=10*10**-2 #radius in m\n",
- "epsilon0=8.85*10**-12 #C**2/N-m**2\n",
- "P=10**5 #N/m**2\n",
- "dr=(q**2)/(96*((math.pi)**2)*(r**3)*epsilon0*P)\n",
- "print \"increase in radius is dr=\",\"{:.2e}\".format(dr),\"m\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "increase in radius is dr= 2.75e-09 m\n"
- ]
- }
- ],
- "prompt_number": 9
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Ex12.8:pg-340"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#in page no.340,numbering is done wrongly,it should be like ex-8,ex-9,ex-10,ex-11,ex-12,ex-13,ex-14\n",
- "import math\n",
- "#to calculate average values of intensities of electric and magnetic fields of radiation\n",
- "#energy of lamp=1000 J/s\n",
- "#area illuminated =4*math.pi*r**2=16*math.pi m**2\n",
- "#energy radiated per unit area per second =1000/16*math.pi\n",
- "#from poynting theorem |s|=|E*H|=E*H eq(1)\n",
- "s=1000/(16*math.pi)\n",
- "muo=4*math.pi*10**-7 #permeability of free space\n",
- "epsilon0=8.85*10**-12 #permittivity in free space\n",
- "#E/H=math.sqrt(muo/epsilon0) eq(2)\n",
- "#from eq(1) and eq(2),we get\n",
- "E=math.sqrt(s*math.sqrt(muo/epsilon0))\n",
- "H=s/E\n",
- "print \"average value of intensity of electric fields of radiation is E=\",round(E,2),\"V/m\"\n",
- "print \"average value of intensity of magnetic fields of radiation is H=\",round(H,2),\"ampere-turn/m\"\n",
- "#answer is given wrong in the book E=48.87 V/m,solution of magnetic fields is not given in the book .\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "average value of intensity of electric fields of radiation is E= 86.58 V/m\n",
- "average value of intensity of magnetic fields of radiation is H= 0.23 ampere-turn/m\n"
- ]
- }
- ],
- "prompt_number": 10
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Ex12.9:pg-340"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "#to calculate amplitudes of electric and magnetic fields of radiation\n",
- "#energy received by an electromagnetic wave per sec per unit area is given by poynting vector |s|=|E*H|=E*H*sin 90 (becoz E is perpendicular to H)\n",
- "#it is given that energy received by earth's surface is\n",
- "s=1400.0 #|s|=2 cal min**-1 cm**-2\n",
- "muo=4.0*math.pi*10**-7 #permittivity of free space\n",
- "epsilon0=8.85*10**-12 #permeability of free space\n",
- "#E*H=1400 eq(1)\n",
- "#E/H=math.sqrt(muo/epsilon0) eq(2)\n",
- "#from eq(1) and eq(2) ,we get\n",
- "E=math.sqrt(s*math.sqrt(muo/epsilon0))\n",
- "#from eq(1) ,we get\n",
- "H=1400.0/E\n",
- "Eo=E*math.sqrt(2) # at distance 2 m\n",
- "Ho=H*math.sqrt(2) # at distance 2 m\n",
- "print \"amplitude of electric field is Eo=\",int(Eo),\"V/m\"\n",
- "print \"amplitude of magnetic field is Ho=\",round(Ho,3),\"amp-turn/m\"\n",
- "\n",
- "# The answers in the textbook are slightly different due to approximation\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "amplitude of electric field is Eo= 1027 V/m\n",
- "amplitude of magnetic field is Ho= 2.726 amp-turn/m\n"
- ]
- }
- ],
- "prompt_number": 16
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Ex12.10:pg-341"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "# no value to be found out , only equation to be written hence skipped following the TBC guidelines"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [],
- "prompt_number": 29
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Ex12.11:pg-341"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "#to calculate skin depth \n",
- "f=10**8 #frequency\n",
- "sigma=3*10.0**7 #conductivity of the medium\n",
- "muo=4*math.pi*10**-7 #permeability of free space\n",
- "Del=math.sqrt(2/(2*math.pi*f*sigma*muo))\n",
- "print \"skin depth is Del=\",\"{:.1e}\".format(Del),\"cm\"\n",
- "\n",
- "# answer in book is wrong"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- " skin depth is Del= 9.2e-06 cm\n"
- ]
- }
- ],
- "prompt_number": 15
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Ex12.12:pg-341"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "#to calculate frequency \n",
- "muo=4*math.pi*10**-7 #permeability of free space\n",
- "sigma=4.3 # in mhos/m\n",
- "Del=0.1 #skin depth in m\n",
- "f=2/(2*math.pi*muo*Del**2)\n",
- "print \"frequency is f=\",\"{:.2e}\".format(f),\"Hz\"\n",
- "#value of frequency is given incorrect in the book \n",
- "#show that for frequencies less than 10**8 ,it can be considered as good conductor\n",
- "epsilon=80*8.854*10**-12\n",
- "f=10**8 #frequency in Hz\n",
- "sigma=4.3\n",
- "#formula is sigma/(omega*epsilon)>4.3/(2*math.pi*10**8*80*epsilon)\n",
- "sigma1=sigma/(2*math.pi*f*epsilon) #where sigma1=sigma/(omega*epsilon)\n",
- "print \"sigma1=\",round(sigma1,2),\"unitless\"\n",
- "#the ocean water to be good conductor ,the value of sigma/(omega*epsilon) should be greater than 1\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "frequency is f= 2.53e+07 Hz\n",
- "sigma1= 9.66 unitless\n"
- ]
- }
- ],
- "prompt_number": 16
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Ex12.13:pg-342"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "#to show that for frequency <10**9 Hz ,a sample of silicon will act like a good conductor\n",
- "sigma=200 #in mhos/m\n",
- "omega=2*math.pi*10**9 \n",
- "epsilon0=8.85*10**-12 #permittivity in free space\n",
- "epsilon=12*epsilon0 \n",
- "sigma1=sigma/(omega*epsilon) #sigma1=sigma/(omega*epsilon)\n",
- "print \"sigma1=\",round(sigma1),\"unitless\"\n",
- "#if sigma/(omega*epsilon) is greater than 1 , silicon is a good conductor at frequency <10**9 Hz\n",
- "#to calculate penetration depth\n",
- "f=10**6 #frequency in Hz\n",
- "muo=4*math.pi*10**-7 #permeability of free space\n",
- "sigma=200\n",
- "Del=math.sqrt(2/(2*math.pi*f*muo*sigma))\n",
- "print \"penetration depth is del=\",round(Del*100,1),\"m\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "sigma1= 300.0 unitless\n",
- "penetration depth is del= 3.6 m\n"
- ]
- }
- ],
- "prompt_number": 25
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Ex12.14:pg-343"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "#to calculate conduction current and displacement current densities \n",
- "sigma=10**-3 #conductivity in mhos/m\n",
- "E=4*10**-6 #where E=4*10**-6*math.sin(9*10**9t) v/m\n",
- "J=sigma*E\n",
- "print \"conduction current density is J=\",J,\"math.sin(9*10**9t) A/m\"\n",
- "epsilon0=8.85*10**-12 #permittivity in free space\n",
- "epsilonr=2.45 #relative permittivity\n",
- "#formula is epsilon0*epsilonr*(delE/delt)\n",
- "#delE/delt=4*10**-6*9*10**9*math.cos(9*10**9*t)\n",
- "Jd=epsilon0*epsilonr*4*10**-6*9*10**9\n",
- "print \"displacement current density is Jd=\",round(Jd,8),\"math.cos(9*10**9*t) A/m**2\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "conduction current density is J= 4e-09 math.sin(9*10**9t) A/m\n",
- "displacement current density is Jd= 7.8e-07 math.cos(9*10**9*t) A/m**2\n"
- ]
- }
- ],
- "prompt_number": 23
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Engineering_Physics/chapter13.ipynb b/Engineering_Physics/chapter13.ipynb deleted file mode 100755 index 5c4d543b..00000000 --- a/Engineering_Physics/chapter13.ipynb +++ /dev/null @@ -1,101 +0,0 @@ -{
- "metadata": {
- "name": "",
- "signature": "sha256:54e553aa602b5b46f788e4b29d82727b714210cbc37d72a1b8625b974bd8b58b"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "heading",
- "level": 1,
- "metadata": {},
- "source": [
- "Chapter13:SUPERCONDUCTIVITY"
- ]
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "\n",
- "Ex13.1:pg-357"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#example 1\n",
- "#to calculate value of Temperature\n",
- "import math\n",
- "Bc=105.0*10**3 #magnetic field in amp/m\n",
- "Bo=150.0*10**3 #critical field of the metal in amp/m \n",
- "Tc=9.2 #critical temperature of the metal in K\n",
- "T=Tc*math.sqrt(1-(Bc/Bo))#temperature\n",
- "print\"\\n the value of temperature is\",round(T,3),\"K\"\n",
- "\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- " the value of temperature is 5.039 K\n"
- ]
- }
- ],
- "prompt_number": 9
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Ex13.2:pg-357"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#example 2\n",
- "#to calculate temperature\n",
- "import math\n",
- "Tc=7.18 #critical temperature in K\n",
- "Bc=4.5*10**3 #critical field in A/m\n",
- "Bo=6.5*10**3 #critical magnetic field in A/m\n",
- "T=Tc*math.sqrt(1-(Bc/Bo))# temperature\n",
- "print\"\\n the temperature is\",round(T,2),\"K\"\n",
- "#to calculate critical current density at that temperature\n",
- "r=1*10**-3 #diameter of the wire in mm\n",
- "TJc=(Bc*2*math.pi*r)/(math.pi*r**2)# critical current density\n",
- "print\"\\n the critical current density at that temperature is\",\"{:.1e}\".format(TJc),\"A/m^2\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- " the temperature is 3.98 K\n",
- "\n",
- " the critical current density at that temperature is 9.0e+06 A/m^2\n"
- ]
- }
- ],
- "prompt_number": 2
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Engineering_Physics/chapter2.ipynb b/Engineering_Physics/chapter2.ipynb deleted file mode 100755 index 8c10ecf8..00000000 --- a/Engineering_Physics/chapter2.ipynb +++ /dev/null @@ -1,1113 +0,0 @@ -{
- "metadata": {
- "name": "",
- "signature": "sha256:2b54094513f6b50d68fe3b6c91cd7d7cafc36254b75abea2a0d6c88b266fd509"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "heading",
- "level": 1,
- "metadata": {},
- "source": [
- "Chapter2:INTERFERENCE"
- ]
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Ex2.2:pg-41"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "#to calculate ratio of intensity\n",
- "#I1/I2=1/25\n",
- "#formula is a1/a2=math.sqrt(I1/I2)=1/5\n",
- "a2=5 #a2=5*a1\n",
- "a1=1\n",
- "I=((1+5)**2)/((1-5)**2)\n",
- "print \"ratio of intensity at the maxima and minima in the interference pattern is Imax/Imin=((a1+a2)**2)/((a1-a2)**2)=\",((a1+a2)**2)/((a1-a2)**2),\"unitless\"\n",
- "#answer is given in terms of ratio\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "ratio of intensity at the maxima and minima in the interference pattern is Imax/Imin=((a1+a2)**2)/((a1-a2)**2)= 2 unitless\n"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Ex2.3:pg-42"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "#to calculate ratio of intensity at this point to that at the centre of a bright fringe\n",
- "#the intensity at any pont is I=a1**2+a2**2+2*a1*a2*cos del\n",
- "#let a1=a2=a\n",
- "#phase difference del is 0\n",
- "#then I0=a**2+a**2+2*a*a*cos 0\n",
- "#we get I0=4a**2\n",
- "I0=4 #intensity\n",
- "#path difference is lemda/8\n",
- "#phase difference =2*math.pi/lemda*path difference=math.pi/4\n",
- "#I1=a**2+a**2+2a*a*cos math.pi/4\n",
- "#I1=3.414a**2\n",
- "I1=3.414\n",
- "intensity=I1/I0\n",
- "print \" ratio of intensity =\",intensity,\"unitless\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- " ratio of intensity = 0.8535 unitless\n"
- ]
- }
- ],
- "prompt_number": 2
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Ex2.4:pg-42"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "#to calculate ratio of maximum intensity to minimum intensity\n",
- "#formula is I1/I2=a1**2/a2**2=100/1\n",
- "#a1/a2=10/1\n",
- "a1=10 #a1=10*a2\n",
- "a2=1\n",
- "print \"the ratio of maximum intensity to minmum intensity in the interference pattern Imax/Imin=((a1+a2)**2)/((a1-a2)**2)=\",((a1+a2)**2)/((a1-a2)**2),\"unitless\" \n",
- "#answer is given in terms of ratio in the book\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "the ratio of maximum intensity to minmum intensity in the interference pattern Imax/Imin=((a1+a2)**2)/((a1-a2)**2)= 1 unitless\n"
- ]
- }
- ],
- "prompt_number": 3
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Ex2.5:pg-43"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "#to calculate relative intensities\n",
- "#Imax/Imin=(a1+a2)**2/(a1-a2)**2+105/95\n",
- "#(a1+a2)/(a1-a2)=1.051\n",
- "#we get a1/a2=40 \n",
- "a1=40 #a1=40*a2\n",
- "a2=1\n",
- "print \"the ratio of the intensities of interfering sources is I1/I2=a1**2/a2**2=\",a1**2/a2**2,\"unitless\"\n",
- "#answer is given in terms of ratio in the book\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "the ratio of the intensities of interfering sources is I1/I2=a1**2/a2**2= 1600 unitless\n"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Ex2.7:pg-52"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "#to calculate distance between the two coherent sources\n",
- "lamda=5890*10**-10 #wavelength in m\n",
- "omega=9.424*10**-4 #width of the fringes in m\n",
- "D=.80 #distance in m\n",
- "twod=D*lamda/omega\n",
- "print \"the distance between two coherent sources is twod=\",twod,\"m\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "the distance between two coherent sources is twod= 0.0005 m\n"
- ]
- }
- ],
- "prompt_number": 5
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Ex2.8:pg-53"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "#to calculate fringe width \n",
- "mu=1.5 #refractive index (unitless)\n",
- "alpha=math.pi/180 #refracting angle in radian\n",
- "Y1=20*10**-2 #distance between the source and the biprism in m\n",
- "Y2=80*10**-2 #distance in m\n",
- "D=Y1+Y2 # distance in m\n",
- "lamda=6900*10**-10 #wavelength in m\n",
- "twod=2*(mu-1)*alpha*Y1 \n",
- "omega=D*lamda/twod\n",
- "print \"the fringe width is omega=\",round(omega,7),\"m\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "the fringe width is omega= 0.0001977 m\n"
- ]
- }
- ],
- "prompt_number": 2
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Ex2.9:pg-53"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "#to calculate wavelength of light\n",
- "omega=(1.888*10**-2)/20 #in (m)\n",
- "D=1.20 #distance of eye piece from the source in m\n",
- "twod=0.00075 #distance between two virtual sources in m\n",
- "lamda=(omega*twod/D)/(10e-11)\n",
- "print \"the wavelength of light is lamda=\",lamda,\"Angstrom\"\n",
- "\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "the wavelength of light is lamda= 5900.0 Angstrom\n"
- ]
- }
- ],
- "prompt_number": 44
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Ex2.10:pg-54"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "#to calculate thickness of glass plate\n",
- "n=3\n",
- "mu=1.5 #refractive index (unitless)\n",
- "lamda=5450*10**-10 #wavelength in m\n",
- "t=n*lamda/(mu-1)\n",
- "print \"the thickness of glass plate is t=\",t,\"m\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "the thickness of glass plate is t= 3.27e-06 m\n"
- ]
- }
- ],
- "prompt_number": 4
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Ex2.11:pg-54"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "#to calculate refractive index of the sheet\n",
- "t=6.3*10**-6 #thickness of thin sheet of transparent material in m\n",
- "lamda=5460*10**-10 #wavelength in m\n",
- "n=6\n",
- "mu=(n*lamda/t)+1\n",
- "print \"the refractive index of the sheet is mu=\",mu,\"unitless\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "the refractive index of the sheet is mu= 1.52 unitless\n"
- ]
- }
- ],
- "prompt_number": 9
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Ex2.12:pg-54"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "#to calculate refractive index of mica\n",
- "t=1.2*10**-8 #thickness of thin sheet of mica in m\n",
- "n=1 \n",
- "lamda=6*10**-7 #wavelength in m\n",
- "mu=(n*lamda/t)+1\n",
- "print \"the refractive index of mica is mu=\",mu,\"unitless\"\n",
- "#answer is given wrong in the book=1.50\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "the refractive index of mica is mu= 51.0 unitless\n"
- ]
- }
- ],
- "prompt_number": 11
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Ex2.13:pg-54"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "#to calculate intensity\n",
- "mu=1.5 #refractive index(unitless)\n",
- "t=1.5*10**-6 #thickness of thin glass plate in m\n",
- "pathdifference=(mu-1)*t # in m\n",
- "lamda=5*10**-7 #wavelength in m\n",
- "#del=2*math.pi*pathdifference/lamda\n",
- "#del=3*math.pi\n",
- "a1=1\n",
- " #where a1=a2=a\n",
- "a2=1\n",
- "#formula is I=a1**2+a2**2+2*a1*a2*cos del\n",
- "# where cos 3math.pi=-1\n",
- "I=a1**2+a2**2+2*a1*a2*(-1) \n",
- "print \"the intensity at the centre of the screen is I=\",I,\"unitless\" \n",
- "#to calculate lateral shift\n",
- "D=1 #distance in m\n",
- "twod=5*10**-4 #distance between two slits in m\n",
- "mu=1.5 #refractive index (unitless)\n",
- "t=1.5*10**-6 #thickness of thin glass plate in m\n",
- "x0=D*(mu-1)*t/twod\n",
- "print \"the lateral shift of the central maximum is x0=\",x0,\"m\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "the intensity at the centre of the screen is I= 0 unitless\n",
- "the lateral shift of the central maximum is x0= 0.0015 m\n"
- ]
- }
- ],
- "prompt_number": 8
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Ex2.14:pg-55"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "#to calculate spacing between the slits\n",
- "lamda=6*10**-5 #wavelength in cm\n",
- "omegatheta=0.1*math.pi/180 #angular width of a fringe in radians\n",
- "twod=lamda/omegatheta\n",
- "print \"the spacing between the slits is twod=\",\"{:.2e}\".format(twod),\"cm\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "the spacing between the slits is twod= 3.44e-02 cm\n"
- ]
- }
- ],
- "prompt_number": 9
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Ex2.15:pg-55"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "#to calculate distance of the third bright fringe on the screen from the central maximum \n",
- "lamda=6.5*10**-5 #wavelength in cm\n",
- "twod=0.2 #distance between the slits in cm\n",
- "D=120 #distance between the plane of the slits and the screen in cm\n",
- "n=3 \n",
- "X3=D*n*lamda/twod\n",
- "print \"the distance of the third bright fringe from the central maximum is X3=\",X3,\"cm\"\n",
- "#to calculate the least distance from the central maximum \n",
- "lamda1=6.5*10**-5 #wavelength in cm\n",
- "lamda2=5.2*10**-5 #wavelength in cm\n",
- "#Xn=Dnlamda1/2d=D(n+1)lamda2/2d\n",
- "#we get,\n",
- "n=lamda2/(lamda1-lamda2)\n",
- "print \"n=\",n,\"unitless\"\n",
- "Xn=D*n*lamda1/twod\n",
- "print \"the distance from the central maximum when the bright fringes due to both wavelengths coincide is Xn=\",Xn,\"cm\"\n",
- " \n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "the distance of the third bright fringe from the central maximum is X3= 0.117 cm\n",
- "n= 4.0 unitless\n",
- "the distance from the central maximum when the bright fringes due to both wavelengths coincide is Xn= 0.156 cm\n"
- ]
- }
- ],
- "prompt_number": 17
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Ex2.16:pg-56"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "#to calculate refractive index \n",
- "D=10 #distance in cm\n",
- "twod=0.2 #distance detween the slits in cm\n",
- "t=0.05 #thickness of transparent plate in cm\n",
- "deltaX=0.5 #in cm\n",
- "mu=(deltaX*twod/(D*t))+1\n",
- "print \"the refractive index of the transparent plate is mu=\",mu,\"unitless\"\n",
- "#to calculate order\n",
- "n=10 \n",
- "lamda=7000*10**-8 #wavelength in cm\n",
- "#path difference =n*lamda\n",
- "n1=n*lamda/(5000*10**-8) \n",
- "print \"the order will be visible is n1=\",n1,\"unitless\"\n",
- "#to calculate distance between the two coherent sources\n",
- "D=100 #distance in m \n",
- "lamda=6000*10**-8 #wavelength in cm\n",
- "omega=0.05 #distance between two consecutive bright fringes on the screen in cm\n",
- "twod=D*lamda/omega\n",
- "print \"the distance between the coherent sources is twod=\",twod,\"cm\"\n",
- "#to calculate wavelength\n",
- "Xn=1 #distance of fourth bright fringe from the central fringe in cm\n",
- "twod=0.02 #distance between the two coherent sources in cm\n",
- "n=4 \n",
- "D=100 #distance in cm \n",
- "lamda=Xn*twod/(n*D)\n",
- "print \"the wavelength of light is lamda=\",lamda,\"cm\"\n",
- "#to calculate wavelength\n",
- "#position of nth bright fringe from the centre of the central fringe is Xn=D*n*lamda/2d----eq(1)\n",
- "#fringe width umega=D*lamda/2d---------------------eq(2)\n",
- "#from eq(1) and eq(2) we get, Xn=n*omega\n",
- "#for 11th bright fringe X11=11*omega\n",
- "#position for nth dark fringe Xn'=(2n+1)D*lamda/4d\n",
- "#X4'=(7/2)*omega\n",
- "#distance between 11th and 4th dark fringe =0.8835 cm\n",
- "#we get \n",
- "omega=0.1178 #in cm\n",
- "twod=0.05 #distance between slis in cm\n",
- "D=100 # distance in cm\n",
- "lamda=omega*twod/D \n",
- "print \"the wavelength of light is lamda=\",lamda,\"cm\"\n",
- "#to calculate changed fringe width\n",
- "#X10-X0=10*omega\n",
- "#given that X10-X0=14.73-12.34=2.39mm \n",
- "omega=0.239 #in mm\n",
- "lamda=6000 #wavelength in angstrom\n",
- "lamda1=5000 #lamda'=5000 angstrom\n",
- "omega1=omega*lamda1/lamda\n",
- "print \"the changed fringe width is omega1=\",omega1,\"mm\"\n",
- "#to calculate thickness of mica sheet\n",
- "n=3\n",
- "mu=1.6 #refractive index(unitless)\n",
- "lamda=5.89*10**-5 #wavelength in cm\n",
- "t=n*lamda/(mu-1)\n",
- "print \"the thickness of mica sheet is t=\",t,\"cm\"\n",
- "#answer of thickness is given wrong in the book =0.002945 cm\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "the refractive index of the transparent plate is mu= 1.2 unitless\n",
- "the order will be visible is n1= 14.0 unitless\n",
- "the distance between the coherent sources is twod= 0.12 cm\n",
- "the wavelength of light is lamda= 5e-05 cm\n",
- "the wavelength of light is lamda= 5.89e-05 cm\n",
- "the changed fringe width is omega1= 0.199166666667 mm\n",
- "the thickness of mica sheet is t= 0.0002945 cm\n"
- ]
- }
- ],
- "prompt_number": 18
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Ex2.17:pg-75"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "#to calculate the smallest thickness of the plate\n",
- "mu=1.5 #refractive index(unitless)\n",
- "r=60*math.pi/180 #angle of refraction in radians\n",
- "lamda=5890*10**-10 #wavelength in m\n",
- "n=1\n",
- "#formula is t=n*lamda/(2*mu*cosr) where cosr=0.5\n",
- "t=n*lamda/(2*mu*0.5)\n",
- "print \"the smallest thickness of the plate which will appear dark by reflection is t=\",\"{:.3e}\".format(t),\"m\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "the smallest thickness of the plate which will appear dark by reflection is t= 3.927e-07 m\n"
- ]
- }
- ],
- "prompt_number": 10
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Ex2.18:pg-75"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "#to calculate least thickness of the film\n",
- "lamda=5893*10**-10#wavelength in m\n",
- "r=0 #in degree\n",
- "mu=1.42 #refractive index\n",
- "n=1\n",
- "#the formula is t=n*lamda/(2*mu*cosr), where cos0=1\n",
- "t=n*lamda/(2*mu*1)\n",
- "print \"the least thickness of the film that will appear black is t=\",t,\"m\"\n",
- "t=(2*n-1)*lamda/(2*mu*1*2)\n",
- "print \"the least thickness of the film that will appear bright is t=\",t,\"m\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "the least thickness of the film that will appear black is t= 2.075e-07 m\n",
- "the least thickness of the film that will appear bright is t= 1.0375e-07 m\n"
- ]
- }
- ],
- "prompt_number": 20
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Ex2.19:pg-76"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "#to calculate thickness of the film\n",
- "lamda1=6.1*10**-7 #wavelength in m\n",
- "lamda2=6*10**-7 # wavelength in m\n",
- "#the two dark consecutive fringes are overlapping for the wavelength lamda1 and lamda2 respectively\n",
- "#then, n*lamda1=(n+1)*lamda2\n",
- "#we get,\n",
- "n=lamda2/(lamda1-lamda2)\n",
- "sini=4.0/5\n",
- "mu=4.0/3\n",
- "#formula is mu=sini/sinr\n",
- "sinr=0.6\n",
- "cosr=math.sqrt(1-(sinr)**2)\n",
- "t=n*lamda1/(2*mu*cosr)\n",
- "print \"the thickness of the film is t=\",\"{:.2e}\".format(t),\"m\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "the thickness of the film is t= 1.72e-05 m\n"
- ]
- }
- ],
- "prompt_number": 11
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Ex2.20:pg-77"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "#to calculate thickness of the film\n",
- "mu=1.33 #refractive index of soap film (unitless)\n",
- "i=45*math.pi/180\n",
- "#the formula is mu=sini/sinr \n",
- "sinr=0.5317 \n",
- "cosr=math.sqrt(1-(sinr)**2)\n",
- "#for destructive interference\n",
- "lamda=5890*10**-10 #wavelength in m\n",
- "n=1\n",
- "t=n*lamda/(2*mu*cosr)\n",
- "print \"the thickness of the film is t=\",\"{:.3e}\".format(t),\"m\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "the thickness of the film is t= 2.614e-07 m\n"
- ]
- }
- ],
- "prompt_number": 12
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Ex2.21:pg-77"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "#to calculate angle of the wedge\n",
- "lamda=6000*10**-10 #wavelength in m\n",
- "mu=1.4 #refractive index in unitless\n",
- "omega=2*10.0**-3 #distance in m\n",
- "theta=lamda/(2*mu*omega)\n",
- "print \"the angle of the wedge is theta =\",\"{:.2e}\".format(theta),\"radians\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "the angle of the wedge is theta = 1.07e-04 radians\n"
- ]
- }
- ],
- "prompt_number": 13
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Ex2.22:pg-77"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "#to calculate wavelength of light \n",
- "theta=10*22.0/(7*60*60*180.0) #angle of wedge in radians\n",
- "omega=5*10.0**-2 #distance between the successive fringes in m\n",
- "mu=1.4 #refractive index\n",
- "lamda=2.0*mu*theta*omega/(10e-10)\n",
- "print \"the wavelength of light is lamda=\",round(lamda),\"Angstrom\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "the wavelength of light is lamda= 6790.0 Angstrom\n"
- ]
- }
- ],
- "prompt_number": 23
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Ex2.23:pg-78"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "#to calculate wavelength of the light\n",
- "D15=0.590*10**-2 #diamater of 15th ring in m\n",
- "D5=0.336*10**-2 #diameter of 5th ring in m\n",
- "p=1 # in m\n",
- "R=1 #radius of plano convex lens in m\n",
- "#formula is lamda=Dn+p**2-Dn**2/4pR\n",
- "lamda=((D15**2)-(D5**2))/(4*p*R)/(10e-10)\n",
- "print \"the wavelength of the monochromatic light is lamda=\",int(lamda),\"Angstrom\"\n",
- "\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "the wavelength of the monochromatic light is lamda= 5880 Angstrom\n"
- ]
- }
- ],
- "prompt_number": 14
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Ex2.24:pg-78"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "#to calculate refractive index of the liquid\n",
- "n=6\n",
- "lamda=6000*10**-10 #wavelength in m\n",
- "R=1 #radius of curvature of the curved surface in m\n",
- "Dn=3.1*10**-3 #diameter of 6th bright ring in m\n",
- "mu=2*(2*n-1)*lamda*R/Dn**2\n",
- "print \"the refractive index of the liquid is mu=\",round(mu,3),\"unitless\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "the refractive index of the liquid is mu= 1.374 unitless\n"
- ]
- }
- ],
- "prompt_number": 28
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Ex2.25:pg-78"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "#to calculate radius of curvature\n",
- "lamda=5900*10**-10 #wavelength in m\n",
- "n=10\n",
- "Dn=5*10**-3 # diameter of 10th dark ring in m\n",
- "R=Dn**2/(4*n*lamda)\n",
- "print \"the radius of curvature of the lens is R=\",round(R,3),\"m\"\n",
- "#to calculate thichness\n",
- "t=n*lamda/2\n",
- "print \"the thickness of the air film is t=\",t,\"m\"\n",
- "\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "the radius of curvature of the lens is R= 1.059 m\n",
- "the thickness of the air film is t= 2.95e-06 m\n"
- ]
- }
- ],
- "prompt_number": 15
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Ex2.26:pg-79"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "#to calculate the distance from the apex of the wedge at which the maximum due to the two wavelengths first coincide\n",
- "#condition for maxima for normal incidence air film is 2t=(2n+1)lamda/2\n",
- "#let nth order maximum due to lamda1 coincides with (n+1)th order maximum due to lamda2 \n",
- "#we get , n=(3lamda2-lamda1)/2(lamda1-lamda2)\n",
- "# we also get, 2t=lamda1*lamda2/(lamda1-lamda2)\n",
- "#t=X*theta\n",
- "lamda1=5896.0*10**-8 #wavelength in cm\n",
- "lamda2=5890.0*10**-8 #wavelength in cm\n",
- "theta=0.3*math.pi/180 #angle of wedge\n",
- "X=lamda1*lamda2/(2*(lamda1-lamda2)*theta)\n",
- "print \"the distance from the apex of the wedge is X=\",round(X,3),\"cm\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "the distance from the apex of the wedge is X= 5.527 cm\n"
- ]
- }
- ],
- "prompt_number": 17
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Ex2.27:pg-80"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "#to calculate radius of curvature \n",
- "n=10.0\n",
- "Dn=0.50 #diameter of 10th ring in cm\n",
- "lamda=6000*10**-8 #wavelength in cm\n",
- "R=Dn**2/(4*n*lamda)\n",
- "print \"the radius of curvature of the lens is R=\",R,\"cm\"\n",
- "#answer is given wrong in the book =106 cm\n",
- "#to calculate thickness of the film\n",
- "t=Dn**2/(8*R)\n",
- "print \"the thickness of the film is t=\",\"{:.1e}\".format(t),\"cm\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "the radius of curvature of the lens is R= 104.166666667 cm\n",
- "the thickness of the film is t= 3.0e-04 cm\n"
- ]
- }
- ],
- "prompt_number": 19
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Ex2.28:pg-80"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "#to calculate diameter\n",
- "#the difference of (n+p)th and nth dark ring is Dn+p**2-Dn**2=4nRlamda\n",
- "N=12 #where N=n+p\n",
- "n=4\n",
- "D12=0.7 #diameter of 12th dark ring in cm\n",
- "D4=0.4 #diameter of 4th dark ring in cm\n",
- "#D12**2-D4**2=4pRlamda where p=8 ----eq(1)\n",
- "#D20**2-D4**2=4pRlamda where p=16 -----eq(2)\n",
- "#divide eq(2) by eq(1) ,we get\n",
- "D20=math.sqrt((2*D12**2)-D4**2)\n",
- "print \"the diameter of 20th dark ring is D20=\",round(D20,2),\"cm\"\n",
- "\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "the diameter of 20th dark ring is D20= 0.91 cm\n"
- ]
- }
- ],
- "prompt_number": 21
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Ex2.29:pg-80"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "#to calculate diameter \n",
- "lamda1=6*10**-5 #wavelength in cm\n",
- "lamda2=4.5*10**-5 #wavelength in cm\n",
- "R=90 #radius of curvature of the curved surface in cm\n",
- "#Dn**2=4nRlamda1 -------eq(1)\n",
- "#Dn+1**2=4(n+1)Rlamda2-------eq(2)\n",
- "#the nth dark ring due to lamda1 coincides with (n+1)th dark ring due to lamda2\n",
- "#from eq(1) and eq(2)-4nRlamda1=4(n+1)Rlamda2\n",
- "# we get,\n",
- "n=lamda2/(lamda1-lamda2)\n",
- "Dn=math.sqrt(4*n*R*lamda1)\n",
- "print \"the diameter of nth dark ring for lamda1 is Dn=\",round(Dn,3),\"cm\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "the diameter of nth dark ring for lamda1 is Dn= 0.255 cm\n"
- ]
- }
- ],
- "prompt_number": 22
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Ex2.30:pg-81"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "#to calculate the difference of square of diameters for nth and (n+p)th ring when light of wavelength lamda is changed to lamda'\n",
- "lamda=6*10**-5 #wavelength in cm\n",
- "lamda1=4.5*10**-5 #wavelength in cm\n",
- "#Let D=(D**2-Dn**2)=0.125 cm**2 \n",
- "D=0.125\n",
- "#formula is D'(n+p)**2-D'n**2=lamda'*(D(n+p)**2-Dn**2)/lamda\n",
- "print \"the difference of square of diameters is D1(n+p)**2-D1n**2=(lamda1*D)/lamda=\",(lamda1*D)/lamda,\"cm**2\"\n",
- "#to calculate difference of square of diamaters when liqquid of refractive index mu' is introduced\n",
- "mu=1 #refractive index (unitless)\n",
- "mu1=1.33 # mu'=1.33\n",
- "#formula is D'(n+p)**2-D'n**2=(mu/mu')*(D(n+p)**2-Dn**2)\n",
- "print \"the difference of square of diameters is D1(n+p)**2-D1n**2=(mu*D)/mu1=\",round((mu*D)/mu1,3),\"cm**2\"\n",
- "#to calculate difference of square of diameters when radius of curvature of convex surface of the plano convex lens is doubled\n",
- "R1=2 #radius of curvature is R'=2R\n",
- "R=1\n",
- "#formula is D'(n+p)**2-D'n**2=(R'/R)*(D(n+p)**2-Dn**2)\n",
- "print \"the difference of square of diameters is D1(n+p)**2-D1n**2=(R1*D)/R=\",round((R1*D)/R,2),\"cm**2\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "the difference of square of diameters is D1(n+p)**2-D1n**2=(lamda1*D)/lamda= 0.09375 cm**2\n",
- "the difference of square of diameters is D1(n+p)**2-D1n**2=(mu*D)/mu1= 0.094 cm**2\n",
- "the difference of square of diameters is D1(n+p)**2-D1n**2=(R1*D)/R= 0.25 cm**2\n"
- ]
- }
- ],
- "prompt_number": 27
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Engineering_Physics/chapter3.ipynb b/Engineering_Physics/chapter3.ipynb deleted file mode 100755 index 1c763114..00000000 --- a/Engineering_Physics/chapter3.ipynb +++ /dev/null @@ -1,1365 +0,0 @@ -{
- "metadata": {
- "name": "",
- "signature": "sha256:f8c5b92f0cb0a1787021af8dd46e58ab42d75991e4d7ba39c4a4693a16608b3e"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "heading",
- "level": 1,
- "metadata": {},
- "source": [
- "Chapter3:DIFFRACTION"
- ]
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Ex3.1:pg-97"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "#to calculate angular width and linear width \n",
- "lamda=6*10**-5\n",
- "e=0.01 #width of slit in cm\n",
- "#position of minima is given by\n",
- "sintheta=lamda/e #sintheta=m*lamda/e ,where m=1,2,3,......\n",
- "print \"sintheta=\",sintheta,\" m\"\n",
- "#since theta is very small,so sintheta is approximately equal to theta\n",
- "theta=sintheta\n",
- "theta1=2*theta\n",
- "print \"total angular width of central maximum is theta1=\",theta1,\" m radians \"\n",
- "d=100 #distance in cm\n",
- "Y=theta*d\n",
- "Y1=2*Y\n",
- "print \"linear width of central maximum on the screen is Y1=\",Y1,\" m cm\"\n",
- "print \"values of m =1,2,3,............ gives the directions of first, second .............minima\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "sintheta= 0.006 m\n",
- "total angular width of central maximum is theta1= 0.012 m radians \n",
- "linear width of central maximum on the screen is Y1= 1.2 m cm\n",
- "values of m =1,2,3,............ gives the directions of first, second .............minima\n"
- ]
- }
- ],
- "prompt_number": 2
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Ex3.2:pg-97"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "#to calculate wavelength of light\n",
- "#in a diffraction pattern due to single slit, minima is given by e*sintheta=m*lamda\n",
- "#since theta is very small, sintheta is approximately equal to theta\n",
- "#theta=Y/d\n",
- "e=0.014 #width of slit in cm\n",
- "d=200 #distance in cm\n",
- "m=2 \n",
- "Y=1.6 #in cm\n",
- "lamda=Y*e/(d*m)\n",
- "print \"wavelength of light is lamda=\",lamda,\"cm\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "wavelength of light is lamda= 5.6e-05 cm\n"
- ]
- }
- ],
- "prompt_number": 5
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Ex3.3:pg-98"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "#to calculate width of slit\n",
- "#direction of minima in fraunhofer diffraction due to single slit is given by math.pi/lamda*e*siuntheta=+m*math.pi,where m=1,2,3\n",
- "#angular spread of the central maximum on either side of the incident light is sintheta=lamda/e,where m=1,position of first minima\n",
- "lamda=5000*10**-8\n",
- "e=lamda/math.sin(math.pi/6)\n",
- "print \"width of slit is e=\",e,\"cm\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "width of slit is e= 0.0001 cm\n"
- ]
- }
- ],
- "prompt_number": 6
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Ex3.4:pg-98"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "#to calculate wavelength of incident light\n",
- "#direction of minima is given by e*sintheta=+m*lamda\n",
- "#for first minima m=1,i.e. e*sintheta=lamda,sintheta is approximately equal to theta,then we can write it as e*theta=lamda ...........eq(1)\n",
- "#theta=Y/d........................eq(2) \n",
- "e=0.02 #in cm\n",
- "Y=0.5 #position of first minima from the central maxima in cm\n",
- "d=200 #distance of screen from the slit in cm\n",
- "#from eq(1) and eq(2),we get\n",
- "lamda=e*Y/d\n",
- "print \"wavelength of incident light is lamda=\",lamda,\"cm\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "wavelength of incident light is lamda= 5e-05 cm\n"
- ]
- }
- ],
- "prompt_number": 7
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Ex3.6:pg-99"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "#to calculate values of lamda1 and lamda2\n",
- "#in fraunhofer diffraction pattern ,the direction of minima is given by e*sintheta=+m*lamda,where m=1,2,.......\n",
- "#direction of fourth minima (m=4) for wavelength lamda1 is given by e*sintheta1=4*lamda1..........eq(1)\n",
- "#similarly, e*sintheta2=5*lamda2..........eq(2)\n",
- "#from eq(1) and eq(2),we get e*sintheta=4*lamda1=5*lamda2....eq(3)\n",
- "y=0.5 #in cm\n",
- "f=100 #in cm\n",
- "theta=y/f #in radian\n",
- "sintheta=theta #theta is very small\n",
- "e=0.05 #width of slit in cm\n",
- "lamda1=e*sintheta/4\n",
- "print \"lamda1=\",lamda1,\"cm\"\n",
- "#from eq(3) we get,\n",
- "lamda2=4*lamda1/5\n",
- "print \"lamda2=\",lamda2,\"cm\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "lamda1= 6.25e-05 cm\n",
- "lamda2= 5e-05 cm\n"
- ]
- }
- ],
- "prompt_number": 8
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Ex3.7:pg-100"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "#to calculate half angular width \n",
- "e=1.2*10**-4 #width of slit in cm\n",
- "y=6*10**-5 #wavelength of monochromatic light in cm\n",
- "theta=y/e\n",
- "print \"half angular width of central bright maxima is theta=\",theta,\"radian\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "half angular width of central bright maxima is theta= 0.5 radian\n"
- ]
- }
- ],
- "prompt_number": 9
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Ex3.8:pg-100"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "#to calculate angle\n",
- "lamda=6000.0*10**-8 #wavelength of light in cm\n",
- "e=0.03 #width of slit in cm\n",
- "#e*sintheta=m*lamda,where m=1\n",
- "theta=math.degrees(math.asin(lamda/e)) \n",
- "print \"angle at which the first dark band are formed in the fraunhofer diffraction pattern is theta=\",round(theta,4),\"degree\"\n",
- "theta1=math.degrees(math.sin(3*lamda/(2*e)))\n",
- "print \"angle at which the next bright band are formed in the fraunhofer diffraction pattern is theta1=\",round(theta1,4),\"degree\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "angle at which the first dark band are formed in the fraunhofer diffraction pattern is theta= 0.1146 degree\n",
- "angle at which the next bright band are formed in the fraunhofer diffraction pattern is theta1= 0.1719 degree\n"
- ]
- }
- ],
- "prompt_number": 17
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Ex3.9:pg-101"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "#to calculate distances of first dark band and of next bright band on either side of the central maximum\n",
- "#formula is e*sintheta=m*lamda,where m=1\n",
- "lamda=5890.0*10**-8 #wavelength of light in cm\n",
- "e=0.03 #width of slit in cm\n",
- "sintheta=lamda/e \n",
- "theta=sintheta #becoz theta is very small,so sintheta is approximately equal to theta\n",
- "f=50.0\n",
- "y=f*theta\n",
- "print \"linear distance of first minimum from the central maximum is y=\",y,\"cm\"\n",
- "sintheta1=3*lamda/(2*e)\n",
- "theta1=sintheta1\n",
- "y1=f*theta1\n",
- "print \"linear distance of first secondary maxima is y1=\",y1,\"cm\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "linear distance of first minimum from the central maximum is y= 0.0981666666667 cm\n",
- "linear distance of first secondary maxima is y1= 0.14725 cm\n"
- ]
- }
- ],
- "prompt_number": 13
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Ex3.10:pg-105"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "#to calculate wavelength of light and missing orders\n",
- "omega=0.25 #fringe width in cm\n",
- "D=170 #distance in cm\n",
- "twod=0.04 # distance in cm\n",
- "lamda=omega*twod/D\n",
- "print \"wavelength of light is lamda=\",lamda,\"cm\"\n",
- "e=0.08 #width of slit in mm\n",
- "d=0.4 #in mm\n",
- "m=1\n",
- "n=m*(e+d)/e \n",
- "print \"missing order is n=\",n,\"unitless\"\n",
- "#we can also find order for m=2,3,....\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "wavelength of light is lamda= 5.88235294118e-05 cm\n",
- "missing order is n= 6.0 unitless\n"
- ]
- }
- ],
- "prompt_number": 14
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Ex3.11:pg-112"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "#to calculate wavelength\n",
- "n=2.0 #order of spectrum\n",
- "theta=math.pi/6 #in radians\n",
- "E=1.0/5000 #let (e+d)=E\n",
- "lamda=E*math.sin(math.pi/6)/n\n",
- "print \"the wavelength of the spectral line is lamda=\",lamda,\"cm\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "the wavelength of the spectral line is lamda= 5e-05 cm\n"
- ]
- }
- ],
- "prompt_number": 16
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Ex3.12:pg-112"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "#to calculate difference in deviations \n",
- "lamda=5*10**-5 #wavelength of light in cm\n",
- "eplusd=1/6000.0 #where eplusd=e+d\n",
- "theta1=math.degrees(math.asin(lamda/eplusd)) #for first order spectrum\n",
- "theta3=math.degrees(math.asin(3*lamda/eplusd)) #for second order spectrum\n",
- "difference=theta3-theta1\n",
- "print \"difference in deviations in first and third order spectra is difference =\",round(difference,1),\"degree\"\n",
- "\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "difference in deviations in first and third order spectra is difference = 46.7 degree\n"
- ]
- }
- ],
- "prompt_number": 15
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Ex3.13:pg-112"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "#to calculate orders\n",
- "#let E=(e+d) \n",
- "#formula is (e+d)*sin thita=n*lamda\n",
- "#for maximum order to be possible thita=90 degree\n",
- "#sin theta=1\n",
- "E=2.54/2620 #in cm\n",
- "lamda=5*10**-5 #wavelength of the incident light in cm\n",
- "n=E/lamda\n",
- "print \"the orders will be visible is n=\",n,\"unitless\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "the orders will be visible is n= 19.3893129771 unitless\n"
- ]
- }
- ],
- "prompt_number": 18
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Ex3.14:pg-113"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "#to calculate number of lines in the grating\n",
- "#theta1=theta2=30 degree\n",
- "#sin theta1=sin theta2=1/2\n",
- "lamda1=6*10**-5 \n",
- " #wavelength in cm\n",
- "lamda2=4.5*10**-5\n",
- "#let (e+d)=E \n",
- "#formula is (e+d)*sin theta1=n*lamda1----------eq(1)\n",
- "#(e+d)*sin theta2=(n+1)*lamda2----------eq(2)\n",
- "#we get,\n",
- "n=lamda2/(lamda1-lamda2) #order of spectrum\n",
- "E=n*lamda1/math.sin(math.pi/6)\n",
- "number=1/E\n",
- "print \"number of lines is number=\",number,\"unitless\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "number of lines is number= 2777.77777778 unitless\n"
- ]
- }
- ],
- "prompt_number": 19
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Ex3.15:pg-113"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "#to calculate order when visible light of wavelength in the range 4000 to 7500 angstrom\n",
- "#let E=(e+d)\n",
- "E=1.0/4000 #in cm\n",
- "lamda1=4*10**-5 \n",
- " #wavelength in cm\n",
- "lamda2=7.5*10**-5\n",
- "n1=E*math.sin(math.pi/2)/lamda1\n",
- "n2=E*math.sin(math.pi/2)/lamda2\n",
- "print \"order when wavelength of 4000 angstrom is n1=\",n1,\"unitless\"\n",
- "print \"order when wavelength of 7500 angstrom is n2=\",n2,\"unitless\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "order when wavelength of 4000 angstrom is n1= 6.25 unitless\n",
- "order when wavelength of 7500 angstrom is n2= 3.33333333333 unitless\n"
- ]
- }
- ],
- "prompt_number": 21
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Ex3.17:pg-114"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "#to calculate angle of diffraction \n",
- "n=1 #order\n",
- "lamda=5000*10**-8 #wavelength of light in cm\n",
- "eplusd=1/5000.0 # in cm\n",
- "theta=math.degrees(math.asin(n*lamda/(eplusd)))\n",
- "minu=(theta-int(theta))*60 # minute\n",
- "print \"angle of diffraction for maximum intensity in the first order is theta=\",int(theta),\"degree\",round(minu),\"minutes\" \n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "angle of diffraction for maximum intensity in the first order is theta= 14 degree 29.0 minutes\n"
- ]
- }
- ],
- "prompt_number": 21
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Ex3.18:pg-115"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "#to calculate number of lines in one centimeter of the grating\n",
- "#let E=(e+d)\n",
- "#formula for grating equation for principal maxima is (e+d)*sin theta=n*lamda\n",
- "n=2 #order of spectrum\n",
- "lamda=5*10**-5 #wavelength in cm\n",
- "E=n*lamda/math.sin(math.pi/6)\n",
- "number=1/E\n",
- "print \"number of lines is number=\",number,\"unitless\"\n",
- "#answer is given wrong in the book ,number of lines=1000\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "number of lines is number= 5000.0 unitless\n"
- ]
- }
- ],
- "prompt_number": 23
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Ex3.19:pg-115"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "#to calculate which spectral line in 5th order will overlap with 4th order line of 5890 angstrom\n",
- "#the grating equation for principal maxima is (e+d)*sin theta =n*lamda\n",
- "n1=5 \n",
- " #order of spectrum \n",
- "n2=4\n",
- "lamda2=5890*10**-8 #wavelength of 4th order spectrum in cm\n",
- "#(e+d)*sin theta=5*lamda-------------eq(1)\n",
- "#(e+d)*sin theta=4*5890*10**-8-----------------eq(2)\n",
- "#from eq(1) and eq(2) ,we get\n",
- "lamda1=n2*lamda2/n1\n",
- "print \"wavelength of 5th order spectrum is lamda1=\",lamda1,\"cm\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "wavelength of 5th order spectrum is lamda1= 4.712e-05 cm\n"
- ]
- }
- ],
- "prompt_number": 24
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Ex3.20:pg-115"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "#to calculate grating element\n",
- "#grating equation for principal maxima is given by (e+d)*sintheta=n*lamda\n",
- "#let nth order spectrum for yellow line (lamda=6000 angstrom) coincide with (n+1)th order spectrum for blue line (lamda=4800 angstrom) \n",
- "#(e+d)*sintheta=n*6000*10**-8..eq(1)\n",
- "#(e+d)*sintheta=(n+1)*4800*10**-8.....eq(2)\n",
- "#from eq(1) and eq(2),we get n=4\n",
- "n=4\n",
- "lamda=6000*10**-8 #wavelength in cm\n",
- "sintheta=3.0/4 \n",
- "eplusd=n*lamda/sintheta\n",
- "print \"grating element is eplusd=\",eplusd,\"cm\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "grating element is eplusd= 0.00032 cm\n"
- ]
- }
- ],
- "prompt_number": 26
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Ex3.21:pg-116"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "#to calculate angle of diffraction for third order spectrum and absent spectra if any\n",
- "n=3\n",
- "lamda=6000*10**-8\n",
- "eplusd=1/200.0\n",
- "theta=math.degrees(math.asin(n*lamda/eplusd))\n",
- "minu=(theta-int(theta))*60 # minute\n",
- "print \"angle of refraction is theta=\",int(theta),\"degree\",round(minu),\"minutes\"\n",
- "d=0.0025\n",
- "e=eplusd-d #width of wire in cm\n",
- "m=1\n",
- "n=eplusd*m/e\n",
- "print \"order of absent spectrum is n=\",n,\"unitless\"\n",
- "print \"here,m=1 is considered because the higher values of m result the order of absent spectrum more than the given order 3\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "angle of refraction is theta= 2 degree 4.0 minutes\n",
- "order of absent spectrum is n= 2.0 unitless\n",
- "here,m=1 is considered because the higher values of m result the order of absent spectrum more than the given order 3\n"
- ]
- }
- ],
- "prompt_number": 24
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Ex3.22:pg-116"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "#to calculate difference in the two wavelengths\n",
- "#grating equation for principal maxima is (e+d)*sintheta=n*lamda...............eq(1)\n",
- "#differentiate both sides ,we get dtheta=n*dlamda/((e+d)*costheta)...........eq(2)\n",
- "lamda=5000 #mean value of wavelengths in angstrom\n",
- "cottheta=1.732 #cot30degree=1.732\n",
- "dtheta=0.01 #in radian\n",
- "#put the value of n from eq(2),we can write eq(2) \n",
- "dlamda=lamda*dtheta*cottheta\n",
- "print \"difference in two wavelengths is dlamda=\",dlamda,\"angstrom\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "difference in two wavelengths is dlamda= 86.6 angstrom\n"
- ]
- }
- ],
- "prompt_number": 2
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Ex3.23:pg-117"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "#to calculate dispersive power \n",
- "#differentiate grating equation ,we get dtheta/dlamda=n/((e+d)*costheta)\n",
- "n=2 #order \n",
- "eplusd=1/4000.0\n",
- "lamda=5000.0*10**-8 #wavelength in cm\n",
- "sintheta=n*lamda/(eplusd)\n",
- "costheta=math.sqrt(1-(sintheta)**2)\n",
- "dtheta=n/((eplusd)*costheta) #where dispersive power dtheta/dlamda=dtheta\n",
- "print \"dispersive power of grating in the second order spectrum is dtheta=\",int(dtheta),\"unitless\"\n",
- "\n",
- "# answer is slightly different due to approximation\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "dispersive power of grating in the second order spectrum is dtheta= 8728 unitless\n"
- ]
- }
- ],
- "prompt_number": 27
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Ex3.24:pg-117"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "#to calculate orders\n",
- "eplusd=1/4000.0\n",
- "lamda1=5*10.0**-5 #wavelengh in cm\n",
- "lamda2=7.5*10**-5\n",
- "nmax1=eplusd/lamda1\n",
- "nmax2=eplusd/lamda2\n",
- "print \"orders will be observed by a grating ,if it is illuminated by light of wavelength of 5000 angstrom is nmax1=\",nmax1,\"unitless \"\n",
- "print \"orders will be observed ,if it is illuminated by light of wavelength of 7500 angstrom is nmax2=\",round(nmax2,1),\"unitless\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "orders will be observed by a grating ,if it is illuminated by light of wavelength of 5000 angstrom is nmax1= 5.0 unitless \n",
- "orders will be observed ,if it is illuminated by light of wavelength of 7500 angstrom is nmax2= 3.3 unitless\n"
- ]
- }
- ],
- "prompt_number": 30
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Ex3.25:pg-118"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "#to calculate difference in wavelengths of two lines\n",
- "#let E=(e+d)=1/5000\n",
- "#we get\n",
- "E=2*10**-4 #in cm\n",
- "n=2 #order of spectrum\n",
- "lamda=5893*10**-8 #wavelength in cm\n",
- "#dtheta=2.5'=(2.5/60)*(3.14/180),we get\n",
- "dtheta=7.27*10**-4 #in radian\n",
- "dlamda=math.sqrt(((E/n)**2)-lamda**2)*dtheta\n",
- "print \"the difference in wavelengths of two lines is dlamda=\",dlamda,\"cm\" \n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "the difference in wavelengths of two lines is dlamda= 5.87353693335e-08 cm\n"
- ]
- }
- ],
- "prompt_number": 5
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Ex3.26:pg-123"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "#to calculate aperture of the objective of a telescope\n",
- "lamda=6*10**-5 #wavelength of light in cm\n",
- "dtheta=4.88*10**-6 # in radians\n",
- "a=1.22*lamda/dtheta\n",
- "print \"the aperture of the objective of a telescope is a=\",a,\"cm\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "the aperture of the objective of a telescope is a= 15.0 cm\n"
- ]
- }
- ],
- "prompt_number": 6
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Ex3.27:pg-123"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "#to calculate separation of two points on the moon\n",
- "lamda=5.5*10**-5 #wavelength of light in cm\n",
- "a=500 #diameter in cm\n",
- "dtheta=1.22*lamda/a #limit of resolution of telescope in radians\n",
- "R=3.8*10**8 #distance between earth and moon in m\n",
- "X=R*dtheta\n",
- "print \"the separation of two points on the moon is X=\",X,\"m\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "the separation of two points on the moon is X= 50.996 m\n"
- ]
- }
- ],
- "prompt_number": 7
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Ex3.28:pg-124"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "#to calculate numerical aperture of the objective\n",
- "lamda=5.461*10**-5 #wavelength in cm\n",
- "S=5.55*10**-5 #distance in cm\n",
- "NA=1.22*lamda/(2*S)\n",
- "print \"the numerical aperture of the objective is NA=\",NA,\"unitless\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "the numerical aperture of the objective is NA= 0.600218018018 unitless\n"
- ]
- }
- ],
- "prompt_number": 8
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Ex3.29:pg-124"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "#to calculate resolving power of microscope\n",
- "NA=0.12 #numerical aperture\n",
- "lamda=6*10**-5 #wavelength of light in cm\n",
- "RP=2*NA/lamda #RP=resolving power\n",
- "print \"the resolving power of microscope is RP=\",RP,\"unitless\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "the resolving power of microscope is RP= 4000.0 unitless\n"
- ]
- }
- ],
- "prompt_number": 9
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Ex3.30:pg-124"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "#to calculate maximum resolving power\n",
- "lamda=5*10**-5 #wavelength of light in cm\n",
- "N=40000 #total number of lines on grating\n",
- "#(e+d)=12.5*10**-5 cm\n",
- "#formula is nmax=(e+d)/lamda\n",
- "#we get\n",
- "nmax=2 #order of spectrum\n",
- "RP=nmax*N #RP=resolving power\n",
- "print \"the maximum resolving power is RP=\",RP,\"unitless\" \n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "the maximum resolving power is RP= 80000 unitless\n"
- ]
- }
- ],
- "prompt_number": 10
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Ex3.31:pg-124"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "#to calculate minimum number of lines in a grating\n",
- "lamda1=5890 \n",
- " #wavelengh in angstrom\n",
- "lamda2=5896\n",
- "dlamda=6 #smallest wavelength difference in angstrom\n",
- "n=2 #order of spectrum \n",
- "lamda=(lamda1+lamda2)/2 #average wavelength in angstrom\n",
- "RP=lamda/dlamda #RP=resolving power\n",
- "N=RP/n\n",
- "print \"minimum number of lines in a grating is N=\",N,\"unitless\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "minimum number of lines in a grating is N= 491 unitless\n"
- ]
- }
- ],
- "prompt_number": 11
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Ex3.32:pg-125"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "#will the telescope be able to observe the wiremesh\n",
- "a=3 #aperture in cm\n",
- "lamda=5.5*10**-5 #wavelength of light in cm\n",
- "#limit of resolution of telescope is given by\n",
- "theta=1.22*lamda/a\n",
- "#alpha=spacing of wire-mesh/distance of objective from wire-mesh\n",
- "alpha=0.2/(80*10**2)\n",
- "print \"theta=\",theta,\"radian\"\n",
- "print \"alpha=\",alpha,\"radian\"\n",
- "print \"if alpha>theta then telescope will be able to observe the wire-mesh\"\n",
- "#value of alpha is given wrong in the book 2.25*10**-5 radian\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "theta= 2.23666666667e-05 radian\n",
- "alpha= 2.5e-05 radian\n",
- "if alpha>theta then telescope will be able to observe the wire-mesh\n"
- ]
- }
- ],
- "prompt_number": 12
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Ex3.33:pg-125"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "#distance between the centres of images of two stars\n",
- "lamda=5500*10**-8 #wavelength of light in cm\n",
- "f=4*10**2 #focal length of telescope objective in cm\n",
- "a=0.01*10**2 #diameter in cm\n",
- "X=1.22*lamda*f/a\n",
- "print \"distance between the centres of images of two stars is X=\",X,\"cm \"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "distance between the centres of images of two stars is X= 0.02684 cm \n"
- ]
- }
- ],
- "prompt_number": 13
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Ex3.34:pg-126"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "#to calculate diameter of a telescope \n",
- "lamda=5.0*10**-5 #wavelength in cm\n",
- "theta=(math.pi/180)*(1.0/1000) #in radians\n",
- "a=1.22*lamda/theta\n",
- "print \"the diameter of a telescope is a=\",a,\"cm\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "the diameter of a telescope is a= 3.4950425503 cm\n"
- ]
- }
- ],
- "prompt_number": 16
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Ex3.35:pg-126"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "#to calculate smallest angle between two stars\n",
- "lamda=5*10**-5 #wavelength in cm\n",
- "a=100*2.54 #diameter in cm\n",
- "theta=1.22*lamda/a\n",
- "print \"the smallest angle between two stars is thita=\",theta,\"radians\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "the smallest angle between two stars is thita= 2.40157480315e-07 radians\n"
- ]
- }
- ],
- "prompt_number": 17
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Ex3.36:pg-126"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "#to calculate limit of resolution of the telescope\n",
- "lamda=5890*10**-8 #wavelength in cm\n",
- "a=1 #diameter in cm\n",
- "theta=1.22*lamda/a\n",
- "print \"the limit of resolution of the telescope is theta=\",theta,\"radians \"\n",
- "\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "the limit of resolution of the telescope is theta= 7.1858e-05 radians \n"
- ]
- }
- ],
- "prompt_number": 21
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Ex3.37:pg-126"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "#to calculate resolving limit of microscope\n",
- "lamda=5.5*10**-5 #wavelengh in cm\n",
- "theta=math.pi/6 #in radians\n",
- "s=1.22*lamda/(2*math.sin(math.pi/6))\n",
- "print \"resolving limit of microscope is s=\",s,\"cm\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "resolving limit of microscope is s= 6.71e-05 cm\n"
- ]
- }
- ],
- "prompt_number": 20
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Ex3.38:pg-127"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "#to calculate resolving power of grating \n",
- "N=15000 #total number of lines on grating\n",
- "lamda=6*10**-5 #wavelength in cm\n",
- "n=2 #order of spectrum\n",
- "RP=n*N\n",
- "print \"resolving power is RP =\",RP,\"unitless\"\n",
- "#to calculate smallest wavelength difference that can be resolved with a light of wavelength 6000angstrom in the second order\n",
- "dlamda=lamda/(n*N)\n",
- "print \"smallest wavelength difference dlamda=\",dlamda,\"cm\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "resolving power is RP = 30000 unitless\n",
- "smallest wavelength difference dlamda= 2e-09 cm\n"
- ]
- }
- ],
- "prompt_number": 19
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Ex3.39:pg-127"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "#to calculate resolving power in the second order\n",
- "N=6*10**4 #N=total number of lines on grating\n",
- "n=2 #order of spectrum\n",
- "RP=n*N #RP=resoling power\n",
- "print \"the resolving power is RP=\",RP,\"unitless\"\n",
- "#to calculate smallest wavelength\n",
- "lamda=6000*10**-8 #wavelength in cm\n",
- "n=3 #order of spectrum\n",
- "dlamda=lamda/(n*N)\n",
- "print \"smallest wavelength that can be resolved in the third order in 6000angstrom wavelength region is dlamda=\",dlamda,\"cm\" \n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "the resolving power is RP= 120000 unitless\n",
- "smallest wavelength that can be resolved in the third order in 6000angstrom wavelength region is dlamda= 3.33333333333e-10 cm\n"
- ]
- }
- ],
- "prompt_number": 18
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Engineering_Physics/chapter4.ipynb b/Engineering_Physics/chapter4.ipynb deleted file mode 100755 index 3e350b0f..00000000 --- a/Engineering_Physics/chapter4.ipynb +++ /dev/null @@ -1,541 +0,0 @@ -{
- "metadata": {
- "name": "",
- "signature": "sha256:7c369c2593aa95d4aac8153f9063553bb327aa743223d1fff41c98aa45d77ebe"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "heading",
- "level": 1,
- "metadata": {},
- "source": [
- "Chapter4:POLARISATION"
- ]
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Ex4.1:pg-147"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "# compare the intensities of ordinary and extraordinary rays\n",
- "#intensity of ordinary rays is given by Io=a**2 *(sin theta)**2\n",
- "#where theta=30 degree\n",
- "#we get Io=a**2/4\n",
- "Io=1.0/4\n",
- "#intensity of extraordinary ray is given by IE=(a*cos theta)**2\n",
- "#we get IE=3*a**2/4\n",
- "IE=3.0/4\n",
- "I=IE/Io\n",
- "print \"the intensities of ordinary and extraordinary rays is I=\",I,\"unitless\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "the intensities of ordinary and extraordinary rays is I= 3.0 unitless\n"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Ex4.2:pg-147"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "#to calculate angle of refraction\n",
- "#according to brewster's law mu=tan ip\n",
- "mu=1.732 #refractive index\n",
- "ip=math.degrees(math.atan(mu)) #polarising angle in degree\n",
- "r=90-ip\n",
- "print \"angle of refraction of ray is r=\",round(r),\"degree\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "angle of refraction of ray is r= 30.0 degree\n"
- ]
- }
- ],
- "prompt_number": 11
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Ex4.3:pg-147"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "#to calculate polarising angle and angle of refraction\n",
- "mu=1.345 #refractive index, mu=1/sinc=1/sin48degree=1/0.7431 \n",
- "ip=math.degrees(math.atan(mu))\n",
- "r=90-ip\n",
- "print \"polarising angle is ip=\",round(ip,3),\"degree\"\n",
- "print \"angle of refraction is r=\",round(r,3),\"degree\" \n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "polarising angle is ip= 53.369 degree\n",
- "angle of refraction is r= 36.631 degree\n"
- ]
- }
- ],
- "prompt_number": 10
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Ex4.4:pg-147"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "#to calculate thickness of a half wave plate of quartz\n",
- "lamda=5*10**-5 #wavelength in cm\n",
- "mue=1.553 \n",
- " #refractive index (unitless)\n",
- "muo=1.544\n",
- "#for a half plate of positive crystal\n",
- "t=lamda/(2*(mue-muo))\n",
- "print \"thickness of a half wave plate of quartz is t=\",\"{:.2e}\".format(t),\"cm\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "thickness of a half wave plate of quartz is t= 2.78e-03 cm\n"
- ]
- }
- ],
- "prompt_number": 9
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Ex4.5:pg-148"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "#to calculate thickness of quarter wave plate\n",
- "lamda=5.890*10**-5 #wavelength of light in cm\n",
- "mue=1.553\n",
- " #refractive index\n",
- "muo=1.544\n",
- "t=lamda/(4*(mue-muo)) \n",
- "print \"thickness of quarter wave plate is t=\",\"{:.3e}\".format(t),\"cm\" \n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "thickness of quarter wave plate is t= 1.636e-03 cm\n"
- ]
- }
- ],
- "prompt_number": 8
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Ex4.6:pg-148"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "#to calculate thickness of a doubly refracting plate\n",
- "lamda=5.890*10**-5 #wavelength in cm\n",
- "muo=1.53 \n",
- " #refractive index\n",
- "mue=1.54\n",
- "t=lamda/(4*(mue-muo))\n",
- "print \"thickness of a plate is t=\",\"{:.2e}\".format(t),\"cm\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "thickness of a plate is t= 1.47e-03 cm\n"
- ]
- }
- ],
- "prompt_number": 6
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Ex4.7:pg-152"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "#to calculate angle of rotation\n",
- "alpha=66 #specific rotation of cane sugar in degree\n",
- "c=15.0/100 #concentration of the solution in gm/cc\n",
- "l=20 #length of tube in cm\n",
- "theta=alpha*l*c/10\n",
- "print \"the angle of rotation of the plane of polarisation is theta=\",theta,\"degree\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "the angle of rotation of the plane of polarisation is theta= 19.8 degree\n"
- ]
- }
- ],
- "prompt_number": 5
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Ex4.8:pg-153"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "#to calculate specific rotation \n",
- "theta=26.4 #in degree\n",
- "l=20 #length in cm\n",
- "c=0.2 #gm/cm**3\n",
- "alpha=10*theta/(l*c)\n",
- "print \"the specific rotation is alpha=\",alpha,\"degree\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "the specific rotation is alpha= 66.0 degree\n"
- ]
- }
- ],
- "prompt_number": 10
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Ex4.9:pg-153"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "#to calculate strength of solution\n",
- "theta=11 #degree\n",
- "l=20.0 #length in cm\n",
- "alpha=66 #specific rotation of sugar in degree\n",
- "c=10*theta/(l*alpha)\n",
- "print \"strength of solution is c=\",round(c,4),\"gm/cm**3\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "strength of solution is c= 0.0833 gm/cm**3\n"
- ]
- }
- ],
- "prompt_number": 4
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Ex4.10:pg-153"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "#to calculate difference in the refractive indices\n",
- "#specific rotation is theta/d=29.73 degree/mm\n",
- "theta=29.73 #where theta=theta/d\n",
- "lamda=5.086*10**-4 #wavelength in mm\n",
- "#optical rotation is given by theta=math.pi*d*(mul-mur)/lamda\n",
- "#where mul and mur are refractive indices for anti-clockwise and clockwise polarised lights\n",
- "mu=theta*lamda/180 #where mu=mul-mur\n",
- "print \"difference in refractive indices is mu=\",\"{:.1e}\".format(mu),\"unitless\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "difference in refractive indices is mu= 8.4e-05 unitless\n"
- ]
- }
- ],
- "prompt_number": 3
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Ex4.11:pg-153"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "#to calculate optical rotation\n",
- "#let theta' be the optical rotation by a solution of strength c' in a tube of length l' then\n",
- "#we get 10*theta'/l'*c'=10*theta/l*c\n",
- "c=1.0/3 #it is given that solution is 1/3 of its previous concentration i.e. c'/c=1/3,where c=c'/c\n",
- "l1=30 #where l1=l'\n",
- " #length in cm \n",
- "l=20.0\n",
- "theta=13 #degree\n",
- "#formula is theta'=l'*c'*theta/(l*c)\n",
- "theta1=l1*c*theta/l\n",
- "print \"optical rotation is theta1=\",theta1,\"degree\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "optical rotation is theta1= 6.5 degree\n"
- ]
- }
- ],
- "prompt_number": 5
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Ex4.12:pg-154"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "#to calculate specific rotation\n",
- "theta=52.8 #optical rotation in degree\n",
- "l=20.0 #length of the solution in cm\n",
- "c=20/50.0 #concentration of the solution in gm/cc\n",
- "alpha=10*theta/(l*c)\n",
- "print \"the specific rotation is alpha=\",alpha,\"degree\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "the specific rotation is alpha= 66.0 degree\n"
- ]
- }
- ],
- "prompt_number": 2
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Ex4.13:pg-154"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "#to calculate length \n",
- "l=40 #length in cm\n",
- "c=5.0/100 #concentration in percentage\n",
- "theta1=35 #optical rotation in degree ,where theta1=theta'\n",
- "c1=10.0/100 #concentration in % ,where c1=c'\n",
- "theta=20\n",
- "#formula of specific rotation is alpha=10*theta/l*c\n",
- "l1=l*c*theta1/(c1*theta)\n",
- "print \"length is l1=\",l1,\"cm\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "length is l1= 35.0 cm\n"
- ]
- }
- ],
- "prompt_number": 6
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Ex4.14:pg-155"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "#to calculate rotation of plane of polarisation of light\n",
- "mur=1.53914\n",
- " #refractive index\n",
- "mul=1.53920\n",
- "lamda=6.5*10**-5 #wavelength in cm\n",
- "d=0.02 #distance in cm\n",
- "thetaR=180*(mul-mur)*d/lamda\n",
- "print \"rotation of plane of polarisation of light is thetaR=\",round(thetaR,3),\"degree\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "rotation of plane of polarisation of light is thetaR= 3.323 degree\n"
- ]
- }
- ],
- "prompt_number": 12
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Ex4.15:pg-155"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "#to calculate % purity of the sugar sample\n",
- "theta=9.9 #optical rotation in degree\n",
- "alpha=66 #specific roation of pure sugar solution in dm**-1(gm/cc)**-1\n",
- "l=20 #length of tube in cm\n",
- "c=10*theta/(l*alpha) #concentration of solution in gm/c.c\n",
- "#it is given that 80 gm of impure sugar is dissolved in a litre of water\n",
- "per=(c*100*10**3)/80 #here c is in gm/litre\n",
- "print \"percentage of the sugar sample is per=\",per,\"%\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "percentage of the sugar sample is per= 93.75 %\n"
- ]
- }
- ],
- "prompt_number": 17
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Engineering_Physics/chapter5.ipynb b/Engineering_Physics/chapter5.ipynb deleted file mode 100755 index bf9aee93..00000000 --- a/Engineering_Physics/chapter5.ipynb +++ /dev/null @@ -1,177 +0,0 @@ -{
- "metadata": {
- "name": "",
- "signature": "sha256:141e2987b2be679ae1fc9e807cf81c12805438f836d3be09701fc88866bf9bb5"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "heading",
- "level": 1,
- "metadata": {},
- "source": [
- "Chapter5:LASERS"
- ]
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Ex5.1:pg-164"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "#to calculate area of the spot on the moon\n",
- "lamda=6*10**-7 #wavelength in m\n",
- "d=2 #diameter in m\n",
- "dtheta=lamda/d #angular spread in radian\n",
- "D=4*10**8 #distance of the moon\n",
- "A=(D*dtheta)**2\n",
- "print \"the areal spread is A=\",\"{:.2e}\".format(A),\"m**2\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "the areal spread is A= 1.44e+04 m**2\n"
- ]
- }
- ],
- "prompt_number": 11
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Ex5.2:pg-164"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "#to calculate angular spread of the beam\n",
- "lamda=8*10**-7 #wavelength in m\n",
- "d=5*10**-3 #aperture in m\n",
- "dtheta=lamda/d \n",
- "print \"the angular spread of the beam is dtheta=\",\"{:.1e}\".format(dtheta),\"radian\"\n",
- "#to calculate the areal spread when it reaches the moon\n",
- "D=4*10**8 #distance of the moon in m\n",
- "A=(D*dtheta)**2\n",
- "print \"the areal spread is A=\",\"{:.3e}\".format(A),\"m**2\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "the angular spread of the beam is dtheta= 1.6e-04 radian\n",
- "the areal spread is A= 4.096e+09 m**2\n"
- ]
- }
- ],
- "prompt_number": 10
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Ex5.3:pg-165"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "#to calculate number of oscillations corresponding to the coherence length\n",
- "L=2.945*10**-2 #coherence length in m\n",
- "lamda=5890*10**-10 #wavelength of sodium light in m\n",
- "n=L/lamda\n",
- "print \"the number of oscillations is n=\",\"{:.1e}\".format(n),\"unitless\"\n",
- "#to calculate coherence time\n",
- "c=3*10**8 #light speed in m\n",
- "Time=L/c #coherence time\n",
- "print \"the coherence Time=\",\"{:.2e}\".format(Time),\"s\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "the number of oscillations is n= 5.0e+04 unitless\n",
- "the coherence Time= 9.82e-11 s\n"
- ]
- }
- ],
- "prompt_number": 7
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Ex5.4:pg-165"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "#to calculate area and intensity of the image\n",
- "lamda=7200*10**-10 #wavelength in m\n",
- "d=5*10**-3 #aperture in m\n",
- "dtheta=lamda/d #angular spread in radian \n",
- "f=0.1 #focal length in m\n",
- "arealspread=(dtheta*f)**2\n",
- "print \"areal spread =\",\"{:.3e}\".format(arealspread),\"m**2\"\n",
- "power=50*10**-3\n",
- "I=power/arealspread\n",
- "print \"intensity of the image is I=\",\"{:.3e}\".format(I),\"watts/m**2\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "areal spread = 2.074e-10 m**2\n",
- "intensity of the image is I= 2.411e+08 watts/m**2\n"
- ]
- }
- ],
- "prompt_number": 4
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [],
- "language": "python",
- "metadata": {},
- "outputs": []
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Engineering_Physics/chapter6.ipynb b/Engineering_Physics/chapter6.ipynb deleted file mode 100755 index 26000b31..00000000 --- a/Engineering_Physics/chapter6.ipynb +++ /dev/null @@ -1,230 +0,0 @@ -{
- "metadata": {
- "name": "",
- "signature": "sha256:7bd6880823e892a54fce1336eeffd037b53706d16df72c924dbe7c27da815476"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "heading",
- "level": 1,
- "metadata": {},
- "source": [
- "Chapter6:FIBER OPTICS AND HOLOGRAPHY"
- ]
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Ex6.1:pg-189"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "#to calculate critical angle for core-cladding interface\n",
- "n1=1.5\n",
- "n2=1.45\n",
- "thetac=math.degrees(math.asin(n2/n1))\n",
- "theta1=90-thetac\n",
- "\n",
- "print \"critical angle for core-cladding interface is theta1=\",round(theta1),\"degree\"\n",
- "#to calculate acceptance angle in air for fibre and corresponding angle of obliquences\n",
- "na=1\n",
- "thetaa=math.degrees(math.asin(n1*0.26/na))\n",
- "\n",
- "print \"acceptance angle thetaa=\",round(thetaa),\"degree\"\n",
- "#to calculate numerical aperture\n",
- "NA=((n1+n2)*(n1-n2))**(1/2.0)\n",
- "print \"numerical aperture of fibre is NA=\",round(NA,4),\"unitless\"\n",
- "#to calculate % of light\n",
- "per=(NA)**2*100\n",
- "print \"light collected is per=\",round(per,2),\"%\"\n",
- "\n",
- "\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "critical angle for core-cladding interface is theta1= 15.0 degree\n",
- "acceptance angle thetaa= 23.0 degree\n",
- "numerical aperture of fibre is NA= 0.3841 unitless\n",
- "light collected is per= 14.75 %\n"
- ]
- }
- ],
- "prompt_number": 8
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Ex6.2:pg-190"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "#to calculate numerical aperture\n",
- "Del=0.02 #relative refractive index difference between the core and the cladding of the fibre i.e. del=(n1-n2)/n1\n",
- "n1=1.46 #refractive index of core of W-step index fibre \n",
- "n2=n1-Del*n1\n",
- "NA=((n1+n2)*(n1-n2))**(1/2.0)\n",
- "print \"numerical aperture is NA=\",round(NA,2),\"unitless\"\n",
- "#to calculate critical angle at the core cladding interface within the fibre\n",
- "thetac=math.degrees(math.asin(n2/n1))\n",
- "print \"thetac=\",int(thetac),\"degree\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "numerical aperture is NA= 0.29 unitless\n",
- "thetac= 78 degree\n"
- ]
- }
- ],
- "prompt_number": 10
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Ex6.3:pg-190"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "#to calculate refractive index of the cladding\n",
- "a=35.0*10**-6 #core diameter in micrometre\n",
- "#formula is del=(n1-n2)/n1\n",
- "#we get\n",
- "Del=1.5/100 \n",
- "n1=1.46 #refractive index of the fibre \n",
- "lamda=0.85*10**-6 #wavelength in micrometer\n",
- "n2=n1-Del*n1\n",
- "print \"refractive index is n2=\",round(n2,3),\"unitless\"\n",
- "#to calculate normalised frequency V number of the fibre\n",
- "V=round((2*math.pi*a*n1*0.173)/lamda,1)\n",
- "print \"normalised frequency V number of the fibre is V=\",V,\"unitless\"\n",
- "#to calculate total number of guided modes in the fibre\n",
- "M=(V**2)/2.0\n",
- "print \"total number of guided modes in the fibre is M=\",int(M),\"modes\"\n",
- "\n",
- "\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "refractive index is n2= 1.438 unitless\n",
- "normalised frequency V number of the fibre is V= 65.3 unitless\n",
- "total number of guided modes in the fibre is M= 2132 modes\n"
- ]
- }
- ],
- "prompt_number": 15
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Ex6.4:pg-191"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "#to calculate cut-off wavelength of the fibre\n",
- "#(2*del)**(1/2)=(2*(n1-n2)/n1)**(1/2)=(0.005)**(1/2)=0.071\n",
- "a=5*10**-6 #radius in micrometre\n",
- "n1=1.46 #core refractive index in micrometre\n",
- "Vc=2.405 #cut-off value of V parametre for single mode operation\n",
- "#formula is LAMBDAc=(2*math.pi*a*n1*(2*del)**(1/2))/Vc \n",
- "lamdac=(2*math.pi*a*n1*0.071)/Vc\n",
- "print \"cut-off wavelength is LAMBDAc=\",round(lamdac/1e-6,2),\"micrometre\"\n",
- "\n",
- "# answer is slightly different due to approximation in book\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "cut-off wavelength is LAMBDAc= 1.35 micrometre\n"
- ]
- }
- ],
- "prompt_number": 5
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Ex6.5:pg-191"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "#to calculate maximum and minimum value of phase constant\n",
- "lamda=0.8*10**-6 #wavelength in micrometre\n",
- "n1=1.6*10**-6 \n",
- " #refractive indices in micrometre\n",
- "n2=1.44*10**-6\n",
- "maximum=(2*math.pi*n1)/lamda\n",
- "minimum=(2*math.pi*n2)/lamda\n",
- "print \"maximum value of phase constant is maximum=\",round(maximum,3),\"radian/micrometre\"\n",
- "print \"minimum value of phase constant is minimum=\",round(minimum,2),\"radian/micrometre\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "maximum value of phase constant is maximum= 12.566 radian/micrometre\n",
- "minimum value of phase constant is minimum= 11.31 radian/micrometre\n"
- ]
- }
- ],
- "prompt_number": 6
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Engineering_Physics/chapter7.ipynb b/Engineering_Physics/chapter7.ipynb deleted file mode 100755 index b30ad5a1..00000000 --- a/Engineering_Physics/chapter7.ipynb +++ /dev/null @@ -1,912 +0,0 @@ -{
- "metadata": {
- "name": "",
- "signature": "sha256:43ad060be6803a5e6c90770bf46ae3612188f9380f800bb70a03161cb97405cb"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Chapter7:WAVE MECHANICS"
- ]
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Ex7.1:pg-200"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "#to calculate de Broglie wavelength \n",
- "v=1.5*10**7 #velocity of proton =(1/20)*velocity of light i.e.3*10**8 in m/s\n",
- "m=1.67*10**-27 #mass of the proton in kg\n",
- "h=6.6*10**-34 #plank's constant \n",
- "lamda=h/(m*v)\n",
- "print \"the de Broglie wavelength is lamda=\",\"{:.3e}\".format(lamda),\"m\" \n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "the de Broglie wavelength is lamda= 2.635e-14 m\n"
- ]
- }
- ],
- "prompt_number": 4
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Ex7.2:pg-200"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "#to calculate de Broglie wavelength\n",
- "#mo*c**2=1.507*10**-10/1.6*10**-19=941.87 Mev\n",
- "#since 12.8 Mev is very small compared to rest mass energy hence relavistic consideration may be ignored\n",
- "m=1.67*10**-27 #mass in kg\n",
- "h=6.62*10**-34 #plank's constant\n",
- "E=12.8*10**6 #energy in Mev\n",
- "lamda=h/math.sqrt(2*m*E*1.6*10**-19)/(1e-10)\n",
- "print \"the de Broglie wavelength is lamda=\",round(lamda,5),\"angstrom\"\n",
- "\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "the de Broglie wavelength is lamda= 8e-05 angstrom\n"
- ]
- }
- ],
- "prompt_number": 11
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Ex7.4:pg-201"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "#to calculate wavelength\n",
- "h=6.6*10**-34 #plank's constant\n",
- "m=9.1*10**-31 #mass of electron in kg\n",
- "E=1.25*10**3 #pottential difference keV\n",
- "lamda=h/math.sqrt(2*m*E*1.6*10**-19)\n",
- "print \"the wavelength is lamda=\",\"{:.2e}\".format(lamda),\"m\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "the wavelength is lamda= 3.46e-11 m\n"
- ]
- }
- ],
- "prompt_number": 6
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Ex7.5:pg-201"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "#to calculate kinetic energy of an electron \n",
- "h=6.63*10**-34 #plank's constant\n",
- "mo=9.1*10**-31 #rest mass of an electron in kg\n",
- "lamda=5896*10**-10 #wavelength in angstrom\n",
- "K=(h**2)/(2*mo*(lamda**2)*1.6*10**-19) \n",
- "print \"kinetic energy of an electron is K=\",\"{:.2e}\".format(K),\"eV\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "kinetic energy of an electron is K= 4.34e-06 eV\n"
- ]
- }
- ],
- "prompt_number": 7
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Ex7.6:pg-202"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "#to calculate the wavelength of an electron of kinetic energy\n",
- "mo=9.1*10**-31 #mass of an electron in kg\n",
- "c=3*10**8 #speed of light in m/s \n",
- "K=1*10**6#kinetic energy in eV\n",
- "h=6.62*10**-34 #planck's constant in J-s\n",
- "#E=moc**2=81.9*10**-15/1.6*10**-19 eV=0.51MeV\n",
- "E=0.51*10**6\n",
- "lamda=(h*c)/(math.sqrt(K*(K+2*E))*1.6*10**-19)\n",
- "print \"wavelength of an electron of kinetic energy is lamda=\",round(lamda,14),\"m\"\n",
- "\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "wavelength of an electron of kinetic energy is lamda= 8.7e-13 m\n"
- ]
- }
- ],
- "prompt_number": 16
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Ex7.7:pg-203"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "#to calculate de Broglie wavelength\n",
- "V=100 #potential difference in volts\n",
- "lamda=12.25/math.sqrt(V)\n",
- "print \"de Broglie wavelength of any electron is lamda=\",lamda,\"angstrom\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "de Broglie wavelength of any electron is lamda= 1.225 angstrom\n"
- ]
- }
- ],
- "prompt_number": 7
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Ex7.9:pg-203"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "#to calculate energy of the neutron\n",
- "h=6.60*10**-34 #plank's constant in J/s\n",
- "m=1.674*10**-27 #mass of the neutron in kg\n",
- "lamda=10**-10 #de Broglie wavelength in m\n",
- "E=(h**2)/(2*m*(lamda**2)*1.6*10**-19)\n",
- "print \"energy of the neutron is E=\",\"{:.2e}\".format(E),\"eV\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "energy of the neutron is E= 8.13e-02 eV\n"
- ]
- }
- ],
- "prompt_number": 8
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Ex7.10:pg-204"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "#to calculate wavelength\n",
- "h=6.6*10**-34 #plank's constant in J/sec\n",
- "m=9.1*10**-31 #mass of electron in kg\n",
- "c=3*10**8 #light speed in m/s\n",
- "lamda=h/(m*c)/(1e-10) # in angstrom\n",
- "print \"wavelength of quantum of radiant energy is lamda=\",round(lamda,4),\"angstrom\"\n",
- "#to calculate number of photons \n",
- "power=12 #power emitted by the lamp =150*(8/100) in watts\n",
- "E=12.0 #energy emitted per second\n",
- "lamda=4500*10.0**-10\n",
- "energy=(h*c)/lamda #energy contained in one photon in J\n",
- "number=E/energy\n",
- "print \"number of photons emitted per sec is number=\",round(number,-16),\"unitless\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "wavelength of quantum of radiant energy is lamda= 0.0242 angstrom\n",
- "number of photons emitted per sec is number= 2.727e+19 unitless\n"
- ]
- }
- ],
- "prompt_number": 22
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Ex7.11:pg-209"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "#to calculate uncertainity in position\n",
- "#actual formula is (delx)min*(delp)max=h/2*math.pi-------------eq(1)\n",
- "#(delp)max=p(momentum of the electron)\n",
- "#mv=mov/math.sqrt(1-(v/c)**2)---------------------eq(2)\n",
- "mo=9*10**-31 #mass of an electron in m/s\n",
- "c=3*10**8 #light speed in m/s\n",
- "v=3*10**7 #velocity in m/s \n",
- "h=6.6*10**-34 #plank's constant in J/s\n",
- "#from eq(1) and eq(2),we get\n",
- "delxmin=(h*math.sqrt(1-(v/c)**2))/(2*math.pi*mo*v)\n",
- "print \"smallest possible uncertainity in the position of an electron is delxmin=\",round(delxmin/1e-10,4),\"angstrom\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "smallest possible uncertainity in the position of an electron is delxmin= 0.0389 angstrom\n"
- ]
- }
- ],
- "prompt_number": 12
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Ex7.12:pg-209"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "#to calculate minimum uncertainity in the velocity\n",
- "delxmax=10**-8 #maximum uncertainity in position in m\n",
- "h=6.626*10**-34 #planck's constant\n",
- "delpmin=h/(2*math.pi*delxmax) #minimum uncertainity in momentum in kg-m/s**2 \n",
- "m=9*10**-31 #mass of an electron in kg\n",
- "delvmin=delpmin/m\n",
- "print \"minimum uncertainity in the velocity is delvmin=\",\"{:.2e}\".format(delvmin),\"m/s\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "minimum uncertainity in the velocity is delvmin= 1.17e+04 m/s\n"
- ]
- }
- ],
- "prompt_number": 13
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Ex7.13:pg-209"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "#to calculate uncertainity in the momentum of the parcticle \n",
- "h=6.626*10**-34 #planck's constant J-s\n",
- "delx=0.01*10**-2 #uncertainity in position in m\n",
- "delp=h/(2*math.pi*delx)\n",
- "print \"uncertainity in the momentum of the parcticle is delp=\",\"{:.2e}\".format(delp),\"kg-m/s**2\"\n",
- "#to calculate uncertainity in the velocity of an electron\n",
- "m=9*10**-31 #mass of an electron in kg\n",
- "delx=5*10**-10 \n",
- "delv=h/(2*math.pi*m*delx)\n",
- "print \"uncertainity in the velocity of an electron is delv=\",\"{:.3e}\".format(delv),\"m/s\"\n",
- "#to calculate uncertainity in the velocity of alpha particle \n",
- "m=4*1.67*10**-27 #mass of alpha particle in kg\n",
- "delx=5*10**-10\n",
- "delv=h/(2*math.pi*m*delx)\n",
- "print \"uncertainity in the velocity of an electron is delv=\",round(delv,2),\"m/s\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "uncertainity in the momentum of the parcticle is delp= 1.05e-30 kg-m/s**2\n",
- "uncertainity in the velocity of an electron is delv= 2.343e+05 m/s\n",
- "uncertainity in the velocity of an electron is delv= 31.57 m/s\n"
- ]
- }
- ],
- "prompt_number": 15
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Ex7.14:pg-210"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "#to calculate uncertainity in position\n",
- "m=9.11*10**-31 #mass of electron in kg\n",
- "delv=40 #uncertainity in velocity in m/s\n",
- "h=6.6*10**-34 #plank's constant \n",
- "delx=h/(2*math.pi*m*delv)\n",
- "print \"uncertainity in the position of the electron is delx=\",\"{:.2e}\".format(delx),\"m\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "uncertainity in the position of the electron is delx= 2.88e-06 m\n"
- ]
- }
- ],
- "prompt_number": 16
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Ex7.15:pg-210"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "#to calculate uncertainity in frequency\n",
- "#delE*delt=h/2*math.pi----eq(1)\n",
- "#delE=h*delv-----------eq(2)\n",
- "delt=10**-8 #uncertainity in time in s\n",
- "#from eq(1) and eq(2),we get\n",
- "delnu=1/(2*math.pi*delt)\n",
- "print \"minimum uncertainity in the frequency of the photon is delv=\",\"{:.3e}\".format(delnu),\"sec**-1\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "minimum uncertainity in the frequency of the photon is delv= 1.592e+07 sec**-1\n"
- ]
- }
- ],
- "prompt_number": 17
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Ex7.16:pg-211"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "#to calculate uncertainity in the energy\n",
- "h=6.63*10**-34 #plank's constant in J-s\n",
- "delt=2.5*10**-14 #uncertainity in time in s\n",
- "delE=h/(2*math.pi*delt*1.6*10**-19)\n",
- "print \"minimum error with which the energy of the state can be measured is delE=\",round(delE,3),\"ev\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "minimum error with which the energy of the state can be measured is delE= 0.026 ev\n"
- ]
- }
- ],
- "prompt_number": 18
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Ex7.17:pg-211"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "#to calculate time required for the atomic system \n",
- "#delE=h*c*dellamda/lamda**2 -----eq(1)\n",
- "#delE*delt=h/2*math.pi----------eq(2)\n",
- "dellamda=10**-14\n",
- "c=3*10**8\n",
- "lamda=6*10**-7\n",
- "#from eq(1)and eq(2),we get\n",
- "delt=(lamda**2)/(2*math.pi*c*dellamda)\n",
- "print \"time required for the atomic system to retain rotational energy is delt=\",\"{:.1e}\".format(delt),\"s\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "time required for the atomic system to retain rotational energy is delt= 1.9e-08 s\n"
- ]
- }
- ],
- "prompt_number": 19
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Ex7.18:pg-211"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "#to calculate minimum uncertainity in the momentum \n",
- "delxmax=5*10**-14 #uncertainity in position in m\n",
- "h=6.626*10**-34 #plank's constant in Js\n",
- "delpmin=h/(2*math.pi*delxmax)\n",
- "print \"minimum uncertainity in the momentum of the nucleon is delpmin=\",\"{:.2e}\".format(delpmin),\"kg m/s\"\n",
- "m=1.675*10**-27 #mass in kg\n",
- "Emin=(delpmin**2)/(2*m*1.6*10**-19)\n",
- "print \"minimum kinetic energy of the nucleon is Emin=\",round(Emin,2),\"eV\"\n",
- "#the answer is given wrong in the book Emin=0.039 eV\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "minimum uncertainity in the momentum of the nucleon is delpmin= 2.11e-21 kg m/s\n",
- "minimum kinetic energy of the nucleon is Emin= 8299.24 eV\n"
- ]
- }
- ],
- "prompt_number": 9
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Ex7.19:pg-212"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "#to calculate uncertainity in velocity\n",
- "delx=1.1*10**-8 #uncertainity in velocity in m\n",
- "h=6.626*10**-34 #plank's constant\n",
- "m=9.1*10**-31 #mass of electron in kg\n",
- "delv=h/(2*math.pi*m*delx)\n",
- "print \"minimum uncertainity in velocity is delv=\",\"{:.2e}\".format(delv),\"m/s\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "minimum uncertainity in velocity is delv= 1.05e+04 m/s\n"
- ]
- }
- ],
- "prompt_number": 14
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Ex7.20:pg-212"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "#to calculate uncertainity in frequency\n",
- "delt=10**-8 #uncertainity in time\n",
- "delnu=1/(2*math.pi*delt) \n",
- "print \"minimum uncertainity in the frequency of a photon is delnu=\",\"{:.2e}\".format(delnu),\"sec**-1\"\n",
- "#to use the uncertainity principle to place a lower limit on the energy an electron must have if it is to be part of a nucleus\n",
- "delx=5*10**-15 #uncertainity in position\n",
- "delp=h/(2*2*math.pi*delx) #uncertainbity in momentum\n",
- "c=3*10**8 #/speed of light in m/s\n",
- "E=delp*c\n",
- "print \"energy of an electron is E=\",\"{:.2e}\".format(E),\"J\"\n",
- "\n",
- "# the answer is slightlty different due to approximation in textboook\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "minimum uncertainity in the frequency of a photon is delnu= 1.59e+07 sec**-1\n",
- "energy of an electron is E= 3.16e-12 J\n"
- ]
- }
- ],
- "prompt_number": 15
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Ex7.22:pg-223"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "#to calculate probability of finding the particle\n",
- "a=25*10**-10#width in angstrom \n",
- "#wave function of the particle is chi(x)=math.sqrt(2/a)*math.sin(n*math.pi*x/a),for the particle in the least energy state n=1\n",
- "chix=math.sqrt(2/a)*math.sin(math.pi*(a/2)/a)\n",
- "delx=5*10**-10 #interval in angstrom\n",
- "P=delx*chix**2\n",
- "print \"probability of finding the particle is P=\",P,\"unitless\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "probability of finding the particle is P= 0.4 unitless\n"
- ]
- }
- ],
- "prompt_number": 9
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Ex7.24:pg-224"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "#to calculate energy of an electron\n",
- "n=1 #least energy of the particle \n",
- "h=6.63*10**-34 #planck's constant in Js\n",
- "m=9.11*10**-31 #mass of electron in kg\n",
- "a=10**-10 #width in angstrom\n",
- "E=(n**2)*(h**2)/(8*m*(1.602*10**-19)*a**2)\n",
- "print \"energy of an electron moving in one dimension in an infinitely high potential box is E=\",round(E,2),\"eV\"\n",
- "#the answer is given wrong in the book E=5.68 eV\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "energy of an electron moving in one dimension in an infinitely high potential box is E= 37.65 eV\n"
- ]
- }
- ],
- "prompt_number": 16
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Ex7.26:pg-225"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "#to calculate probability\n",
- "x1=0.45 #x1=0.45*L\n",
- "x2=0.55 #x2=0.55*L\n",
- "n=1 #for ground state \n",
- "#formula is P=integrate('(2/L)*math.sin(n*math.pi*x)**2),'x',x1,x2)\n",
- "from scipy.integrate import quad\n",
- "def integrand(x):\n",
- " return 2*(math.sin(n*math.pi*x)**2)\n",
- "P1 ,er=quad(integrand,x1,x2)\n",
- "\n",
- "print \"P1=\",round(P1,3),\"unitless\"\n",
- "probability1=P1*100\n",
- "print \"probability for the ground states is probability1 =\",round(probability1,1),\"%\"\n",
- "n=2 #for first excited state\n",
- "P2, er=quad(integrand,x1,x2)\n",
- "print \"P2=\",round(P2,4),\"unitless\"\n",
- "probability2=P2*100 \n",
- "print \"probability for first excited states is probability2=\",round(probability2,2),\"%\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "P1= 0.198 unitless\n",
- "probability for the ground states is probability1 = 19.8 %\n",
- "P2= 0.0065 unitless\n",
- "probability for first excited states is probability2= 0.65 %\n"
- ]
- }
- ],
- "prompt_number": 5
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Ex7.28:pg-226"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "#to calculate energy of a neutron\n",
- "#consider nucleus as a cubical box of size 10**-14m\n",
- "#x=y=z=a=10**-14=l\n",
- "#for neutron to be in the lowest energy state nx=ny=nz=1\n",
- "#formula is E=(math.pi**2*h**2/8*math.pi**2*m)*((nx/lx)**2+(ny/ly)**2+(nz/lz)**2)\n",
- "h=6.626*10**-34 #planck's constant in Js\n",
- "m=1.6*10**-27 #mass in kg\n",
- "l=10**-14 #in m\n",
- "E=(math.pi**2)*(h**2)*3/(4*(math.pi**2)*2*m*(1.6*10**-19)*l**2)\n",
- "print \"lowest energy of a neutron is E=\",round(E/(1e6),2),\"MeV\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "lowest energy of a neutron is E= 6.43 MeV\n"
- ]
- }
- ],
- "prompt_number": 13
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Ex7.29:pg-226"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "#to calculate zero point energy of a linear harmonic oscillator\n",
- "h=6.63*10**-34 #planck's constant in Js\n",
- "nu=50 #frequency in Hz\n",
- "zeropointenergy=(h*nu)/2\n",
- "print \"zeropointenergy=\",\"{:.2e}\".format(zeropointenergy),\"J\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "zeropointenergy= 1.66e-32 J\n"
- ]
- }
- ],
- "prompt_number": 21
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Ex7.30:pg-226"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "#to calculate zero point energy\n",
- "nu=1 #frequency in Hz\n",
- "h=6.63*10**-34 #planck's constant in Js\n",
- "zeropointenergy=(h*nu)/2\n",
- "print \"zeropointenergy=\",zeropointenergy,\"J\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "zeropointenergy= 3.315e-34 J\n"
- ]
- }
- ],
- "prompt_number": 14
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Ex7.31:pg-226"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "#to calculate frequency of vibration\n",
- "En=0.1*1.6*10**-19 #energy of a linear harmonic oscillator in eV\n",
- "n=3.0 #third excited state\n",
- "h=6.63*10**-34 #planck's constant\n",
- "nu=En/((n+(1/2.0))*h)\n",
- "print \"the frequency of vibration is nu=\",round(nu,-9),\"Hz\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "the frequency of vibration is nu= 6.895e+12 Hz\n"
- ]
- }
- ],
- "prompt_number": 16
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Engineering_Physics/chapter8.ipynb b/Engineering_Physics/chapter8.ipynb deleted file mode 100755 index 106ed7f9..00000000 --- a/Engineering_Physics/chapter8.ipynb +++ /dev/null @@ -1,620 +0,0 @@ -{
- "metadata": {
- "name": "",
- "signature": "sha256:2b8beebd2e19262e31711f6c9b2785f35ddc221da552f5b3fe2ed984a495062e"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "heading",
- "level": 1,
- "metadata": {},
- "source": [
- "Chapter8:X-RAY"
- ]
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Ex8.1:pg-240"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "#to calculate value of planck's constant\n",
- "e=1.6*10**-19 #in C\n",
- "V=100*10**3 #voltage in KV\n",
- "c=3*10**8 #light speed in m/s\n",
- "lamdamin=12.35*10**-12 #wavelength in m\n",
- "h=e*V*lamdamin/c\n",
- "print \"the value of plancks constant is h=\",\"{:.2e}\".format(h),\"J-s\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "the value of plancks constant is h= 6.59e-34 J-s\n"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Ex8.2:pg-240"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "#to calculate maximum frequency\n",
- "h=6.6*10**-34 #planck's constant in J-s\n",
- "c=3.0*10**8 #light speed in m/s\n",
- "Ve=50000 #accelerating potential in V\n",
- "lamdamin=h*c/Ve #wavelength in m\n",
- "numax=c/lamdamin\n",
- "print \"maximum frequency present in the radiation from an X-ray tube is numax=\",\"{:.2e}\".format(numax),\"Hz\"\n",
- "#answer is given in thec book is incorrect =1.2*10**19 Hz\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "maximum frequency present in the radiation from an X-ray tube is numax= 7.58e+37 Hz\n"
- ]
- }
- ],
- "prompt_number": 2
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Ex8.3:pg-240"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "#to calculate number of electrons \n",
- "I=2*10**-3 #current in mA\n",
- "e=1.6*10**-19 \n",
- "n=I/e\n",
- "print \"number of electrons striking the target per second is n=\",n,\"unitless\"\n",
- "#to calculate speed\n",
- "m=9.1*10**-31 #mass of electron in kg\n",
- "V=12.4*10**3 #potential difference in V\n",
- "v=math.sqrt(2*V*e/m)\n",
- "print \"the speed with which electrons strike the target is v=\",\"{:.1e}\".format(v),\"m/s\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "number of electrons striking the target per second is n= 1.25e+16 unitless\n",
- "the speed with which electrons strike the target is v= 6.6e+07 m/s\n"
- ]
- }
- ],
- "prompt_number": 3
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Ex8.4:pg-240"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "#to calculate wavelength\n",
- "n=2 #second order for longest wavelength\n",
- "d=2.82*10**-10 # spacing in angstrom\n",
- "sintheta=1 \n",
- "lamdamax=2*d*sintheta/n\n",
- "print \"the longest wavelength that can be analysed by a rock salt crystal is lamdamax=\",lamdamax,\"m\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "the longest wavelength that can be analysed by a rock salt crystal is lamdamax= 2.82e-10 m\n"
- ]
- }
- ],
- "prompt_number": 5
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Ex8.5:pg-241"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "#to calculate spacing of the crystal\n",
- "h=6.62*10**-34 #planck's constant in J-s\n",
- "m=9.1*10**-31 #mass of electron in kg\n",
- "V=344 #voltage in V\n",
- "e=1.6*10**-19\n",
- "lamda=h/math.sqrt(2*m*e*V) #wavelength in m\n",
- "#according to Bragg's law\n",
- "n=1\n",
- "#formula is 2*d*sintheta=n*lamda\n",
- "d=n*lamda/(2*math.sin(math.pi/6))\n",
- "print \"the spacing of the crystal is d=\",round(d/1e-10,2),\"angstrom\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "the spacing of the crystal is d= 0.66 angstrom\n"
- ]
- }
- ],
- "prompt_number": 5
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Ex8.6:pg-241"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "#to calculate wavelength of Kalpha line for an atom\n",
- "R=1.1*10**5\n",
- "z=92\n",
- "#Ka line is emitted when electron jumps from l shell(n2=2) to k shell(n1=1)\n",
- "#formula is 1/alphaa=R*(z-b)*((1/n1**2)-(1/n2)**2)\n",
- "alphaa=4/(3*R*(z-1)**2)\n",
- "print \"wavelength of Kalpha line for an atom is alphaa=\",\"{:.3e}\".format(alphaa),\"cm\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "wavelength of Kalpha line for an atom is alphaa= 1.464e-09 cm\n"
- ]
- }
- ],
- "prompt_number": 7
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Ex8.7:pg-241"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "#to calculate thickness\n",
- "#mass absorption coefficient mum of an absorber is related with linear absorption coefficient mu and density of the material rho is given by\n",
- "#mu=rho*mum=2.7*0.6=1.62 cm**-1\n",
- "mu=1.62\n",
- "#if initial intensity Io of the X-ray beam is reduced to I in traversing a distance x in absorber I=Io*e**-mu*x\n",
- "#where I/Io=20\n",
- "#put above values in the below equation , we get\n",
- "x=(2.3026*(math.log(20)/math.log(10)))/1.62\n",
- "print \"thickness is x=\",round(x,2),\"cm\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "thickness is x= 1.85 cm\n"
- ]
- }
- ],
- "prompt_number": 8
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Ex8.8:pg-242"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "#to calculate atomic number of the element\n",
- "#equation for balmer series in hydrogen spectrum is 1/lamda=R*((1/2**2)-(1/n**2))\n",
- "#for series limit n=infinity ,R=4/lamdainfinity i.e. R=4/364.6nm\n",
- "#X-ray wavelength of K series is 1/lamda=R*(z-1)**2*((1/1**2)-(1/n**2))\n",
- "lamda=0.1*10**-9\n",
- "R=4/(364.6*10**-9)\n",
- "#for n=infinity ,minimum wavelength of k series is given by\n",
- "z=math.sqrt(1/(lamda*R))+1\n",
- "print \"atomic number is z=\",int(z),\"unitless\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "atomic number is z= 31 unitless\n"
- ]
- }
- ],
- "prompt_number": 9
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Ex8.9:pg-242"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "#to calculate wavelength\n",
- "d=1.87*10**-10 #spacing in angstrom\n",
- "n=2 \n",
- "#formula is lamda=2*d*sintheta/n\n",
- "lamda=2*d*math.sin(math.pi/6)/n\n",
- "print \"the wavelength of X-rays is lamda=\",round(lamda/1e-10,3),\"angstrom\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "the wavelength of X-rays is lamda= 0.935 angstrom\n"
- ]
- }
- ],
- "prompt_number": 11
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Ex8.10:pg-242"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "#to calculate wavelength of second X-ray beam\n",
- "#from bragg's law\n",
- "#lamda=(d*math.sin(math.pi/3))/n eq(1)\n",
- "#it is given that,theta=60,n=3,lamda=1.97 angstrom\n",
- "#from eq(1) we get,2*d*sin60degree=3*0.97 eq(2)\n",
- "#let lamda' be the second X-ray beam \n",
- "#we get 2*d'*sin theta'=n'*lamda' eq(3)\n",
- "#from eq(2) and eq(3),we get\n",
- "lamda1=math.sin(math.pi/6)*3*0.97/math.sin(math.pi/3) #where lamda1=lamda'\n",
- "print \"wavelength of X-ray is lamda1=\",round(lamda1,2),\"angstrom\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "wavelength of X-ray is lamda1= 1.68 angstrom\n"
- ]
- }
- ],
- "prompt_number": 12
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Ex8.11:pg-243"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "#to calculate wavelength\n",
- "d=2.82*10**-10 #spacing in m\n",
- "n=1 \n",
- "lamda=2*d*math.sin(10*math.pi/180)/n\n",
- "print \"wavelength of X-ray is lamda=\",round(lamda/1e-10,3),\"angstrom\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "wavelength of X-ray is lamda= 0.979 angstrom\n"
- ]
- }
- ],
- "prompt_number": 13
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Ex8.12:pg-243"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "#deduce possible spacing of the set of planes\n",
- "#for first order , 2*d*sintheta1=1*lamda...eq(1)\n",
- "#for second order ,2*d*sintheta2=2*lamda..eq(2)\n",
- "#for third order, 2*d*sintheta3=3*lamda......eq(3)\n",
- "#for fourth order, 2*d*sintheta4=4*lamda..............eq(4)\n",
- "#divide eq(2) by eq(1),we get sintheta2=2*sintheta1\n",
- "#similarly,sintheta3=3*sintheta1,sintheta4=4*sintheta1\n",
- "lamda=1.32*10**-10\n",
- "sintheta1=0.1650\n",
- "d1=lamda/(2*sintheta1)#for first order n=1,d1=d/n\n",
- "d2=lamda/(2*2*sintheta1) #for second order n=2,d2=d/n\n",
- "d3=lamda/(2*3*sintheta1) #for third order n=3,d3=d/n\n",
- "d4=lamda/(2*4*sintheta1) #for fourth order n=4,d4=d/n\n",
- "print \"d1=\",d1,\"m\"\n",
- "print \"d2=\",d2,\"m\"\n",
- "print \"d3=\",round(d3,2),\"m\"\n",
- "print \"d4=\",d4,\"m\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "d1= 4e-10 m\n",
- "d2= 2e-10 m\n",
- "d3= 0.0 m\n",
- "d4= 1e-10 m\n"
- ]
- }
- ],
- "prompt_number": 14
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Ex8.13:pg-248"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "#to calculate compton shift and wavelength\n",
- "h=6.63*10**-34 #planck's constant in J-s\n",
- "m0=9.11*10**-31 #mass of electron\n",
- "c=3*10**8 #light speed in m/s\n",
- "dellamda=h*(1-(1/math.sqrt(2)))/(m0*c)\n",
- "lamda0=2*10**-10\n",
- "lamda=dellamda+lamda0\n",
- "print \"compton shift is dellamda=\",round(dellamda/1e-10,4),\"angstrom\"\n",
- "print \"wavelength of the scattered X-rays is lamda=\",round(lamda/1e-10,4),\"angstrom\"\n",
- "#to calculate fraction of energy lost by the photon in the collision\n",
- "#energy lost =E0-E/E0=(hc/lamda0)-(hc/lamda)/(ha/lamda0)\n",
- "#we get,\n",
- "energylost=dellamda/lamda\n",
- "print \"energylost =\",round(energylost,5),\"unitless\" \n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "compton shift is dellamda= 0.0071 angstrom\n",
- "wavelength of the scattered X-rays is lamda= 2.0071 angstrom\n",
- "energylost = 0.00354 unitless\n"
- ]
- }
- ],
- "prompt_number": 18
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Ex8.14:pg-249"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "#to calculate wavelength and energy\n",
- "#formula is lamda'-lamda=h*(1-cos phi)/(m0*c),where phi=90 degree, lamda'=2lamda ---------------eq(1)\n",
- "#dellamda=2lamda-lamda=lamda ----------------------------eq(2)\n",
- "h=6.62*10**-34 #planck's constant\n",
- "c=3*10**8 #light speed in m.s\n",
- "m0=9*10**-31 #mass of electron in kg\n",
- "#from eq(1) and eq(2) ,we get\n",
- "lamda=h/(m0*c)\n",
- "print \"wavelength is lamda=\",round(lamda/1e-10,4),\"angstrom\"\n",
- "E=h*c/lamda\n",
- "print \"energy of the incident photon is E=\",E,\"J\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "wavelength is lamda= 0.0245 angstrom\n",
- "energy of the incident photon is E= 8.1e-14 J\n"
- ]
- }
- ],
- "prompt_number": 19
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Ex8.15:pg-249"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "#to calculate wavelength of radiation and direction of emission\n",
- "h=6.6*10**-34 #planck's constant in J-s\n",
- "c=3*10**8 #speed of light in m/s\n",
- "energy=510*10**3 #energy of photon in eV\n",
- "lamda=h*c/(energy*1.6*10**-19)\n",
- "mo=9.1*10**-31 #mass of electron in Kg\n",
- "lamda1=lamda+h*(1-math.cos(math.pi/2))/(mo*c)\n",
- "print \"wavelength of radiation is lamda1=\",\"{:.2e}\".format(lamda1),\"m\"\n",
- "theta=math.degrees(math.atan(lamda*math.sin(math.pi/2)/(lamda1-lamda*math.cos(math.pi/2))))\n",
- "print\"direction of emission of electron is theta=\",round(theta,2),\"degree\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "wavelength of radiation is lamda1= 4.84e-12 m\n",
- "direction of emission of electron is theta= 26.61 degree\n"
- ]
- }
- ],
- "prompt_number": 21
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Ex8.16:pg-249"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "#to calculate wavelength of two X-rays\n",
- "h=6.6*10**-34 #planck's constant in J-s\n",
- "c=3.0*10.0**8 #light speed in m/s\n",
- "mo=9.1*10**-31 #mass of electron in kg\n",
- "lamda=10.0*10**-12 #wavelength in pm\n",
- "lamda1=lamda+((h/(mo*c))*(1-(1/math.sqrt(2))))\n",
- "print \"wavelength of two X-rays is lamda1=\",round(lamda1*(1e12),1),\"picometer\"\n",
- "#to calculate maximum wavelength\n",
- "lamda2=lamda+((2*h)/(mo*c))\n",
- "print \"maximum wavelength present in the scattered X-rays is lamda2=\",round(lamda2*(1e12),2),\"picometer\"\n",
- "#to calculate maximum kinetic energy \n",
- "Kmax=(h*c)*((1/lamda)-(1/lamda2))/(1.6*10**-19)\n",
- "print \"maximum kinetic energy of the recoil electrons is Kmax=\",round(Kmax/1000.0,1),\"KeV\"\n",
- "\n",
- "# the answer is slightly different due to approximation\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "wavelength of two X-rays is lamda1= 10.7 picometer\n",
- "maximum wavelength present in the scattered X-rays is lamda2= 14.84 picometer\n",
- "maximum kinetic energy of the recoil electrons is Kmax= 40.3 KeV\n"
- ]
- }
- ],
- "prompt_number": 6
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Engineering_Physics/chapter9.ipynb b/Engineering_Physics/chapter9.ipynb deleted file mode 100755 index d92ff5fa..00000000 --- a/Engineering_Physics/chapter9.ipynb +++ /dev/null @@ -1,130 +0,0 @@ -{
- "metadata": {
- "name": "",
- "signature": "sha256:5dfcc3b8a36b2e69e76da78cbe3cbbcd618d029b3ed8c322ee0560bc514169c3"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "heading",
- "level": 1,
- "metadata": {},
- "source": [
- "Chapter9:DIELECTRIC PROPERTIESOF MATERIALS"
- ]
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Ex9.1:pg-269"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "#to calculate dielectric constant of the liquid\n",
- "#capacitance of the air filled dielectric Cair=Q/Vo-----------eq(1)\n",
- "#when dielectric is filled between the plates ,Cliquid=Q/V\n",
- "#then Cliquid=epsilonr*Q/Vo------eq(2)\n",
- "Vo=60\n",
- " #voltage in volts\n",
- "V=30\n",
- "#from eq(1) and eq(2),we get\n",
- "epsilon0=Vo/V\n",
- "print \"the dielectric constant of the liquid is epsilon0=\",epsilon0,\"unitless\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "the dielectric constant of the liquid is epsilon0= 2 unitless\n"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Ex9.2:pg-269"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "#to calculate charge on the capacitance\n",
- "epsilon0=8.854*10**-12 #permittivity\n",
- "epsilonr=6 #relative permittivity\n",
- "V=100 #voltage in volts\n",
- "d=1.5*10**-3 #distance in m\n",
- "A=4*10**-4#area in m**2\n",
- "Q=epsilon0*epsilonr*A*V/d\n",
- "print \"the charge on the capacitance is Q=\",\"{:.2e}\".format(Q),\"Coulomb\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "the charge on the capacitance is Q= 1.42e-09 Coulomb\n"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Ex9.3:pg-269"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "#to calculate voltage\n",
- "A=6.50*10**-4 #area in m**2\n",
- "Q=2*10**-10 #charge in C\n",
- "d=4*10**-3 #plate separation in m\n",
- "epsilon0=8.854*10**-12\n",
- "epsilonr=3.5 #dielectric constant\n",
- "V=Q*d/(epsilon0*epsilonr*A)\n",
- "print \"the resultant voltage across the capacitor is V=\",round(V,1),\"volt\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "the resultant voltage across the capacitor is V= 39.7 volt\n"
- ]
- }
- ],
- "prompt_number": 2
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Engineering_Physics/screenshots/1.png b/Engineering_Physics/screenshots/1.png Binary files differdeleted file mode 100755 index 4282275d..00000000 --- a/Engineering_Physics/screenshots/1.png +++ /dev/null diff --git a/Engineering_Physics/screenshots/11.png b/Engineering_Physics/screenshots/11.png Binary files differdeleted file mode 100755 index 0fc56fcd..00000000 --- a/Engineering_Physics/screenshots/11.png +++ /dev/null diff --git a/Engineering_Physics/screenshots/1_converted.png b/Engineering_Physics/screenshots/1_converted.png Binary files differdeleted file mode 100755 index 6f902c6e..00000000 --- a/Engineering_Physics/screenshots/1_converted.png +++ /dev/null diff --git a/Engineering_Physics/screenshots/2.png b/Engineering_Physics/screenshots/2.png Binary files differdeleted file mode 100755 index 9e1e8cf6..00000000 --- a/Engineering_Physics/screenshots/2.png +++ /dev/null diff --git a/Engineering_Physics/screenshots/22.png b/Engineering_Physics/screenshots/22.png Binary files differdeleted file mode 100755 index 3a440c46..00000000 --- a/Engineering_Physics/screenshots/22.png +++ /dev/null diff --git a/Engineering_Physics/screenshots/2_converted.png b/Engineering_Physics/screenshots/2_converted.png Binary files differdeleted file mode 100755 index 5b3697bc..00000000 --- a/Engineering_Physics/screenshots/2_converted.png +++ /dev/null diff --git a/Engineering_Physics/screenshots/3.png b/Engineering_Physics/screenshots/3.png Binary files differdeleted file mode 100755 index 662e6b45..00000000 --- a/Engineering_Physics/screenshots/3.png +++ /dev/null diff --git a/Engineering_Physics/screenshots/33.png b/Engineering_Physics/screenshots/33.png Binary files differdeleted file mode 100755 index fc148b77..00000000 --- a/Engineering_Physics/screenshots/33.png +++ /dev/null diff --git a/Engineering_Physics/screenshots/3_converted.png b/Engineering_Physics/screenshots/3_converted.png Binary files differdeleted file mode 100755 index 47f7a27c..00000000 --- a/Engineering_Physics/screenshots/3_converted.png +++ /dev/null diff --git a/Engineering_Physics/screenshots/U3.png b/Engineering_Physics/screenshots/U3.png Binary files differdeleted file mode 100755 index fc148b77..00000000 --- a/Engineering_Physics/screenshots/U3.png +++ /dev/null diff --git a/Engineering_Physics/screenshots/Untitled.png b/Engineering_Physics/screenshots/Untitled.png Binary files differdeleted file mode 100755 index ef585cf3..00000000 --- a/Engineering_Physics/screenshots/Untitled.png +++ /dev/null diff --git a/Engineering_Physics/screenshots/Untitled2.png b/Engineering_Physics/screenshots/Untitled2.png Binary files differdeleted file mode 100755 index 3a440c46..00000000 --- a/Engineering_Physics/screenshots/Untitled2.png +++ /dev/null diff --git a/Engineering_Physics/screenshots/area_of_the_spot_on_the_moon.png b/Engineering_Physics/screenshots/area_of_the_spot_on_the_moon.png Binary files differdeleted file mode 100755 index d72ce215..00000000 --- a/Engineering_Physics/screenshots/area_of_the_spot_on_the_moon.png +++ /dev/null diff --git a/Engineering_Physics/screenshots/electric_flux.png b/Engineering_Physics/screenshots/electric_flux.png Binary files differdeleted file mode 100755 index ba949261..00000000 --- a/Engineering_Physics/screenshots/electric_flux.png +++ /dev/null diff --git a/Engineering_Physics/screenshots/velocity_of_rocket.png b/Engineering_Physics/screenshots/velocity_of_rocket.png Binary files differdeleted file mode 100755 index d86b5e4c..00000000 --- a/Engineering_Physics/screenshots/velocity_of_rocket.png +++ /dev/null |
