diff options
Diffstat (limited to 'Applied_Physics-II')
| -rwxr-xr-x | Applied_Physics-II/chapter1.ipynb | 1353 | ||||
| -rwxr-xr-x | Applied_Physics-II/chapter2.ipynb | 1480 | ||||
| -rwxr-xr-x | Applied_Physics-II/chapter3.ipynb | 836 | ||||
| -rwxr-xr-x | Applied_Physics-II/chapter4.ipynb | 163 | ||||
| -rwxr-xr-x | Applied_Physics-II/chapter5.ipynb | 1166 | ||||
| -rwxr-xr-x | Applied_Physics-II/chapter6.ipynb | 1183 | ||||
| -rwxr-xr-x | Applied_Physics-II/screenshots/Ex2_2.png | bin | 0 -> 28955 bytes | |||
| -rwxr-xr-x | Applied_Physics-II/screenshots/Ex2_2_1.png | bin | 0 -> 28955 bytes | |||
| -rwxr-xr-x | Applied_Physics-II/screenshots/Ex2_2_2.png | bin | 0 -> 28955 bytes |
9 files changed, 6181 insertions, 0 deletions
diff --git a/Applied_Physics-II/chapter1.ipynb b/Applied_Physics-II/chapter1.ipynb new file mode 100755 index 00000000..c4b1a860 --- /dev/null +++ b/Applied_Physics-II/chapter1.ipynb @@ -0,0 +1,1353 @@ +{ + "metadata": { + "celltoolbar": "Raw Cell Format", + "name": "", + "signature": "sha256:fad8e22fb99cc3e157ab8315172e2ff6ddae35bb6f49be764a49d2d1d3f70fcc" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Chapter 1: Interference" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 1.1, Page number 1-16" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Given Data:\n", + "i=45*math.pi/180 #angle of incidence\n", + "u=1.33 #Refractive index of a soap film\n", + "lamda=5.896*10**-7 #wavelength of required yellow light\n", + "\n", + "#Calculations:\n", + "#u=sin i/sin r #Snell's law .So,\n", + "r=math.asin(math.sin(i)/u) #angle of reflection\n", + "\n", + "#Now, condition for bright fringe is\n", + "#2ut*cos r=(2n-1)lamda/2\n", + "#Here n=1\n", + "t=lamda/(2*2*u*math.cos(r)) #minimum thickness of film at which light will appear bright yellow\n", + "print\"Minimum thickness of film at which light will appear bright yellow of required wavelength is =\",t,\"m\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Minimum thickness of film at which light will appear bright yellow of required wavelength is = 1.30853030399e-07 m\n" + ] + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 1.2, Page number 1-16" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Given Data:\n", + "theta=40./3600*math.pi/180 #angle of wedge in radians\n", + "B=0.12*10**-2 #fringe spacing\n", + "\n", + "#Calculations:\n", + "#We know, B=lam/(2*u*theta). Here u=1\n", + "lamda=2*B*theta #wavelength of light used\n", + "print\"Wavelength of light used is =\",lamda,\"m\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Wavelength of light used is = 4.65421133865e-07 m\n" + ] + } + ], + "prompt_number": 4 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 1.3, Page number 1-17" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Given Data:\n", + "i=30*math.pi/180 #angle of incidence\n", + "u=1.46 #Refractive index of a oil\n", + "lamda=5.890*10**-7 #wavelength of required yellow light\n", + "n=8 #eighth dark band\n", + "\n", + "#Calculations:\n", + "#u=sin i/sin r #Snell's law .So,\n", + "r=math.asin(math.sin(i)/u) #angle of reflection\n", + "\n", + "#Now, condition for dark fringe is\n", + "#2ut*cos r=n*lamda\n", + "t=n*lamda/(2*u*math.cos(r)) #thickness of film\n", + "print\"Thickness of the film is =\",t,\"m\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Thickness of the film is = 1.71755887917e-06 m\n" + ] + } + ], + "prompt_number": 6 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 1.4, Page number 1-17" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Given Data:\n", + "\n", + "B=0.1*10**-2 #fringe spacing\n", + "lamda=5.893*10**-7 #Wavelength of light\n", + "u=1.52 #Refractive index of wedge\n", + "\n", + "#Calculations:\n", + "#We know, B=lamda/(2*u*theta). Here u=1\n", + "theta1=lamda/(2*u*B) #angle of wedge in radians\n", + "theta=theta1*3600*180/math.pi #angle of wedge in seconds\n", + "print\"Angle of wedge is =\",theta,\"seconds of an arc\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Angle of wedge is = 39.9841612899 seconds of an arc\n" + ] + } + ], + "prompt_number": 7 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 1.5, Page number 1-18" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Given Data:\n", + "t=0.2/(100)**2*10**-2 #thickness of film in meter\n", + "lamda=5.5*10**-7 #wavelength of light in meter\n", + "r=0 #normal incidence\n", + "n=1 #first band\n", + "\n", + "#Calculations:\n", + "\n", + "#Condition for dark fringe is\n", + "#2ut*cos r =n*lamda\n", + "u=n*lamda/(2*t*math.cos(r)) #Refractive index of a oil\n", + "print\"Refractive index of a oil is =\",u\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Refractive index of a oil is = 1.375\n" + ] + } + ], + "prompt_number": 9 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 1.6, Page number 1-18" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Given Data:\n", + "lamda=5.893*10**-7 #Wavelength of light\n", + "u=1.42 #Refractive index of a soap film\n", + "r=0 #normal incidence\n", + "n=1 #first band\n", + "\n", + "#Calculations:\n", + "\n", + "#(i)\n", + "#Condition for dark fringe is\n", + "#2ut*cos r=n*lamda\n", + "t1=n*lamda/(2*u*cos(r)) #thickness of film for dark black fringe\n", + "print\"Thickness of the film for dark black fringe is =\",t1,\"m\"\n", + "\n", + "#(ii)\n", + "#Now, condition for bright fringe is\n", + "#2ut*cos r=(2n-1)lamda/2\n", + "t2=lamda/(2*2*u*math.cos(r)) #Thickness of film for bright fringe\n", + "print\"Thickness of film for bright fringe is =\",t2,\"m\"\n", + "\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Thickness of the film for dark black fringe is = 2.075e-07 m\n", + "Thickness of film for bright fringe is = 1.0375e-07 m\n" + ] + } + ], + "prompt_number": 4 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 1.7, Page number 1-19" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math \n", + "#Given Data:\n", + "i=30*math.pi/180 #angle of incidence\n", + "u=1.43 #Refractive index of a soap film\n", + "lamda=6*10**-7 #wavelength of light\n", + "n=1 #For minimum thickness\n", + "\n", + "#Calculations:\n", + "#u=sin i/sin r #Snell's law .So,\n", + "r=(math.asin(math.sin(i)/u)) #angle of reflection\n", + "\n", + "#Now, condition of minima in transmitted system is\n", + "#2ut*cos(r)=(2n-1)lamda/2\n", + "t=lamda/(2*2*u*math.cos(r)) #minimum thickness of film\n", + "print\"Minimum thickness of film is \",t,\"m\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Minimum thickness of film is 1.11962124395e-07 m\n" + ] + } + ], + "prompt_number": 49 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 1.8, Page number 1-19" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Given Data:\n", + "\n", + "lamda = 5893*10**-10 #Wavelength of light\n", + "theta = 1 #assuming value of theta\n", + "\n", + "#We know, B=lamda/(2*u*theta). Here u=1\n", + "B = lamda/(2*theta) #fringe spacing\n", + "n=20 #interference fringes\n", + "\n", + "#Calculations:\n", + "#t=n*B*tan(theta)\n", + "t = 20*B*theta #Thickness of wire\n", + "print\"Thickness of wire is =\",t,\"m\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Thickness of wire is = 5.893e-06 m\n" + ] + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 1.9, Page number 1-20" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Given Data:\n", + "u1=1.3 #Refractive index of oil\n", + "u2=1.5 #Refractive index of glass\n", + "lamda1=7*10**-7 #Wavelength of light\n", + "lamda2=5*10**-7 #Wavelength of light\n", + "\n", + "#Calculations:\n", + "\n", + "#for finding value of n, solve:\n", + "#(2n+1)*lamda1/2=(2(n+1)+1)*lamda2/2\n", + "#We get,n=2\n", + "n=2\n", + "\n", + "toil=(2*n+1)*lamda1/(2*u1*2) #thickness of oil layer\n", + "print\"Thickness of oil layer is =\",toil,\"m\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Thickness of oil layer is = 6.73076923077e-07 m\n" + ] + } + ], + "prompt_number": 5 + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "Example 1.10, Page number 1-21" + ], + "language": "python", + "metadata": {}, + "outputs": [] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Given Data:\n", + "u1=1.2 #Refractive index of drop of oil\n", + "u2=1.33 #Refractive index of water\n", + "lamda=4.8*10**-7 #wavelength of light\n", + "n=3 #order\n", + "r=0 #normal incidence,so r=0\n", + "\n", + "#Calculations:\n", + "t=n*lamda/(2*u1) #Thickness of oil drop\n", + "print\"Thickness of oil drop is =\",t,\"m\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Thickness of oil drop is = 6e-07 m\n" + ] + } + ], + "prompt_number": 6 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 1.11, Page number 1-22" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Given Data:\n", + "i=math.asin(4/5) #angle of incidence\n", + "u=4/3 #Refractive index of a soap film\n", + "lamda1=6.1*10**-7 #wavelength of light\n", + "lamda2=6*10**-7 #wavelength of light\n", + "\n", + "#Calculations:\n", + "#u=sin i/sin r #Snell's law .So,\n", + "r=math.asin(math.sin(i)/u) #angle of reflection\n", + "\n", + "#Now, condition for dark band is\n", + "#2ut*cos r=n*lamda\n", + "#for consecutive bands, n=lamda2/(lamda1-lamda2). hence\n", + "\n", + "t=lamda2*lamda1/((lamda1-lamda2)*2*u*math.sqrt(1-(math.sin(i)/u)**2)) #thickness of film\n", + "print\"Thickness of the film is =\",t,\"m\"\n", + "\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Thickness of the film is = 1.83e-05 m\n" + ] + } + ], + "prompt_number": 55 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 1.12, Page number 1-40" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Given Data:\n", + "n=10 #10th dark ring\n", + "Dn=0.5*10**-2 #Diameter of ring\n", + "lamda=6*10**-7 #wavelength of light\n", + "\n", + "#Calculations:\n", + "#As Dn^2=4*n*R*lamda\n", + "R=Dn**2/(4*n*lamda) #Radius of curvature of the lens\n", + "print\"Radius of curvature of the lens is =\",R,\"m\"\n", + "\n", + "t=Dn**2/(8*R) #thickness of air film\n", + "print\"Thickness of air film is =\",t,\"m\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Radius of curvature of the lens is = 1.04166666667 m\n", + "Thickness of air film is = 3e-06 m\n" + ] + } + ], + "prompt_number": 9 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 1.13, Page number 1-41" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Given Data:\n", + "\n", + "B=0.25*10**-2 #fringe spacing\n", + "lamda=5.5*10**-7 #Wavelength of light\n", + "u=1.4 #Refractive index of wedge\n", + "\n", + "#Calculations:\n", + "#We know, B=lamda/(2*u*theta).\n", + "theta1=lamda/(2*u*B) #angle of wedge in radians\n", + "theta=theta1*3600*180/math.pi #angle of wedge in seconds\n", + "print\"Angle of wedge is =\",theta,\"seconds\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Angle of wedge is = 16.2065204908 seconds\n" + ] + } + ], + "prompt_number": 13 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 1.14, Page number 1-41" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Given Data:\n", + "n=4 #4th dark ring\n", + "m=12 #m=n+p\n", + "D4=0.4*10**-2 #Diameter of 4th ring\n", + "D12=0.7*10**-2 #Diameter of 12th ring\n", + "\n", + "#Calculations:\n", + "\n", + "#(Dn+p)^2-Dn^2=4*p*lamda*R\n", + "#Solving, (D12^2-D4^2)/(D20^2-D4^2)\n", + "#We get above value =1/2. Hence\n", + "D20=math.sqrt(2*D12**2-D4**2) #Diameter of 20th ring\n", + "print\"Diameter of 20th ring is =\",D20,\"m\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Diameter of 20th ring is = 0.00905538513814 m\n" + ] + } + ], + "prompt_number": 14 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 1.15, Page number 1-42" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Given Data:\n", + "n=6 #6th bright ring\n", + "D6=0.31*10**-2 #Diameter of 6th ring\n", + "lamda=6*10**-7 #wavelength of light\n", + "R=1 #Radius of curvature\n", + "\n", + "#Calculations:\n", + "\n", + "#Diameter of nth bright ring is \n", + "#Dn^2=2(2n-1)*lamda*R/u. Hence\n", + "u=2*(2*n-1)*lamda*R/(D6)**2 #Refractive index of liquid\n", + "print\"Refractive index of liquid is =\",u\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Refractive index of liquid is = 1.37356919875\n" + ] + } + ], + "prompt_number": 15 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 1.16, Page number 1-42" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Given Data:\n", + "lamda=6*10**-7 #wavelength of light\n", + "k=0.125*10**-4 #k=D(n+1)^2-Dn^2.\n", + "u=1 #Refractive index of medium between lens and plate\n", + "#Calculations:\n", + "\n", + "#(i)\n", + "lamda1=4.5*10**-7 #new wavelength of light\n", + "#Difference between squres of diameters of successive rings is directly proportional to wavelength.So,\n", + "k1=lamda1/lamda*k #new Difference between squres of diameters of successive rings after changing wavelength\n", + "print\"New Difference between squres of diameters of successive rings after changing wavelength is =\",k1,\"m^2\"\n", + "\n", + "#(ii)\n", + "u2=1.33 #Refractive index of liquid introduced between lens and plate\n", + "#Difference between squres of diameters of successive rings is inversely proportional to Refractive index.so,\n", + "k2=u/u2*k #new Difference between squres of diameters of successive rings after changing refractive index\n", + "print\"New Difference between squres of diameters of successive rings after changing refrective index is =\",k2,\"m^2\"\n", + "\n", + "#(iii)\n", + "#Difference between squres of diameters of successive rings is directly proportional to Radius of curvature.So,\n", + "#after doubling radius of curvature,\n", + "k3=2*k #new Difference between squres of diameters of successive rings after doubling radius of curvature\n", + "print\"New Difference between squres of diameters of successive rings after doubling radius of curvature is =\",k3,\"m^2\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "New Difference between squres of diameters of successive rings after changing wavelength is = 9.375e-06 m^2\n", + "New Difference between squres of diameters of successive rings after changing refrective index is = 9.3984962406e-06 m^2\n", + "New Difference between squres of diameters of successive rings after doubling radius of curvature is = 2.5e-05 m^2\n" + ] + } + ], + "prompt_number": 16 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 1.17, Page number 1-43" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Given Data:\n", + "Dn=0.225*10**-2 #Diameter of nth ring\n", + "Dm=0.45*10**-2 #Diameter of (n+9)th ring\n", + "lamda=6*10**-7 #wavelength of light\n", + "R=0.9 #Radius of curvature\n", + "p=9\n", + "\n", + "#Calculations:\n", + "#(Dn+p)^2-Dn^2=4*p*lamda*R/u\n", + "u=4*p*lamda*R/((Dm)**2-Dn**2) #Refractive index of liquid\n", + "print\"Refractive index of liquid is =\",u\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Refractive index of liquid is = 1.28\n" + ] + } + ], + "prompt_number": 17 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 1.18, Page number 1-44" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Given Data:\n", + "D10=0.5*10**-2 #Diameter of 10th ring\n", + "lamda=5.5*10**-7 #wavelength of light\n", + "u=1.25 #Refractive index of liquid\n", + "\n", + "\n", + "#Calculations:\n", + "#As Dn^2=4*n*R*lamda/u\n", + "#Dn^2 is inversely proportional to refractive index.\n", + "D10n=D10/math.sqrt(u) #new diameter of 10th ring after changing medium between lens and plate\n", + "print\"new diameter of 10th ring after changing medium between lens and plate is =\",D10n,\"m\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "new diameter of 10th ring after changing medium between lens and plate is = 0.004472135955 m\n" + ] + } + ], + "prompt_number": 18 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 1.19, Page number 1-45" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Given Data:\n", + "D5=0.336*10**-2 #Diameter of 5th ring\n", + "D15=0.59*10**-2 #Diameter of 15th ring\n", + "lamda=5.89*10**-7 #wavelength of light\n", + "p=10 #n=5,n+p=15\n", + "\n", + "#Calculations:\n", + "#(Dn+p)^2-Dn^2=4*p*lamdaR/u\n", + "R=((D15)**2-D5**2)/(4*p*lamda) #Radius of curvature of the lens\n", + "print\"Radius of curvature of the lens is =\",R,\"m\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Radius of curvature of the lens is = 0.998319185059 m\n" + ] + } + ], + "prompt_number": 20 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 1.20, Page number 1-45" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Given Data:\n", + "n=10 #10th dark ring\n", + "D10=0.6*10**-2 #Diameter of ring\n", + "lamda=6*10**-7 #wavelength of light\n", + "u=4./3 #Refractive index of water\n", + "\n", + "\n", + "#Calculations:\n", + "#As Dn^2=4*n*R*lamda/u\n", + "R=(D10**2)*u/(4*n*lamda) #Radius of curvature of the lens\n", + "print\"Radius of curvature of the lens is =\",R,\"m\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Radius of curvature of the lens is = 2.0 m\n" + ] + } + ], + "prompt_number": 27 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 1.20.1, Page number 1-52" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Given Data:\n", + "i=45*math.pi/180 #angle of incidence\n", + "u=1.2 #Refractive index of a film\n", + "t=4*10**-7 #thickness of film\n", + "\n", + "#Calculations:\n", + "#u=sin i/sin r #Snell's law .So,\n", + "r=math.asin(math.sin(i)/u) #angle of reflection\n", + "\n", + "#Now, condition for dark fringe is\n", + "#2ut*cos r=n*lamda\n", + "lamda1=2*u*t*math.cos(r)/1 #n=1\n", + "print\"For n=1 wavelength is =\",lamda1,\"m\"\n", + "print\"This is in the visible spectrum and it will remain absent.\" \n", + "\n", + "lamda2=2*u*t*math.cos(r)/2 #n=2\n", + "print\"For n=2 wavelength is =\",lamda2,\"m\"\n", + "print\"This is not in the visible spectrum\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "For n=1 wavelength is = 7.75628777187e-07 m\n", + "This is in the visible spectrum and it will remain absent.\n", + "For n=2 wavelength is = 3.87814388593e-07 m\n", + "This is not in the visible spectrum\n" + ] + } + ], + "prompt_number": 30 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 1.20.2, Page number 1-53" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Given Data:\n", + "r=45*math.pi/180 #angle of refraction\n", + "u=1.45 #Refractive index of a medium\n", + "lamda=5.5*10**-7 #wavelength of required yellow light\n", + "n=1\n", + "\n", + "#Calculations:\n", + "\n", + "#Now, condition for dark fringe is\n", + "#2ut*cos r=n*lamda\n", + "t=n*lamda/(2*u*math.cos(r)) #thickness of thin medium\n", + "print\"Thickness of the thin medium is =\",t,\"m\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Thickness of the thin medium is = 2.68212917002e-07 m\n" + ] + } + ], + "prompt_number": 31 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 1.21, Page number 1-45" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Given Data:\n", + "u=1.33 #Refractive index of a soap film\n", + "r=0 #normal incidence\n", + "t=5*10**-7 #thickness of film\n", + "\n", + "#Calculations:\n", + "\n", + "#Now, condition for maxima is\n", + "#2ut*cos r=(2n-1)lamda/2\n", + "lamda1=4*u*t*math.cos(r)/(2*1-1) #n=1\n", + "print\"For n=1 wavelength is =\",lamda1,\"m\"\n", + "lamda2=4*u*t*math.cos(r)/(2*2-1) #n=2\n", + "print\"For n=2 wavelength is =\",lamda2,\"m\"\n", + "lamda3=4*u*t*math.cos(r)/(2*3-1) #n=3\n", + "print\"For n=3 wavelength is =\",lamda3,\"m\"\n", + "lamda4=4*u*t*math.cos(r)/(2*4-1) #n=4\n", + "print\"For n=4 wavelength is =\",lamda4,\"m\"\n", + "\n", + "print\"Out of these wavelengths wavelength for n=3 lies in the visible spectrum.\"\n", + "print\"Hence, wavelength for n=3 is the most reflected wavelength.\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "For n=1 wavelength is = 2.66e-06 m\n", + "For n=2 wavelength is = 8.86666666667e-07 m\n", + "For n=3 wavelength is = 5.32e-07 m\n", + "For n=4 wavelength is = 3.8e-07 m\n", + "Out of these wavelengths wavelength for n=3 lies in the visible spectrum.\n", + "Hence, wavelength for n=3 is the most reflected wavelength.\n" + ] + } + ], + "prompt_number": 32 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 1.22, Page number 1-46" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Given Data:\n", + "u=1.5 #Refractive index of a oil\n", + "lamda=5.88*10**-7 #wavelength of required yellow light\n", + "n=1 #for smallest thickness\n", + "r=60*math.pi/180 #angle of reflection\n", + "\n", + "#Calculations:\n", + "\n", + "#Now, condition for dark fringe is\n", + "#2ut*cos r=n*lamda\n", + "t=n*lamda/(2*u*math.cos(r)) #thickness of film\n", + "print\"Thickness of the film is =\",t,\"m\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Thickness of the film is = 3.92e-07 m\n" + ] + } + ], + "prompt_number": 33 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 1.23, Page number 1-46" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Given Data:\n", + "theta=20./3600*math.pi/180 #angle of wedge in radians\n", + "B=0.25*10**-2 #fringe spacing\n", + "u=1.4 #Refractive index of film\n", + "\n", + "#Calculations:\n", + "#We know, B=lamda/(2*u*theta).\n", + "lamda=2*B*theta*u #wavelength of light\n", + "print\"Wavelength of light is =\",lamda,\"m\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Wavelength of light is = 6.78739153553e-07 m\n" + ] + } + ], + "prompt_number": 35 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 1.24, Page number 1-47" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Given Data:\n", + "#Dn=2*D40\n", + "\n", + "#Calculations:\n", + "#As Dn^2 = 4*n*R*lamda/u and Dn^2 = 4*D40^2\n", + "#i.e. 4*n*R*lamda/u = 4*4*40*R*lamda/u .hence,\n", + "n=4*40 #order of the required ring\n", + "print\"Order of the dark ring which will have double the diameter of that of 40th ring is =\",n\n", + " \n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Order of the dark ring which will have double the diameter of that of 40th ring is = 160\n" + ] + } + ], + "prompt_number": 36 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 1.25, Page number 1-47" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Given Data:\n", + "lamda1=6*10**-7 #wavelength of light\n", + "lamda2=4.5*10**-7 #wavelength of light\n", + "R=0.9 #Radius of curvature\n", + "\n", + "#Calculations:\n", + "#As Dn^2=4*n*R*lamda.\n", + "#Dn^2=D(n+1)^2 for different wavelengths.we get,\n", + "n=lamda2/(lamda1-lamda2) #nth dark ring due to lam1 which coincides with (n+1)th dark ring due lamda2\n", + "D3=math.sqrt(4*n*R*lamda1) #diameter of 3rd dark ring for lamda1\n", + "print\"Diameter of 3rd dark ring for lam1 is =\",D3,\"m\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Diameter of 3rd dark ring for lam1 is = 0.00254558441227 m\n" + ] + } + ], + "prompt_number": 38 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 1.26, Page number 1-48" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Given Data:\n", + "i=45*math.pi/180 #angle of incidence\n", + "u=4./3 #Refractive index of soap film\n", + "lamda=5*10**-7 #wavelength of light\n", + "t=1.5*10**-6 #thickness of film\n", + "\n", + "#Calculations:\n", + "#u=sin i/sin r #Snell's law .So,\n", + "r=math.asin(math.sin(i)/u) #angle of reflection\n", + "\n", + "#Now, condition for dark band is\n", + "#2ut*cos r=n*lamda\n", + "n=2*u*t*math.cos(r)/lamda #order of band\n", + "print\"order of dark band is =\",n\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "order of dark band is = 6.78232998313\n" + ] + } + ], + "prompt_number": 39 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 1.27, Page number 1-49" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Given Data:\n", + "D5=0.336*10**-2 #Diameter of 5th ring\n", + "D15=0.59*10**-2 #Diameter of 15th ring\n", + "lamda=5.89*10**-7 #wavelength of light\n", + "p=10 #n=5,n+p=15\n", + "\n", + "#Calculations:\n", + "#(Dn+p)^2-Dn^2=4*p*lamda*R/u\n", + "R=((D15)**2-D5**2)/(4*p*lamda) #Radius of curvature of the lens\n", + "print\"Radius of curvature of the lens is =\",R,\"m\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Radius of curvature of the lens is = 0.998319185059 m\n" + ] + } + ], + "prompt_number": 40 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 1.29, Page number 1-50" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#As Dn^2=4*n*R*lamda.\n", + "#thus, Dn is directly proportional to sqaure root of n\n", + "D5=math.sqrt(5) #D5 is directly proportional to sqaure root of 5\n", + "D4=math.sqrt(4) #D4 is directly proportional to sqaure root of 4\n", + "k1=D5-D4\n", + "print\"Separation between D5 and D4 is directly proportional to =\",k1\n", + "\n", + "D80=math.sqrt(80) #D80 is directly proportional to sqaure root of 80\n", + "D79=math.sqrt(79) #D79 is directly proportional to sqaure root of 79\n", + "k2=D80-D79\n", + "print\"Separation between D80 and D79 is directly proportional to =\",k2\n", + "\n", + "print\"Thus, (D80-D79) < (D5-D4). Hence proved.\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Separation between D5 and D4 is directly proportional to = 0.2360679775\n", + "Separation between D80 and D79 is directly proportional to = 0.0560774926836\n", + "Thus, (D80-D79) < (D5-D4). Hence proved.\n" + ] + } + ], + "prompt_number": 41 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 1.30, Page number 1-51" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Given Data:\n", + "D5=0.336*10**-2 #Diameter of 5th ring\n", + "D15=0.59*10**-2 #Diameter of 15th ring\n", + "p=10 #n=5,n+p=15\n", + "R=1 #Radius of curvature\n", + "\n", + "#Calculations:\n", + "#(Dn+p)^2-Dn^2=4*p*lamda*R/u\n", + "lamda=((D15)**2-D5**2)/(4*p*R) #Wavelength of light\n", + "print\"Wavelength of light is =\",lamda,\"m\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Wavelength of light is = 5.8801e-07 m\n" + ] + } + ], + "prompt_number": 42 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 1.31, Page number 1-51" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "\n", + "#Condition for bright band is\n", + "#2ut*cos r = (2n-1)*lamda1\n", + "\n", + "#for consecutive bands, 2n=(lamda1+lamda2)/(lamda1-lamda).\n", + "#thus, 2ut*cos r = lamda2*lamda1/(lamda1-lamda2)\n", + "\n", + "#And, thicknessof film \n", + "#t= lamda2*lamda1/((2*u*cosr)(lamda1-lamda2))\n", + "print\"Hence expression for thickness of film is obtained.\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Hence expression for thickness of film is obtained.\n" + ] + } + ], + "prompt_number": 56 + }, + { + "cell_type": "code", + "collapsed": false, + "input": [], + "language": "python", + "metadata": {}, + "outputs": [] + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file diff --git a/Applied_Physics-II/chapter2.ipynb b/Applied_Physics-II/chapter2.ipynb new file mode 100755 index 00000000..6af9170f --- /dev/null +++ b/Applied_Physics-II/chapter2.ipynb @@ -0,0 +1,1480 @@ +{ + "metadata": { + "celltoolbar": "Raw Cell Format", + "name": "", + "signature": "sha256:1b2e4fb9c2f216cbf762800ac72003e90570cd62834a70860272dd5120eea60f" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Chapter 2: Diffraction of light" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 2.1,Page number 2-30" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Given Data:\n", + "m=4 #order\n", + "N=1./5000*10**-2 #N=(a+b) grating element(cm)\n", + "\n", + "#Calculations:\n", + "\n", + "#We know, (a+b)*sin(theta)=m*lamda\n", + "#for longest wavelength, sin(theta)=1\n", + "lamda=(N/m) #longest wavelength\n", + "print\"The longest wavelength is =\" ,lamda,\"m\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The longest wavelength is = 5e-07 m\n" + ] + } + ], + "prompt_number": 6 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 2.2,Page number 2-30" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Given Data:\n", + "m=1 #order\n", + "lamda=6.5*10**-7 #Wavelength of red light\n", + "theta=30*3.142/180 #angle of diffraction\n", + "\n", + "#Calculations:\n", + "\n", + "#We know, a*sin(theta)=m*lamda\n", + "a=m*lamda/math.sin(theta) #width of slit\n", + "print\"width of slit is = \",a,\"m\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "width of slit is = 1.29984715296e-06 m\n" + ] + } + ], + "prompt_number": 22 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 2.3,Page number 2-31" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Given Data:\n", + "m=1 #order\n", + "lamda=4*10**-7 #Wavelength of light\n", + "a=10**-6 #width of slit\n", + "\n", + "#Calculations:\n", + "\n", + "#We know, a*sin(theta)=m*lamda\n", + "theta=math.asin(m*lamda/a)*180/3.142 #angular position in first minima\n", + "print\"angular position in first minima is =\",theta,\"degrees\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "angular position in first minima is = 23.5751216716 degrees\n" + ] + } + ], + "prompt_number": 7 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 2.4,Page number 2-31" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Given Data:\n", + "m=1 #order\n", + "lamda1=4*10**-7 #Wavelength of light\n", + "lamda2=7*10**-7 #Wavelength of light\n", + "n=1./6000*10**-2 #n=(a+b) grating element\n", + "\n", + "#Calculations:\n", + "\n", + "#We know, (a+b)*sin(theta)=m*lamda\n", + "theta1=math.asin(m*lamda1/n)*(180/3.142) #angle of diffraction\n", + "theta2=math.asin(m*lamda2/n)*(180/3.142) #angle of diffraction\n", + "d=theta2-theta1 #angular breadth of first order visible spectrum\n", + "print\"angular breadth of first order visible spectrum is = \",d,\"degrees\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "angular breadth of first order visible spectrum is = 10.9466277612 degrees\n" + ] + } + ], + "prompt_number": 8 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 2.5,Page number 2-31" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Given Data:\n", + "m=1 #order\n", + "lamda=6.56*10**-7 #Wavelength of red light\n", + "theta=18.25*math.pi/180 #angle of diffraction\n", + "W=2*10**-2 #width of grating\n", + "\n", + "#Calculations:\n", + "\n", + "#We know, (a+b)*sin(theta)=m*lamda\n", + "N=math.sin(theta)/(m*lamda) #N-number of lines per m, N=1/(a+b)\n", + "Tn=N*W #Total number of lines on grating\n", + "print\"Total number of lines on grating is =\",Tn\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Total number of lines on grating is = 9547.67702694\n" + ] + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 2.7,Page number 2-33" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Given Data:\n", + "GE=2.54/15000*10**-2 #GE=(a+b) grating element\n", + "lamda1=4*10**-7 #Wavelength of light\n", + "lamda2=7*10**-7 #Wavelength of light\n", + "\n", + "#Calculations:\n", + "\n", + "#We know, (a+b)*sin(theta)=m*lamda\n", + "theta11=math.asin(1*lamda1/GE)*180/math.pi #angular position of first minima for lamda1\n", + "theta12=math.asin(2*lamda1/GE)*180/math.pi #angular position of second minima for lamda1\n", + "theta13=math.asin(3*lamda1/GE)*180/math.pi #angular position of third minima for lamda1\n", + "\n", + "theta21=math.asin(1*lamda2/GE)*180/math.pi #angular position of first minima for lamda2\n", + "theta22=math.asin(2*lamda2/GE)*180/math.pi #angular position of second minima for lamda2\n", + "theta23=math.asin(1)*180/math.pi #angular position of third minima for lamda2\n", + "\n", + "print\"Thus the angular position for lamda1 and lamda2 are as follows:\"\n", + "print\"First order:\",theta11,\"degrees\"\n", + "print\"\",theta21,\"degrees --Isolated\"\n", + "\n", + "print\"Second order:\",theta12,\"degrees\"\n", + "print\"\",theta22,\"degrees --Overlap\"\n", + "\n", + "print\"Third order: \",theta13,\"degrees\"\n", + "print\"\",theta23,\" degrees --Overlap \"\n", + "\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Thus the angular position for lamda1 and lamda2 are as follows:\n", + "First order: 13.6635763633 degrees\n", + " 24.4177053663 degrees --Isolated\n", + "Second order: 28.1928605617 degrees\n", + " 55.7685229906 degrees --Overlap\n", + "Third order: 45.1261086702 degrees\n", + " 90.0 degrees --Overlap \n" + ] + } + ], + "prompt_number": 13 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 2.8,Page number 2-34" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Given Data:\n", + "lamda=5.893*10**-7 #Wavelength of light\n", + "d=0.01*10**-2 #width of slit (a=d)\n", + "f=1 #distance between screen and slit\n", + "\n", + "#Calculations:\n", + "x=f*lamda/d #separation between central maxima and first minima\n", + "print\"Separation between central maxima and first minima is = \",x,\"m\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " Separation between central maxima and first minima is = 0.005893 m\n" + ] + } + ], + "prompt_number": 16 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 2.9,Page number 2-34" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Given Data:\n", + "m=1 #order\n", + "lamda=6*10**-7 #Wavelength of light\n", + "a=12*10**-7 #width of slit\n", + "\n", + "#Calculations:\n", + "\n", + "#We know, a*sin(theta)=m*lamda\n", + "theta=math.asin(m*lamda/a)*180/math.pi #angular position in first minima\n", + "print\"Half angular width of first maxima is =\",theta,\"Degrees\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Half angular width of first maxima is = 30.0 Degrees\n" + ] + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 2.10,Page number 2-34" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Given Data:\n", + "lamda=6*10**-7 #Wavelength of light\n", + "a=0.02*10**-2 #width of slit (a=d)\n", + "f=2 #distance between screen and slit\n", + "\n", + "#Calculations:\n", + "\n", + "#We know, a*sin(theta)=m*lamda, here m=1\n", + "theta=math.asin(lamda/a)*180*60/math.pi #angular position in first minima (1 degree=60 minutes)\n", + "print\"Total angular width is =\",2*theta,\"minutes\"\n", + "\n", + "x=f*lamda/a #separation between central maxima and first minima\n", + "print\"Linear width is = \",2*x,\"m\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Total angular width is = 20.6265115646 minutes\n", + "Linear width is = 0.012 m\n" + ] + } + ], + "prompt_number": 19 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 2.11,Page number 2-35" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Given Data:\n", + "a=0.14*10**-3 #width of slit\n", + "n=2 #order\n", + "y=1.6*10**-2 #separation between second dark band and central bright band\n", + "D=2 #distance between screen and slit\n", + "\n", + "#Calculations:\n", + "\n", + "theta=y/D #from diagram \n", + "\n", + "#We know, a*sin(theta)=n*lamda\n", + "#here sin(theta)=theta\n", + "lamda=a*theta/n #wavelength of light\n", + "print\"wavelength of light is =\",lamda,\"m\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "wavelength of light is = 5.6e-07 m\n" + ] + } + ], + "prompt_number": 20 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 2.13,Page number 2-36" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Given Data:\n", + "lamda=6.328*10**-7 #Wavelength of light\n", + "N=1./6000*10**-2 #N=(a+b) grating element\n", + "\n", + "#Calculations:\n", + "\n", + "#We know, N*sin(theta)=m*lamda\n", + "theta1=math.asin(1*lamda/N)*180/math.pi #angular position in first order maxima,m=1\n", + "print\"Angular position in first order maxima is =\",theta1,\"Degrees\"\n", + "\n", + "theta2=math.asin(2*lamda/N)*180/math.pi #angular position in second order maxima,m=2\n", + "print\"Angular position in second order maxima is = \",theta2,\"Degrees\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Angular position in first order maxima is = 22.3138625335 Degrees\n", + "Angular position in second order maxima is = 49.4078093436 Degrees\n" + ] + } + ], + "prompt_number": 21 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 2.14,Page number 2-37" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Given Data:\n", + "lamda1=6*10**-7 #wavelength of yellow light\n", + "lamda2=4.8*10**-7 #wavelength of blue light\n", + "theta=(math.asin(3/4)) #angle of diffraction\n", + "\n", + "#Calculations:\n", + "\n", + "#for consecutive bands, n*lamda1=(n+1)*lamda2. thus,\n", + "n=lamda2/(lamda1-lamda2) #order\n", + "\n", + "#We know, (a+b)*sin(theta)=m*lamda\n", + "N=n*lamda1/(3./4) #N=(a+b) grating element\n", + "print\"Grating element (a+b) is =\",N,\"m\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Grating element (a+b) is = 3.2e-06 m\n" + ] + } + ], + "prompt_number": 9 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 2.15.1,Page number 2-54" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Given Data:\n", + "a=0.2*10**-3 #width of slit\n", + "n=1 #order\n", + "y=0.5*10**-2 #separation between first minima and central bright band\n", + "D=2 #distance between screen and slit\n", + "\n", + "#Calculations:\n", + "\n", + "theta=y/D #from diagram \n", + "\n", + "#We know, a*sin(theta)=n*lamda\n", + "#here sin(theta)=theta\n", + "lamda=a*theta/n #wavelength of light\n", + "print\"wavelength of light is = \",lamda,\"m\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "wavelength of light is = 5e-07 m\n" + ] + } + ], + "prompt_number": 7 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 2.15.2,Page number 2-55" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Given Data:\n", + "lamda1=5.4*10**-7 #Wavelength of light\n", + "lamda2=4.05*10**-7 #Wavelength of light\n", + "theta=30*math.pi/180 #angle of diffraction\n", + "\n", + "#Calculations:\n", + "#We know, (a+b)*sin(theta)=n*lamda\n", + "#n*lamda1=(n+1)*lamda2, we get \n", + "n=3\n", + "N=math.sin(theta)/(n*lamda1)*10**-2 #Number of lines per m= 1/(a+b)*10^-2\n", + "print\"Number of lines per cm is = \",N\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Number of lines per cm is = 3086.41975309\n" + ] + } + ], + "prompt_number": 20 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 2.15.4,Page number 2-56" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Given Data:\n", + "GE=1./6000*10**-2 #GE=(a+b) grating element\n", + "lamda1=5.893*10**-7 #Wavelength of light\n", + "lamda2=5.896*10**-7 #Wavelength of light\n", + "m=2 #order\n", + "\n", + "#Calculations:\n", + "theta1=math.asin(m*lamda1/GE)*180/math.pi #angular position in first minima\n", + "theta2=math.asin(m*lamda2/GE)*180/math.pi #angular position in second minima\n", + "\n", + "a_s=(theta2-theta1) #Angular separation in minutes\n", + "print\"Angular separation is =\",a_s,\"Degrees\"\n", + "\n", + "dlamda=lamda2-lamda1 #difference in wavelength\n", + "lamda=(lamda2+lamda1)/2 #Mean wavelength\n", + "\n", + "#We know that R.P.=lamda/dlamda=m*N\n", + "N=lamda/dlamda/m #Number of lines on grating for first order\n", + "print\"Number of lines on grating for first order is =\",N\n", + "print\"But, number of lines per cm on grating is 6000. \\n Which is greater than number of lines per cm needed for resolution.\"\n", + "print\"Hence, both lines will be well resolved in 2nd order.\"\n", + "\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Angular separation is = 0.0291798772234 Degrees\n", + "Number of lines on grating for first order is = 982.416666667\n", + "But, number of lines per cm on grating is 6000. \n", + " Which is greater than number of lines per cm needed for resolution.\n", + "Hence, both lines will be well resolved in 2nd order.\n" + ] + } + ], + "prompt_number": 21 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 2.17,Page number 2-39" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Given Data:\n", + "d=0.04*10**-2 #Separation between slits\n", + "D=1.7 #distance between screen and slit\n", + "B=0.25*10**-2 #Fringe spacing\n", + "\n", + "#Calculations:\n", + "#We know,B=D*lamda/d\n", + "lamda=B*d/D #Wavelength of light\n", + "print\"Wavelength of light is = \",lamda,\"m\"\n", + "\n", + "#The condition for missing order is,\n", + "#(a+b)/a = m/n\n", + "b=0.04*10**-2 #Separation in slits\n", + "a=0.08*10**-3 #Slit width\n", + "n=(a+b)/a #missing orders for m=1,2,3\n", + "\n", + "n1=1*n\n", + "n2=2*n\n", + "n3=3*n\n", + "print\"Missing orders are =\",n1,\",\",n2,\",\",n3\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Wavelength of light is = 5.88235294118e-07 m\n", + "Missing orders are = 6.0 , 12.0 , 18.0\n" + ] + } + ], + "prompt_number": 26 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 2.18,Page number 2-39" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Given Data:\n", + "N=2.54/2620*10**-2 #N=(a+b) grating element\n", + "lamda=5*10**-7 #Wavelength of red light\n", + "\n", + "#Calculations:\n", + "\n", + "#We know, (a+b)*sin(theta)=n*lamda\n", + "#maximum value of sin(theta)=1\n", + "n=N/lamda #Maximum number of orders visible\n", + "print\"Maximum number of orders visible is =\",n\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Maximum number of orders visible is = 19.3893129771\n" + ] + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 2.19,Page number 2-40" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Given Data:\n", + "N=1./4000*10**-2 #N=(a+b) grating element\n", + "lamda1=5*10**-7 #Wavelength of light\n", + "lamda2=7.5*10**-7 #Wavelength of light\n", + "\n", + "#Calculations:\n", + "\n", + "#We know, (a+b)*sin(theta)=n*lamda\n", + "#maximum value of sin(theta)=1\n", + "n1=N/lamda1 #Maximum number of orders visible\n", + "n2=N/lamda2 #Maximum number of orders visible\n", + "print\"The observed number of orders range between =\",n2,\"to\",n1\n", + "\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The observed number of orders range between = 3.33333333333 to 5.0\n" + ] + } + ], + "prompt_number": 5 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 2.20,Page number 2-40" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Given Data:\n", + "n=5 #order\n", + "lamda=6*10**-7 #Wavelength of light\n", + "\n", + "#Calculations:\n", + "#We know, a*sin(theta)=n*lamda\n", + "#n*lamda=n1*lamda1\n", + "lamda1=n*lamda/4 #for n1=4\n", + "print\"For n1=4 wavelength is =\",lamda1,\"m\"\n", + "\n", + "lamda2=n*lamda/5 #for n1=5\n", + "print\"For n1=5 wavelength is =\",lamda2,\"m\"\n", + "\n", + "lamda3=n*lamda/6 #for n1=6\n", + "print\"For n1=5 wavelength is =\",lamda3,\"m\"\n", + "\n", + "lamda4=n*lamda/7 #for n1=7\n", + "print\"For n1=5 wavelength is =\",lamda4,\"m\"\n", + "\n", + "lamda5=n*lamda/8 #for n1=8\n", + "print\"For n1=5 wavelength is =\",lamda5,\"m\"\n", + "\n", + "\n", + "print\"So,in the grating spectrum spectrum lines with wavelengths n1=6 and n1=7 will coincide with fifth order line of 6*10^-7 m\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "For n1=4 wavelength is = 7.5e-07 m\n", + "For n1=5 wavelength is = 6e-07 m\n", + "For n1=5 wavelength is = 5e-07 m\n", + "For n1=5 wavelength is = 4.28571428571e-07 m\n", + "For n1=5 wavelength is = 3.75e-07 m\n", + "So,in the grating spectrum spectrum lines with wavelengths n1=6 and n1=7 will coincide with fifth order line of 6*10^-7 m\n" + ] + } + ], + "prompt_number": 6 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 2.21,Page number 2-41" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Given Data:\n", + "GE=18000*10**-10 #GE=(a+b) grating element\n", + "lamda=5*10**-7 #Wavelength of red light\n", + "\n", + "#Calculations:\n", + "DP1=1./sqrt(GE**2-lamda**2)*10**-10 #Dispersive power\n", + "print\"Dispersive power for first order is =\",DP1,\"rad/Angstrom\"\n", + "\n", + "m=3\n", + "DP2=1/sqrt((GE/m)**2-lamda**2)*10**-10 #Dispersive power\n", + "print\"Dispersive power for second order is =\",DP2,\"rad/Angstrom\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Dispersive power for first order is = 5.78314931966e-05 rad/Angstrom\n", + "Dispersive power for second order is = 0.000301511344578 rad/Angstrom\n" + ] + } + ], + "prompt_number": 13 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 2.22,Page number 2-42" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Given Data:\n", + "N=2.54/15000*10**-2 #N=(a+b) grating element\n", + "lamda=5.9*10**-7 #Wavelength of light\n", + "m=2 #order\n", + "f=25*10**-2 #focal length of lens\n", + "\n", + "#Calculations:\n", + "\n", + "#We know, (a+b)*sin(theta)=m*lamda\n", + "theta=math.asin(m*lamda/N) #angular position in first minima\n", + "\n", + "Ad=m/N/cos(theta) #angular dispersion\n", + "\n", + "ld=f*Ad*10**-8 #linear dispersion (dx/dl) in cm/angstrom\n", + "print\"Linear dispersion in spectrograph is =\",ld,\"cm/angstrom\"\n", + "\n", + "dlamda=(5896-5890) #difference in wavelength\n", + "dx=ld*dlamda*10**-2 #separation between spectral lines in meter\n", + "print\"Separation between spectral lines is =\",dx,\"m\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Linear dispersion in spectrograph is = 0.00411696586101 cm/angstrom\n", + "Separation between spectral lines is = 0.00024701795166 m\n" + ] + } + ], + "prompt_number": 15 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 2.23,Page number 2-47" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Given Data:\n", + "m=1 #order\n", + "lamda1=5.89*10**-7 #Wavelength of light\n", + "lamda2=5.896*10**-7 #Wavelength of light\n", + "\n", + "#Calculations:\n", + "dlamda=lamda2-lamda1 #difference in wavelength\n", + "lamda=(lamda2+lamda1)/2 #Mean wavelength\n", + "\n", + "#We know that R.P.=m*N=lamda/dlamda\n", + "N=lamda/dlamda/m #minimum number of lines which will just resolve\n", + "print\"Minimum number of lines which will just resolve is =\",N\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Minimum number of lines which will just resolve is = 982.166666667\n" + ] + } + ], + "prompt_number": 16 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 2.24,Page number 2-47" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Given Data:\n", + "N=5*5000 #N=W/(a+b) Number of lines on grating\n", + "m=2 #order\n", + "lamda=6*10**-7 #Wavelength of light\n", + "\n", + "#Calculations:\n", + "#(i)\n", + "RP=m*N #Resolving power\n", + "print\"(i)Resolving power is =\",RP\n", + "\n", + "#(ii)\n", + "#We know that R.P.=lamda/dlamda\n", + "dlamda=lamda/RP #Smallest wavelength which can be resolved\n", + "print\"(ii)Smallest wavelength which can be resolved is =\",dlamda,\"m\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "(i)Resolving power is = 50000\n", + "(ii)Smallest wavelength which can be resolved is = 1.2e-11 m\n" + ] + } + ], + "prompt_number": 19 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 2.25,Page number 2-48" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Given Data:\n", + "GE=1./4000*10**-2 #GE=(a+b) grating element\n", + "lamda=5*10**-7 #Wavelength of red light\n", + "m=3 #order\n", + "\n", + "#Calculations:\n", + "\n", + "#We know, (a+b)*sin(theta)=m*lamda\n", + "theta=math.asin(m*lamda/GE) #ngular position in first minima\n", + "\n", + "DP=m/(GE*math.cos(theta))*10**-2 #Dispersive power\n", + "print\"Dispersive power is =\",DP\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Dispersive power is = 15000.0\n" + ] + } + ], + "prompt_number": 22 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 2.26,Page number 2-48" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Given Data:\n", + "m=2 #order\n", + "lamda=6*10**-7 #Wavelength of light\n", + "dlamda=6*10**-10 #difference in wavelength\n", + "W=2*10**-2 #Width of surface\n", + "\n", + "#Calculations:\n", + "\n", + "#We know that R.P.=lamda/dlamda=m*N\n", + "N=lamda/dlamda/m #Number of lines on grating\n", + "GE=W/N #Grating element=(a+b)\n", + "print\"Grating element is =\",GE,\"cm\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Grating element is = 4e-05 cm\n" + ] + } + ], + "prompt_number": 23 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 2.27,Page number 2-49" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Given Data:\n", + "m=2 #order\n", + "lamda1=5.77*10**-7 #Wavelength of light\n", + "lamda2=5.791*10**-7 #Wavelength of light\n", + "GE=1./6000*10**-2 #GE=(a+b) grating element\n", + "\n", + "#Calculations:\n", + "\n", + "#We know, (a+b)*sin(theta)=m*lamda\n", + "theta1=math.asin(m*lamda1/GE)*180/math.pi #angular position in first minima\n", + "theta2=math.asin(m*lamda2/GE)*180/math.pi #angular position in second minima\n", + "\n", + "a_s=(theta2-theta1)*60 #Angular separation in minutes\n", + "print\"Angular separation is = \",a_s,\"minutes\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Angular separation is = 12.0270825521 minutes\n" + ] + } + ], + "prompt_number": 25 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 2.28,Page number 2-49" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Given Data:\n", + "n=1 #order\n", + "lam=5.89*10**-7 #Wavelength of light\n", + "a=0.3*10**-3 #width of slit\n", + "\n", + "#Calculations:\n", + "\n", + "#We know, a*sin(theta)=n*lamda\n", + "theta1=math.asin(n*lamda/a)*180/math.pi*60 #angular position in first dark band in minutes\n", + "print\"Angular position in first dark band is = \",theta1,\"mimutes\"\n", + "\n", + "#We know,for bright band a*sin(theta)=(2n+1)*lamda/2\n", + "theta2=math.asin(1.5*lamda/a)*180/math.pi*60 #angular position in first bright band in minutes\n", + "print\"Angular position in first bright band is =\",theta2,\"minutes\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Angular position in first dark band is = 6.87549812524 mimutes\n", + "Angular position in first bright band is = 10.3132557823 minutes\n" + ] + } + ], + "prompt_number": 26 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 2.29,Page number 2-50" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Given Data:\n", + "GE=2.54/16000*10**-2 #GE=(a+b) grating element\n", + "lamda=6*10**-7 #Wavelength of light\n", + "\n", + "#Calculations:\n", + "\n", + "#We know, (a+b)*sin(theta)=m*lamda\n", + "#maximum value of sin(theta)=1\n", + "m=GE/lamda #Maximum order of spectra\n", + "print\"Maximum order of spectra is =\",m\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Maximum order of spectra is = 2.64583333333\n" + ] + } + ], + "prompt_number": 28 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 2.30,Page number 2-50" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Given Data:\n", + "GE=1./5000*10**-2 #GE=(a+b) grating element\n", + "lamda=5.89*10**-7 #Wavelength of light\n", + "N=3*5000 #N=W/(a+b) Number of lines on grating\n", + " \n", + "#Calculations:\n", + "\n", + "#We know, (a+b)*sin(theta)=m*lamda\n", + "#maximum value of sin(theta)=1\n", + "m=GE/lamda #Maximum order of spectra\n", + "print\"Maximum order of spectra is =\",m\n", + "\n", + "RP=3*N #Resolving power (round of m to 3)\n", + "print\"Resolving power is =\",RP\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Maximum order of spectra is = 3.39558573854\n", + "Resolving power is = 45000\n" + ] + } + ], + "prompt_number": 29 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 2.32,Page number 2-52" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Given Data:\n", + "lamda1=5.89*10**-7 #Wavelength of light\n", + "lamda2=5.896*10**-7 #Wavelength of light\n", + "\n", + "#Calculations:\n", + "dlamda=lamda2-lamda1 #difference in wavelength\n", + "lamda=(lamda2+lamda1)/2 #Mean wavelength\n", + "\n", + "#(i)\n", + "m1=1 #first order\n", + "#We know that R.P.=lamda/dlamda=m*N\n", + "N1=lamda/dlamda/m1 #Number of lines on grating\n", + "print\"(i)Number of lines on grating for first order is =\",N1\n", + "\n", + "#(ii)\n", + "m2=2 #second order\n", + "#We know that R.P.=lamda/dlamda=m*N\n", + "N2=lamda/dlamda/m2 #Number of lines on grating\n", + "print\"(ii)Number of lines on grating for second order is =\",N2\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "(i)Number of lines on grating for first order is = 982.166666667\n", + "(ii)Number of lines on grating for second order is = 491.083333333\n" + ] + } + ], + "prompt_number": 30 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 2.33,Page number 2-52" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Given Data:\n", + "m=1 #order\n", + "lamda=6.553*10**-7 #Wavelength of light\n", + "dlamda=1.8*10**-10 #difference in wavelength\n", + "\n", + "\n", + "#Calculations:\n", + "\n", + "#We know that R.P.=lam/dlam=m*N\n", + "N=lamda/dlamda/m #Number of lines on grating\n", + "print\"Number of lines on grating is =\",N\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Number of lines on grating is = 3640.55555556\n" + ] + } + ], + "prompt_number": 31 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 2.34,Page number 2-53" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Given Data:\n", + "lamda1=5.14034*10**-7 #Wavelength of light\n", + "lamda2=5.14085*10**-7 #Wavelength of light\n", + "\n", + "#Calculations:\n", + "dlamda=lamda2-lamda1 #difference in wavelength\n", + "lamda=(lamda2+lamda1)/2 #Mean wavelength\n", + "\n", + "#We know that R.P.=lamda/dlamda=m*N\n", + "N=lamda/dlamda/1 #Number of lines on grating\n", + "print\"Number of lines on grating for first order is =\",N\n", + "\n", + "#Hence R.P. for second order should be\n", + "RP1=2*N\n", + "print\"Resolving power in second order should be is= \",RP1\n", + "\n", + "#But here,\n", + "\n", + "lamda3=8.03720*10**-7 #Wavelength of light\n", + "lamda4=8.03750*10**-7 #Wavelength of light\n", + "dlamda2=lamda4-lamda3 #difference in wavelength\n", + "lamda2=(lamda4+lamda3)/2 #Mean wavelength\n", + "\n", + "RP2=lamda2/dlamda2\n", + "print\"Resolving power in second order is= \",RP2\n", + "\n", + "print\"So, the grating will not be able to resolve 8.0372*10^-7 and 8.03750*10^-7 in second order.\"\n", + "print\"Because Resolving power is greter than actual Resolving power.\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Number of lines on grating for first order is = 10079.5980392\n", + "Resolving power in second order should be is= 20159.1960784\n", + "Resolving power in second order is= 26791.1666667\n", + "So, the grating will not be able to resolve 8.0372*10^-7 and 8.03750*10^-7 in second order.\n", + "Because Resolving power is greter than actual Resolving power.\n" + ] + } + ], + "prompt_number": 32 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 2.35,Page number 2-53" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#For grating , Condition of maxima is (a+b)sin(theta)=n*lamda\n", + "#Given (a+b) < 2*lamda\n", + "#For maximum order, sin(90)=1\n", + "#So, n must be less than 2\n", + "#i.e. only first order possible if width of grating element is less than twice the wavelength\n", + "print\"Hence, Only first order possible if width of grating element is less than twice the wavelength.\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Hence, Only first order possible if width of grating element is less than twice the wavelength.\n" + ] + } + ], + "prompt_number": 33 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 2.36,Page number 2-54" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Given Data:\n", + "n=1 #order\n", + "lamda=5.89*10**-7 #Wavelength of light\n", + "a=0.3*10**-3 #width of slit\n", + "\n", + "#Calculations:\n", + "\n", + "#We know, a*sin(theta)=n*lamda\n", + "theta1=math.asin(n*lam/a)*180/math.pi #angular position in first dark band\n", + "print\"Angular position in first dark band is =\",theta1,\"Degrees\"\n", + "\n", + "#We know,for bright band a*sin(theta)=(2n+1)*lamda/2\n", + "theta2=math.asin(1.5*lamda/a)*180/math.pi #angular position in first bright band\n", + "print\"Angular position in first bright band is =\",theta2,\"Degrees\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Angular position in first dark band is = 0.112490786047 Degrees\n", + "Angular position in first bright band is = 0.168736314576 Degrees\n" + ] + } + ], + "prompt_number": 35 + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file diff --git a/Applied_Physics-II/chapter3.ipynb b/Applied_Physics-II/chapter3.ipynb new file mode 100755 index 00000000..714dfad3 --- /dev/null +++ b/Applied_Physics-II/chapter3.ipynb @@ -0,0 +1,836 @@ +{ + "metadata": { + "celltoolbar": "Raw Cell Format", + "name": "", + "signature": "sha256:5f637d4fa0aa8bb3bc54a28134ac85a937c06f0d3a591f536de2719824a41f5c" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Chapter 3: Optical Fibre" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 3.1,Page number 3-19" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Given Data:\n", + "n1=1.61 #Core index\n", + "n2=1.55 #Cladding index\n", + "\n", + "#Calculations:\n", + "NA=math.sqrt(n1**2-n2**2) #Formula\n", + "\n", + "print\"Numerical Aperture of Fibre is = \",NA\n", + "\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Numerical Aperture of Fibre is = 0.435430821142\n" + ] + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 3.2,Page number 3-19" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Given Data:\n", + "n1=1.65 #Core index\n", + "n2=1.53 #Cladding index\n", + "\n", + "#Calculations:\n", + "NA=math.sqrt(n1**2-n2**2) #Formula\n", + "\n", + "print\"Numerical Aperture of Fibre is =\",NA\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Numerical Aperture of Fibre is = 0.617737808459\n" + ] + } + ], + "prompt_number": 3 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 3.3,Page number 3-19" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Given Data:\n", + "n1=1.48 #R.I. of Core\n", + "n2=1.39 #R.I. of Cladding\n", + "\n", + "#Calculations:\n", + "NA=math.sqrt(n1**2-n2**2) #Formula to find NA\n", + "phi=math.asin(NA)*180/(3.1472) #Acceptance angle\n", + "\n", + "print\"Numerical Aperture of Fibre is =\",NA\n", + "print\"Acceptance angle of Fibre is =\",phi,\"degrees\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Numerical Aperture of Fibre is = 0.508232230383\n", + "Acceptance angle of Fibre is = 30.491727191 degrees\n" + ] + } + ], + "prompt_number": 8 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 3.4,Page number 3-20" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#given data:\n", + "u1=3.6 #Refractive Index of the Substance at 850 nm\n", + "u2=3.4 #Refractive Index of the Substance at 1300 nm\n", + "Vv=3*10**8 #Velocity of light in free space\n", + "\n", + "#Calculations:\n", + "# i)Finding wavelength at 850 nm\n", + "Vs1=Vv/u1 #Velocity of light in substance at 850 nm\n", + "print\"Velocity of light in substance at 850 nm =\" ,Vs1,\"m/sec\"\n", + "\n", + "lam1=850*10**-9/u1 #Wavelength of light in substance at 850nm\n", + "print\" Wavelength of light in substance at 850nm =\",lam1,\"m\"\n", + "\n", + "\n", + "#ii)Finding wavelength at 1300 nm\n", + "Vs2=Vv/u2 #Velocity of light in substance at 1300 nm\n", + "print\"Velocity of light in substance at 1300 nm =\",Vs2,\" m/sec\"\n", + "\n", + "lam2=1300*10**-9/u2 #Wavelength of light in substance at 1300nm\n", + "print\"Wavelength of light in substance at 1300nm =\" ,lam2,\"m \"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Velocity of light in substance at 850 nm = 83333333.3333 m/sec\n", + " Wavelength of light in substance at 850nm = 2.36111111111e-07 m\n", + "Velocity of light in substance at 1300 nm = 88235294.1176 m/sec\n", + "Wavelength of light in substance at 1300nm = 3.82352941176e-07 m \n" + ] + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 3.5,Page number 3-20" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Given Data:\n", + "u1=1.5 #R.I. of Core\n", + "u2=1.45 #R.I.of Cladding\n", + "delta= (u1-u2)/u1 #Fractional Refractive index\n", + "\n", + "#Calculations:\n", + "NA=u1*sqrt(2*delta) #Formula to find NA\n", + "theta0=math.asin(NA)*180/(3.1472) #Acceptance angle\n", + "thetac=math.asin(u2/u1)*180/(3.1472) #Critical angle\n", + "\n", + "print\"Numerical Aperture of Fibre is =\",NA\n", + "print\"Acceptance angle of Fibre is =\",theta0,\"degrees\" \n", + "print\" Critical angle of Fibre is =\",thetac,\"degrees\" \n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Numerical Aperture of Fibre is = 0.387298334621\n", + "Acceptance angle of Fibre is = 22.7458994397 degrees\n", + " Critical angle of Fibre is = 75.0309676099 degrees\n" + ] + } + ], + "prompt_number": 7 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 3.6,Page number 3-20" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Given Data:\n", + "NA=0.22 # Numerical Aperture of Fibre\n", + "delta=0.012 #Fractional index\n", + "\n", + "#Calculations:\n", + "#Delta=(u1-u2)/u1\n", + "u1=NA/math.sqrt(2*delta) #Formula\n", + "u2=u1-(u1*delta) #Formula\n", + "\n", + "print\"Refractive Index of core of fibre is =\",u1\n", + "print\"Refractive Index of cladding of fibre is =\",u2\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Refractive Index of core of fibre is = 1.42009389361\n", + "Refractive Index of cladding of fibre is = 1.40305276689\n" + ] + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 3.7,Page number 3-21" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Given Data:\n", + "u1=1.466 #R.I. of Core\n", + "u2=1.46 #R.I.of Cladding\n", + "V=2.4 #Cut off parameter\n", + "lamda=0.8*10**-6 #wavelength in meter\n", + "\n", + "#Calculations:\n", + "NA=math.sqrt(u1**2-u2**2) #Formula to find Numerical Aperture\n", + "print\"Numerical Aperture of Fibre is =\",NA\n", + "#(printing mistake in book)printed answer is 1.13 but correct answer is 0.13\n", + "print\"(printing mistake in book)\"\n", + "\n", + "# V = 2*3.142*a*NA / lamda\n", + "a=V*lamda/(2*3.142*NA) #core radius\n", + "print\"Core radius of Fibre is (a) =\",a,\"m\"\n", + "\n", + "#w/a= 1.1\n", + "w=1.1*a #Spot size\n", + "print\"Spot size of Fibre is =\",w,\"m\"\n", + "\n", + "theta=2*lamda*180/3.142/(3.142*w) #Divergence angle\n", + "print\"Divergence angle of Fibre is =\",theta,\"degrees\"\n", + "\n", + "w10=lamda*10/(3.142*w) #Spot size at 10 m\n", + "print\"Spot size at 10 m of Fibre is =\",w10,\"m\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Numerical Aperture of Fibre is = 0.1324990566\n", + "(printing mistake in book)\n", + "Core radius of Fibre is (a) = 2.30596263706e-06 m\n", + "Spot size of Fibre is = 2.53655890076e-06 m\n", + "Divergence angle of Fibre is = 11.5009886523 degrees\n", + "Spot size at 10 m of Fibre is = 1.00378073182 m\n" + ] + } + ], + "prompt_number": 5 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 3.8,Page number 3-21" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Given Data:\n", + "w=98 #Spot size in meter\n", + "d=50*10**-6 #Core diameter in meter\n", + "a=d/2 #core radius\n", + "u1=1.47 #R.I. of Core\n", + "u2=1.45 #R.I.of Cladding\n", + "lamda=0.85*10**-6 #Wavlength in meter\n", + "NA=math.sqrt(u1**2-u2**2) #Formula to find NA\n", + "\n", + "#Calculations:\n", + "V=2*3.142*a*NA/lamda #cut off parameter\n", + "N=(V**2)/2 #Number of modes\n", + "\n", + "print\"Cut off parameter of Fibre is =\",V\n", + "print\"Number of modes of Fibre is =\",N\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Cut off parameter of Fibre is = 44.6646240577\n", + "Number of modes of Fibre is = 997.464321107\n" + ] + } + ], + "prompt_number": 6 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 3.9,Page number 3-21" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Given Data:\n", + "u1=1.47 #R.I. of Core\n", + "u2=1.46 #R.I.of Cladding\n", + "lamda=1.3*10**-6 #wavelength in meter\n", + "\n", + "#Calculations:\n", + "NA=math.sqrt(u1**2-u2**2) #Formula to find Numerical Aperture\n", + "\n", + "#The condition for single mode is V<2.405\n", + "#2*3.142*a*NA/lamda < 2.405\n", + "\n", + "a=2.405*lamda/(2*3.142*NA) #Maximum radius of fibre\n", + "\n", + "print\"Maximum radius of Fibre is =\",a,\"meter\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Maximum radius of Fibre is = 2.90662126448e-06 meter\n" + ] + } + ], + "prompt_number": 7 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 3.10,Page number 3-22" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Given Data:\n", + "u1=1.465 #R.I. of Core\n", + "u2=1.46 #R.I.of Cladding\n", + "lamda=1.25*10**-6 #operating wavelength\n", + "\n", + "#Calculations:\n", + "delta=(u1-u2)/u1 #Fractional Refractive index\n", + "print\"Fractional Refractive index of Fibre is =\",delta\n", + "\n", + "#For single mode propagation codition is \n", + "# a/lamda < 1.4/(3.142*sqrt(u1(u1-u2)))\n", + "\n", + "a=lamda*1.4/(3.142*u1*math.sqrt(delta)) #core radius\n", + "\n", + "u=u1-(math.sqrt(2*delta)/(2*3.142*(a/lamda))) #effective refractive index\n", + "print\"Effective Refractive index for lowest mode propagation is =\",u\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Fractional Refractive index of Fibre is = 0.00341296928328\n", + "Effective Refractive index for lowest mode propagation is = 1.46247461864\n" + ] + } + ], + "prompt_number": 8 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 3.11,Page number 3-22" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Given Data:\n", + "u1=1.54 #R.I. of Core\n", + "u2=1.5 #R.I.of Cladding\n", + "lamda=1.3*10**-6 #wavelength in meter\n", + "a=25*10**-6 #core radius in meter\n", + "\n", + "#Calculations:\n", + "NA=math.sqrt(u1**2-u2**2) #Formula to find Numerical Aperture\n", + "\n", + "V=2*3.142*a*NA/lamda #cut off parameter\n", + "print\"Cut off parameter of Fibre is =\",V\n", + "\n", + "N=(V**2)/2 #Number of modes\n", + "print\" Number of modes of Fibre is =\",N\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Cut off parameter of Fibre is = 42.1404937865\n", + " Number of modes of Fibre is = 887.910608284\n" + ] + } + ], + "prompt_number": 9 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 3.11.1,Page number 3-25" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Given Data:\n", + "u1=1.52 #R.I. of Core\n", + "u2=1.5189 #R.I.of Cladding\n", + "lamda=1.3*10**-6 #wavelength in meter\n", + "d=29*10**-6 #core diameter in meter\n", + "a=d/2\n", + "\n", + "#Calculations:\n", + "NA=math.sqrt(u1**2-u2**2) #Formula to find Numerical Aperture\n", + "V=2*3.142*a*NA/lamda #Normalised frequency\n", + "Nm=(V**2)/2 #Number of modes\n", + "\n", + "print\"Normalised frequency of Fibre is (V)=\",V\n", + "print\"The Maximum Number of modes the Fibre will support is (Nm) =\",Nm\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Normalised frequency of Fibre is (V)= 4.05242861605\n", + "The Maximum Number of modes the Fibre will support is (Nm) = 8.2110888441\n" + ] + } + ], + "prompt_number": 11 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 3.12,Page number 3-22" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Given Data:\n", + "u1=1.5 #R.I. of Core\n", + "d=10*10**-6 #diameter of core\n", + "a=d/2 #core radius\n", + "lamda=1.3*10**-6 #wavelength\n", + "V=2.405 #cut off parameter for single mode\n", + "\n", + "#Calculations:\n", + "\n", + "#We know, V=2*3.142*a*NA/lamda\n", + "NA=V*lamda/(2*3.142*a) #Numerical Aperture\n", + "\n", + "theta=math.asin(NA)*180/3.142 #Acceptance angle\n", + "print\"Acceptance angle of Fibre is =\",theta,\"Degrees\"\n", + "\n", + "#Also, NA=u1*sqrt(2*delta)\n", + "delta=(NA/u1)**2/2 #Fractional index\n", + "print\"Maximum Fractional Refractive index of Fibre is =\",delta\n", + "\n", + "#delta=(u1-u2)/u1\n", + "u2=u1*(1-delta) #R.I.of cladding\n", + "print\"Refractive index of cladding of Fibre is =\",u2\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Acceptance angle of Fibre is = 5.71002346964 Degrees\n", + "Maximum Fractional Refractive index of Fibre is = 0.00220035113094\n", + "Refractive index of cladding of Fibre is = 1.4966994733\n" + ] + } + ], + "prompt_number": 13 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 3.13,Page number 3-23" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Given Data:\n", + "n1=1.5 #R.I. of core\n", + "delta=0.0005 #Fractional index difference\n", + "\n", + "#Calculations:\n", + "#(a):\n", + "#Delta=(u1-u2)/u1\n", + "n2=n1-(n1*delta) #R.I. of cladding\n", + "print\"(a)Refractive Index of cladding of fibre is =\",n2\n", + "\n", + "#(b):\n", + "phi=math.asin(n2/n1)*180/3.142 #Critical internal reflection angle\n", + "print\"(b)Critical internal reflection angle of Fibre is =\",phi,\"degrees\"\n", + "\n", + "#(c):\n", + "theta0=math.asin(math.sqrt(n1**2-n2**2))*180/3.142 #External critical Acceptance angle\n", + "print\"(c)External critical Acceptance angle of Fibre is =\",theta0,\"degrees\"\n", + "\n", + "#(d):\n", + "NA=n1*math.sqrt(2*delta) #Formula to find Numerical Aperture\n", + "print\"(d)Numerical Aperture of Fibre is =\",NA\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "(a)Refractive Index of cladding of fibre is = 1.49925\n", + "(b)Critical internal reflection angle of Fibre is = 88.1766396681 degrees\n", + "(c)External critical Acceptance angle of Fibre is = 2.71810509125 degrees\n", + "(d)Numerical Aperture of Fibre is = 0.0474341649025\n" + ] + } + ], + "prompt_number": 14 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 3.14,Page number 3-24" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Given Data:\n", + "NA1=0.20 #Numerical Aperture of Fibre\n", + "n2=1.59 #R.I. of cladding\n", + "\n", + "#Calculations:\n", + "#NA=sqrt(n1**2-n2**2)\n", + "#In air, n0=1\n", + "n1=math.sqrt(NA1**2+n2**2) #R.I.of core\n", + "\n", + "#Now, in water \n", + "n0=1.33\n", + "NA2=math.sqrt(n1**2-n2**2)/n0 #Numerical Aperture in water\n", + "theta0=math.asin(NA2)*180/3.142 #Acceptance angle of fibre in water\n", + "print\"Acceptance angle of Fibre in water is =\",theta0,\"degrees\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Acceptance angle of Fibre in water is = 8.6475921767 degrees\n" + ] + } + ], + "prompt_number": 15 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 3.15,Page number 3-24" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Given Data:\n", + "n1=1.45 #R.I.of core\n", + "n2=1.40 #R.I. of cladding\n", + "\n", + "#Calculations:\n", + "NA=math.sqrt(n1**2-n2**2) #Numerical Aperture\n", + "print\"Numerical Aperture of Fibre is =\",NA\n", + "\n", + "theta0=math.asin(NA)*180/3.142 #Acceptance angle of fibre\n", + "print\"Acceptance angle of Fibre is =\",theta0,\"degrees\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Numerical Aperture of Fibre is = 0.377491721764\n", + "Acceptance angle of Fibre is = 22.1755250876 degrees\n" + ] + } + ], + "prompt_number": 16 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 3.16,Page number 3-24" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Given Data:\n", + "NA=0.16 #Numerical Aperture of Fibre\n", + "n1=1.45 #R.I. of core\n", + "d=90*10**-6 #Core diameter\n", + "\n", + "#Calculations:\n", + "#NA=sqrt(n1**2-n2**2)\n", + "n2=math.sqrt(n1**2-NA**2) #R.I.of cladding\n", + "print\"(a)Refractive Index of cladding of fibre is =\",n2\n", + "\n", + "theta0=math.asin(NA)*180/3.142 #Acceptance angle of fibre\n", + "print\"(b)Acceptance angle of Fibre is =\",theta0,\"degrees\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "(a)Refractive Index of cladding of fibre is = 1.44114537782\n", + "(b)Acceptance angle of Fibre is = 9.20570258795 degrees\n" + ] + } + ], + "prompt_number": 18 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 3.17,Page number 3-25" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Given Data:\n", + "n1=1.48 #R.I. of core\n", + "delta=0.055 #Realtive R.I.\n", + "lamda=1*10**-6 #Wavelength of light\n", + "a=50*10**-6 #core radius\n", + "\n", + "#Calculations:\n", + "#Delta=(u1-u2)/u1\n", + "n2=n1-(n1*delta) #R.I. of cladding\n", + "NA=n1*math.sqrt(2*delta) #Formula to find Numerical Aperture\n", + "print\"Numerical Aperture of Fibre is =\",NA\n", + "\n", + "\n", + "theta0=math.asin(NA)*180/3.142 #Acceptance angle of fibre\n", + "print\"Acceptance angle of Fibre is =\",theta0,\"degrees\"\n", + "\n", + "V=2*3.142*a*NA/lamda #V number\n", + "N=(V**2)/2 #Number of guided modes\n", + "\n", + "#In book,instead of NA , value of delta is taken into calculation.\n", + "#Thus there is calculation mistake in values of V and N.\n", + "\n", + "print\"V number of Fibre is =\",V\n", + "print\"Number of guided mode of Fibre is =\",N\n", + "print\"(Calculation mistake in book)\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Numerical Aperture of Fibre is = 0.490860468973\n", + "Acceptance angle of Fibre is = 29.3933421943 degrees\n", + "V number of Fibre is = 154.228359351\n", + "Number of guided mode of Fibre is = 11893.1934141\n", + "(Calculation mistake in book)\n" + ] + } + ], + "prompt_number": 19 + }, + { + "cell_type": "code", + "collapsed": false, + "input": [], + "language": "python", + "metadata": {}, + "outputs": [] + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file diff --git a/Applied_Physics-II/chapter4.ipynb b/Applied_Physics-II/chapter4.ipynb new file mode 100755 index 00000000..11b66b2f --- /dev/null +++ b/Applied_Physics-II/chapter4.ipynb @@ -0,0 +1,163 @@ +{ + "metadata": { + "celltoolbar": "Raw Cell Format", + "name": "", + "signature": "sha256:e413b3bea25c7729b7ae1f595bf0f843336be8c682782f6b92de1c93aa46ed9d" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Chapter 4: Laser" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 4.1,Page number 4-27" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Given Data:\n", + "lamda=694.3*10**-9 #Wavelength in meter\n", + "T=300 #Temperature in Kelvin\n", + "\n", + "h=6.63*10**-34 #Planck's Constant\n", + "c=3*10**8 #Velocity of light\n", + "K=1.38*10**-21 #Boltzmann Constant\n", + "\n", + "#Calculations:\n", + "delE= h*c/lamda #Energy difference between two energy states N and N0\n", + "\n", + "#N=N0*e^-delE/(K*T)\n", + "R=math.e**(-delE/(K*T)) #R=Ratio of N and N0 i.e.(R=N/N0)\n", + "\n", + "#(Printing mistake in textbook)\n", + "#instead of e^-.692, it has taken e^-69.2\n", + "\n", + "print\"The ratio of population of two energy states is = \",R\n", + "print\" (calculation mistake in book)\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The ratio of population of two energy states is = 0.500588928485\n", + " (calculation mistake in book)\n" + ] + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 4.2,Page number 4-28" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "import math\n", + "\n", + "#Given Data:\n", + "lamda=6328*10**-10 #Wavelength in meter\n", + "P=4.5*10**-3 #Power in watts\n", + "h=6.63*10**-34 #Planck's Constant\n", + "c=3*10**8 #Velocity of light\n", + "\n", + "#Calculations:\n", + "delE= h*c/lamda #Energy difference\n", + "#N*delE=P\n", + "N=P/delE #number of photons emitted per second\n", + "\n", + "print\"Number of photons emitted per second is =\",N\n", + "\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Number of photons emitted per second is = 1.43167420814e+16\n" + ] + } + ], + "prompt_number": 3 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 4.3,Page number 4-29" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Given Data:\n", + "lamda=780*10**-9 #Wavelength of photon in meter\n", + "P=20*10**-3 #Power of each pulse in watts\n", + "t=10*10**-9 #Duration of each pulse\n", + "h=6.63*10**-34 #Planck's Constant\n", + "c=3*10**8 #Velocity of light\n", + "\n", + "#Calculations:\n", + "delE= h*c/lamda #Energy of each photon\n", + "E=P*t #Energy of each pulse\n", + "\n", + "N=E/delE #Number of photons in each pulse\n", + "print\"Number of photons in each pulse is =\",N\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Number of photons in each pulse is = 784313725.49\n" + ] + } + ], + "prompt_number": 6 + }, + { + "cell_type": "code", + "collapsed": false, + "input": [], + "language": "python", + "metadata": {}, + "outputs": [] + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file diff --git a/Applied_Physics-II/chapter5.ipynb b/Applied_Physics-II/chapter5.ipynb new file mode 100755 index 00000000..e9d81103 --- /dev/null +++ b/Applied_Physics-II/chapter5.ipynb @@ -0,0 +1,1166 @@ +{ + "metadata": { + "celltoolbar": "Raw Cell Format", + "name": "", + "signature": "sha256:fec7a7bbab2881090f60069cf3d4c856415ca45a1545225541a225f28ce72b8e" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Chapter 5: Foundations of Quantum Mechanics" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 5.1,Page number 5-23" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Given Data:\n", + "m=6.68*10**-27 #mass of alpha particle\n", + "V=30*10**3 #potential difference\n", + "e=1.6*10**-19 #charge of an electron\n", + "q=2*e #Charge of alpha particle\n", + "h=6.63*10**-34 #Planck's constant\n", + "\n", + "#Calculations:\n", + "lamda=h/math.sqrt(2*m*q*V) #de Broglie wavelength\n", + "print\"de Broglie wavelength associated with alpha particle is =\" ,lamda,\"m\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "de Broglie wavelength associated with alpha particle is = 5.85429607723e-14 m\n" + ] + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 5.2,Page number 5-23" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Given Data:\n", + "m=1 #mass of given particle in kg\n", + "h=6.63*10**-34 #Planck's constant\n", + "v=1*10**3 #velocity of particle\n", + "\n", + "#Calculations:\n", + "lamda=h/(m*v) #de Broglie wavelength\n", + "print\"de Broglie wavelength associated with particle is =\",lamda,\"m\"\n", + "print\"This wavelength is too small for any practical significance.\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "de Broglie wavelength associated with particle is = 6.63e-37 m\n", + "This wavelength is too small for any practical significance.\n" + ] + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 5.3,Page number 5-24" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Given Data:\n", + "m1=40*10**-3 #mass of bullet in kg\n", + "m2=9.1*10**-31 #mass of electron in kg\n", + "h=6.63*10**-34 #Planck's constant\n", + "v=1100 #velocity of bullet and electron\n", + "\n", + "#Calculations:\n", + "lamda1=h/(m1*v) #de Broglie wavelength\n", + "print\"de Broglie wavelength associated with bullet is =\",lamda1,\"m\"\n", + "\n", + "lamda2=h/(m2*v) #de Broglie wavelength\n", + "print\"de Broglie wavelength associated with electron is =\",lamda2,\"m\"\n", + "\n", + "print\"Wavelength of bullet is too small.Hence it can not be measured with help of diffraction effect.\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " de Broglie wavelength associated with bullet is = 1.50681818182e-35 m\n", + "de Broglie wavelength associated with electron is = 6.62337662338e-07 m\n", + "Wavelength of bullet is too small.Hence it can not be measured with help of diffraction effect.\n" + ] + } + ], + "prompt_number": 4 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 5.4,Page number 5-24" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Given Data:\n", + "V=100 #potential difference\n", + "d=2.15*10**-10 #lattice spacing\n", + "\n", + "#Calculations:\n", + "lamda=12.26*10**-10/(math.sqrt(V)) #wavelength associated with electron in meter\n", + "\n", + "#using bragg's law for first order lamda=2d sin(theta)\n", + "theta=math.degrees(math.asin(lamda/(2*d))) #glancing angle in degrees\n", + "print\"Glancing angle at which first reflection occurs is =\",theta,\"Degrees\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Glancing angle at which first reflection occurs is = 16.5657992687 Degrees\n" + ] + } + ], + "prompt_number": 5 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 5.5,Page number 5-25" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Given Data:\n", + "mn=1.674*10**-27 #mass of neutron\n", + "h=6.63*10**-34 #Planck's constant\n", + "lamda=1*10**-10 #wavelength of neutron\n", + "\n", + "#Calculations:\n", + "\n", + "#we know, lamda=h/sqrt(2*m*E) #de Broglie wavelength\n", + "E1=h**2/(2*mn*lamda**2) #Energy of neutron in joules\n", + "E=E1/(1.6*10**-19) #Energy of neutron in electron-Volts\n", + "\n", + "print\"Energy of neutron is =\",E,\"eV\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Energy of neutron is = 0.0820581317204 eV\n" + ] + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 5.6,Page number 5-25" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Given Data:\n", + "mn=1.67*10**-27 #mass of neutron\n", + "h=6.6*10**-34 #Planck's constant\n", + "lamda=3*10**-10 #wavelength of neutron\n", + "d=3.036*10**-10 #lattice spacing\n", + "\n", + "#Calculations:\n", + "\n", + "#we know, lamda=h/sqrt(2*m*E) #de Broglie wavelength\n", + "E1=h**2/(2*mn*lamda**2) #Energy of neutron in joules\n", + "E=E1/(1.6*10**-19) # Energy of neutron in electron-Volts\n", + "print\"Energy of neutron is =\",E,\"eV\"\n", + "\n", + "#using bragg's law for first order lamda=2d sin(theta)\n", + "theta=math.degrees(math.asin(lamda/(2*d))) #glancing angle in degrees\n", + "print\" Glancing angle at which first orde reflection occurs is =\",theta,\"Degrees\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Energy of neutron is = 0.00905688622754 eV\n", + " Glancing angle at which first orde reflection occurs is = 29.6085193042 Degrees\n" + ] + } + ], + "prompt_number": 4 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 5.7,Page number 5-26" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Given Data:\n", + "m=9.108*10**-31 #mass of electron\n", + "h=6.625*10**-34 #Planck's constant\n", + "lamda=5*10**-7 #wavelength of electron\n", + "\n", + "#Calculations:\n", + "\n", + "#we know, lamda=h/sqrt(2*m*E) #de Broglie wavelength\n", + "E1=h**2/(2*m*lamda**2) #Energy of electron in joules\n", + "E=E1/(1.6*10**-19) #Energy of electron in electron-Volts\n", + "print\"Energy of electron is =\",E,\"eV\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Energy of electron is = 6.02363650088e-06 eV\n" + ] + } + ], + "prompt_number": 5 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 5.8,Page number 5-27" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Given Data:\n", + "mn=1.676*10**-27 #mass of neutron\n", + "me=9.1*10**-31 #mass of electron\n", + "h=6.625*10**-34 #Planck's constant\n", + "\n", + "#Calculations:\n", + "#Part 1:\n", + "En1=0.025 #Energy in eV of neutron\n", + "En=En1*(1.6*10**-19) #Energy in joules\n", + "\n", + "lamda1=h/math.sqrt(2*mn*En) #wavelength of a beam of neutron\n", + "print\"wavelength of a beam of neutron is =\",lamda1,\"m\"\n", + "\n", + "#Part 2:\n", + "lamda2=2*10**-10 #wavelength of electron and photon\n", + "\n", + "#//we know, lamda=h/sqrt(2*m*E) #de Broglie wavelength\n", + "Ee1=h**2/(2*me*lamda2**2) #Energy of electron in joules\n", + "Ee=Ee1/(1.6*10**-19) #Energy of electron in electron-Volts\n", + "print\"Energy of electron is =\",Ee,\"eV\"\n", + "\n", + "p1=h/lamda2 #momentum of electron\n", + "print\" Momentum of electron is =\",p1,\"kg.m/s\"\n", + "\n", + "C=3*10**8 #Velocity of light\n", + "Ep=h*C/lamda2 #Energy of photon in joules\n", + "print\"Energy of photon is =\",Ep,\"Joules\"\n", + "\n", + "p2=h/lamda2 #momentum of photon\n", + "print\"Momentum of photon is =\",p2,\"kg.m/s\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "wavelength of a beam of neutron is = 1.80927208246e-10 m\n", + "Energy of electron is = 37.6808250343 eV\n", + " Momentum of electron is = 3.3125e-24 kg.m/s\n", + "Energy of photon is = 9.9375e-16 Joules\n", + "Momentum of photon is = 3.3125e-24 kg.m/s\n" + ] + } + ], + "prompt_number": 7 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 5.9,Page number 5-28" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Given data:\n", + "#We have alpha particle,neutron,proton and electron.\n", + "\n", + "#To find: shortest wavelength\n", + "\n", + "print\"We know, lamda=h/sqrt(2*m*E) #de Broglie wavelength\"\n", + "\n", + "#Wavelength is inversely proportional to mass of particle for constant energy\n", + "print\"i.e., Wavelength is inversely proportional to mass of particle for constant energy. \"\n", + "\n", + "print\"We have alpha particle,neutron,proton and electron.\"\n", + "\n", + "#AS,alpha particle has highest mass.Thus it will have shortest wavelength.\n", + "print\"Out of above, alpha particle has highest mass.\"\n", + "\n", + "print\"Hence it will have shortest wavelength.\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "We know, lamda=h/sqrt(2*m*E) #de Broglie wavelength\n", + "i.e., Wavelength is inversely proportional to mass of particle for constant energy. \n", + "We have alpha particle,neutron,proton and electron.\n", + "Out of above, alpha particle has highest mass.\n", + "Hence it will have shortest wavelength.\n" + ] + } + ], + "prompt_number": 9 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 5.10,Page number 5-28" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Given Data:\n", + "me=9.108*10**-31 # mass of electron\n", + "mp=1.66*10**-27 # bmass of proton\n", + "h=6.625*10**-34 # Planck's constant\n", + "lamda=1*10**-10 # wavelength of electron and proton\n", + "\n", + "#Calculations:\n", + "\n", + "#we know, lamda=h/sqrt(2*m*E) #de Broglie wavelength\n", + "Ee1=h**2/(2*me*lamda**2) #Energy of electron in joules\n", + "Ee=Ee1/(1.6*10**-19) #Energy of electron in electron-Volts\n", + "print\"Energy of electron is =\",Ee,\"eV\"\n", + "\n", + "Ep1=h**2/(2*mp*lamda**2) #Energy of photon in joules\n", + "Ep=Ep1/(1.6*10**-19) #Energy of photon in electron-Volts\n", + "print\"Energy of photon is =\",Ep,\"eV\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Energy of electron is = 150.590912522 eV\n", + "Energy of photon is = 0.0826254235693 eV\n" + ] + } + ], + "prompt_number": 10 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 5.11,Page number 5-29" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Given Data:\n", + "m1=50*10**-9 #mass of particle in kg\n", + "m2=9.1*10**-31 #mass of electron in kg\n", + "h=6.625*10**-34 #Planck's constant\n", + "v1=1 #velocity of particle\n", + "v2=3*10**6 #velocity of electron\n", + "\n", + "#Calculations:\n", + "lamda1=h/(m1*v1)*10**10 #de Broglie wavelength\n", + "print\"de Broglie wavelength associated with particle is =\",lamda1,\"Angstrom\"\n", + "\n", + "lamda2=h/(m2*v2)*10**10 #de Broglie wavelength\n", + "print\"de Broglie wavelength associated with electron is =\",lamda2,\"Angstrom\"\n", + "\n", + "print\"Wavelength of electron is measurable.\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "de Broglie wavelength associated with particle is = 1.325e-16 Angstrom\n", + "de Broglie wavelength associated with electron is = 2.42673992674 Angstrom\n", + "Wavelength of electron is measurable.\n" + ] + } + ], + "prompt_number": 13 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 5.12,Page number 5-29" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Given Data:\n", + "me=9.1*10**-31 #mass of electron in kg\n", + "h=6.63*10**-34 #Planck's constant\n", + "\n", + "#Calculations:\n", + "\n", + "E1=2*10**3 #Energy in eV of electron\n", + "E=E1*(1.6*10**-19) #Energy in joules\n", + " \n", + "lamda=h/math.sqrt(2*me*E) #wavelength of electron\n", + "print\"Wavelength of electron is =\",lamda,\"m\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Wavelength of electron is = 2.7472794985e-11 m\n" + ] + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 5.13,Page number 5-30" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Given Data:\n", + "me=9.1*10**-31 #mass of electron\n", + "h=6.63*10**-34 #Planck's constant\n", + "lamda=2*10**-10 #wavelength of electron and photon\n", + "\n", + "#Calculations:\n", + "p1=h/lamda #momentum of electron\n", + "print\"Momentum of electron is =\",p1,\"kg.m/s\"\n", + "\n", + "Ee=p1**2/(2*me) #Energy of electron in joules\n", + "print\"Energy of electron is =\",Ee,\"Joules\"\n", + "\n", + "p2=h/lamda #momentum of photon\n", + "print\"Momentum of photon is =\",p2,\"kg.m/s\"\n", + "\n", + "c=3*10**8 #Velocity of light\n", + "Ep=h*c/lamda #Energy of photon in joules\n", + "print\"Energy of photon is =\",Ep,\"Joules\"\n", + "\n", + "\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Momentum of electron is = 3.315e-24 kg.m/s\n", + "Energy of electron is = 6.03803571429e-18 Joules\n", + "Momentum of photon is = 3.315e-24 kg.m/s\n", + "Energy of photon is = 9.945e-16 Joules\n" + ] + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 5.14,Page number 5-31" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Given Data:\n", + "m=1.676*10**-27 #mass of neutron\n", + "h=6.625*10**-34 #Planck's constant\n", + "lamda=1*10**-10 #wavelength of neutron\n", + "\n", + "#Calculations:\n", + "C=3*10**8 #Velocity of light\n", + "Ep1=h*C/lamda #Energy of photon in joules\n", + "E1=Ep1/(1.6*10**-19) #Energy of photon in electron-Volts\n", + "print\"Energy of photon is =\",E1,\"eV\"\n", + "\n", + "#we know, lamda=h/sqrt(2*m*E) #de Broglie wavelength\n", + "En1=h**2/(2*m*lamda**2) #Energy of neutron in joules\n", + "E2=En1/(1.6*10**-19) #Energy of neutron in electron-Volts\n", + "print\"Energy of neutron is =\",E2,\"eV\"\n", + "\n", + "R=E1/E2 #Ratio of energies of proton to neutron\n", + "print\"Ratio of energies of proton to neutron is =\",R\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Energy of photon is = 12421.875 eV\n", + "Energy of neutron is = 0.0818366367094 eV\n", + "Ratio of energies of proton to neutron is = 151788.679245\n" + ] + } + ], + "prompt_number": 3 + }, + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Example 5.14.1,Page number 5-36" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Given Data:\n", + "v=900 #velocity of electron in m/s\n", + "delv=v*0.001/100 #uncertainity in velocity\n", + "h=6.63*10**-34 #Planck's constant\n", + "m=9.1*10**-31 #mass of an electron\n", + "\n", + "#Calculations:\n", + "delp=m*delv #uncertainity in the measured values of momentum\n", + "\n", + "#using heisenberg's uncertainity formula\n", + "delx=h/(2*3.142*delp) #uncertainity in its position\n", + "print\"Uncertainity with which position of electron can be located is >=\",delx,\"m\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Uncertainity with which position of electron can be located is >= 0.0128823012337 m\n" + ] + } + ], + "prompt_number": 4 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 5.14.2,Page number 5-37" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Given Data:\n", + "m=1.6*10**-27 #mass of proton in kg\n", + "h=6.63*10**-34 #Planck's constant\n", + "v=3./20*10**8 #velocity of particle\n", + "\n", + "#Calculations:\n", + "lamda=h/(m*v) #de Broglie wavelength\n", + "print\"de Broglie wavelength associated with proton is =\",lamda,\"m\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "de Broglie wavelength associated with proton is = 2.7625e-14 m\n" + ] + } + ], + "prompt_number": 6 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 5.14.3,Page number 5-37" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Given Data:\n", + "m=1.676*10**-27 #mass of neutron\n", + "h=6.634*10**-34 #Planck's constant\n", + "\n", + "#Calculations:\n", + "E1=0.025 #Energy in eV of neutron\n", + "E=E1*(1.6*10**-19) #Energy in joules\n", + "#As E=m*v**2/2\n", + "v=math.sqrt(2*E/m) #Velocity of neutron beam\n", + "\n", + "lamda=h/(m*v) #wavelength of a beam of neutron\n", + "print\"wavelength of a beam of neutron is =\",lamda,\"m\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "wavelength of a beam of neutron is = 1.81172996152e-10 m\n" + ] + } + ], + "prompt_number": 8 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 5.14.4,Page number 5-37" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Given Data:\n", + "delx=10*10**-9 #uncertainity in position of electron\n", + "h=6.63*10**-34 #Planck's constant\n", + "m=9.1*10**-31 #mass of an electron\n", + "E=10**3*1.6*10**-19 #Energy of electron in joules\n", + "\n", + "#Calculations:\n", + "p=math.sqrt(2*m*E) #momentum of electron\n", + "#using heisenberg's uncertainity formula\n", + "delp=h/(2*math.pi*delx) #uncertainity in the momentum\n", + "\n", + "P=delp/p*100 #percentage of uncertainity in momentum\n", + "print\"Percentage of uncertainity in momentum of electron is =\",P,\"percent\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Percentage of uncertainity in momentum of electron is = 0.0618355139385 percent\n" + ] + } + ], + "prompt_number": 9 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 5.15,Page number 5-31" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Given Data:\n", + "m=1.676*10**-27 #mass of neutron\n", + "h=6.63*10**-34 #Planck's constant\n", + "lamda=2*10**-12 #wavelength of neutron\n", + "c=3*10**8 #Velocity of light\n", + "\n", + "#Calculations:\n", + "p=h/lamda #momentum of neutron\n", + "KE=p**2/(2*m) #Kinetic Energy of neutron in joules\n", + "print\"Kinetic Energy of electron is =\",KE,\"Joules\"\n", + "\n", + "#velocity of particle is same as group velocity. Thus,\n", + "vg=p/m #group velocity\n", + "print\"group velocity of neutron is =\",vg,\"m/s\"\n", + "\n", + "#using, vg*vp=c**2\n", + "vp=c**2/vg #phase velocity\n", + "print\" phase velocity of neutron is =\",vp,\"m/s\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Kinetic Energy of electron is = 3.27840841289e-17 Joules\n", + "group velocity of neutron is = 197792.362768 m/s\n", + " phase velocity of neutron is = 4.55022624434e+11 m/s\n" + ] + } + ], + "prompt_number": 11 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 5.16,Page number 5-32" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Given Data:\n", + "m=1.157*10**-30 #mass of particle in kg\n", + "h=6.63*10**-34 #Planck's constant\n", + "c=3*10**8 #Velocity of light\n", + "\n", + "#Calculations:\n", + "E1=80 #Energy in eV of particle\n", + "E=E1*(1.6*10**-19) #Energy in joules\n", + " \n", + "lamda=h/math.sqrt(2*m*E) #wavelength of particle\n", + "print\"Wavelength of particle is =\",lamda,\"m\"\n", + "\n", + "#Now,\n", + "vg=h/(lamda*m) #group velocity\n", + "print\"Group velocity of particle is =\",vg,\"m/s\"\n", + "\n", + "#using, vg*vp=c**2\n", + "vp=c**2/vg #phase velocity\n", + "print\"Phase velocity of particle is =\",vp,\"m/s\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Wavelength of particle is = 1.21822320075e-10 m\n", + "Group velocity of particle is = 4703848.2563 m/s\n", + "Phase velocity of particle is = 19133270270.7 m/s\n" + ] + } + ], + "prompt_number": 12 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 5.17,Page number 5-33" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Given Data:\n", + "v=400 #velocity of electron in m/s\n", + "delv=0.01/100 #uncertainity in velocity\n", + "h=6.63*10**-34 #Planck's constant\n", + "m=9.11*10**-31 #mass of an electron\n", + "\n", + "#Calculations:\n", + "p=m*v #momentum of an electron\n", + "delp=p*delv #uncertainity in the measured values of momentum\n", + "\n", + "#using heisenberg's uncertainity formula\n", + "delx=h/(2*math.pi*delp) #accuracy in its position\n", + "print\"Accuracy in its position is >=\",delx,\"m\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Accuracy in its position is >= 0.00289571150576 m\n" + ] + } + ], + "prompt_number": 3 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 5.18,Page number 5-33" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "#Given Data:\n", + "delx=10**-8 #maximum uncertainity in position of electron\n", + "h=6.63*10**-34 #Planck's constant\n", + "m=9.1*10**-31 #mass of an electron\n", + "\n", + "#Calculations:\n", + "#using heisenberg's uncertainity formula\n", + "delp=h/(2*math.pi*delx) #minimum uncertainity in the measured values of momentum\n", + "\n", + "delv=delp/m #minimum uncertainity in the velocity of an electron\n", + "print\"Minimum uncertainity in the velocity of an electron is =\",delv,\"m/s\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Minimum uncertainity in the velocity of an electron is = 11595.5744253 m/s\n" + ] + } + ], + "prompt_number": 4 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 5.19,Page number 5-34" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Given Data:\n", + "delv=2*10**4 #uncertainity in velocity\n", + "h=6.63*10**-34 #Planck's constant\n", + "m=9.1*10**-31 #mass of an electron\n", + "\n", + "#Calculations:\n", + "delp=m*delv #uncertainity in the measured values of momentum\n", + "\n", + "#using heisenberg's uncertainity formula\n", + "delx=h/(2*math.pi*delp) #accuracy in its position\n", + "print\"Minimum space required by electron to be confined in an atom is >=\",delx,\"m\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Minimum space required by electron to be confined in an atom is >= 5.79778721263e-09 m\n" + ] + } + ], + "prompt_number": 5 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 5.20,Page number 5-34" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Given Data:\n", + "delt=1.4*10**-10 #uncertainity in time spent by nucleus in excited state\n", + "h=6.63*10**-34 #Planck's constant\n", + "\n", + "#Calculations:\n", + "\n", + "#using, delE*delt>= h/(2*math.pi)\n", + "delE1= h/(2*math.pi*delt) #uncertaininty in its energy in excited state in joules\n", + "delE=delE1/(1.6*10**-19) #uncertaininty in its energy in excited state in eV\n", + "print\"Uncertaininty in its energy in excited state is >=\",delE,\"eV\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Uncertaininty in its energy in excited state is >= 4.71070211026e-06 eV\n" + ] + } + ], + "prompt_number": 6 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 5.21,Page number 5-35" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Given Data:\n", + "a=2*10**-10 #width of potential well in m\n", + "h=6.63*10**-34 #Planck's constant\n", + "m=9.1*10**-31 #mass of an electron\n", + "\n", + "#Calculations:\n", + "#we know equation for energy of an electron\n", + "n0=1\n", + "E01=n0**2*h**2/(8*m*a**2) #Energy in ground state\n", + "E0=E01/(1.6*10**-19) #Energy in eV\n", + "print\"Energy of an electron in ground state is=\",E0,\"eV\"\n", + "\n", + "n1=2\n", + "E11=n1**2*h**2/(8*m*a**2) #Energy in first excited state\n", + "E1=E11/(1.6*10**-19) #Energy in eV\n", + "print\" Energy of an electron in first excited state is=\",E1,\"eV\"\n", + "\n", + "\n", + "n2=3\n", + "E21=n2**2*h**2/(8*m*a**2) #Energy in second excited state\n", + "E2=E21/(1.6*10**-19) #Energy in eV\n", + "print\"Energy of an electron in second excited state is=\",E2,\"eV\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Energy of an electron in ground state is= 9.43443080357 eV\n", + " Energy of an electron in first excited state is= 37.7377232143 eV\n", + "Energy of an electron in second excited state is= 84.9098772321 eV\n" + ] + } + ], + "prompt_number": 8 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 5.22,Page number 5-36" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Given Data:\n", + "a=25*10**-10 #width of well\n", + "delx=5*10**-10 #uncertainity in position of particle\n", + "n=1 #ground state\n", + "\n", + "#calculation:\n", + "x1=a/2\n", + "psi1=math.sqrt(2/a)*math.sin(n*math.pi/a*x1)\n", + "P1=(psi1**2)*delx #Probability of finding particle at distance of x1\n", + "print\"Probability of finding particle at a distance of x1 is =\",P1\n", + "\n", + "x2=a/3\n", + "psi2=math.sqrt(2/a)*math.sin(n*math.pi/a*x2)\n", + "P2=(psi2**2)*delx #Probability of finding particle at distance of x2\n", + "print\"Probability of finding particle at a distance of x2 is =\",P2\n", + "print\"(There is print mistake in book).\"\n", + "\n", + "x3=a\n", + "psi3=math.sqrt(2/a)*math.sin(n*math.pi/a*x3)\n", + "P3=(psi3**2)*delx #Probability of finding particle at distance of x3\n", + "print\"Probability of finding particle at a distance of x3 is =\",P3\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Probability of finding particle at a distance of x1 is = 0.4\n", + "Probability of finding particle at a distance of x2 is = 0.3\n", + "(There is print mistake in book).\n", + "Probability of finding particle at a distance of x3 is = 5.99903913065e-33\n" + ] + } + ], + "prompt_number": 9 + }, + { + "cell_type": "code", + "collapsed": false, + "input": [], + "language": "python", + "metadata": {}, + "outputs": [] + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file diff --git a/Applied_Physics-II/chapter6.ipynb b/Applied_Physics-II/chapter6.ipynb new file mode 100755 index 00000000..dec8353e --- /dev/null +++ b/Applied_Physics-II/chapter6.ipynb @@ -0,0 +1,1183 @@ +{ + "metadata": { + "celltoolbar": "Raw Cell Format", + "name": "", + "signature": "sha256:bc506d3130781232c57938d478c5c5179ee56efb068c1be93d57258ec96fcd79" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Chapter 6: Magnetic Materials and Circuits" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 6.1,Page number 2-26" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Given Data:\n", + "H=198 #Magnetizing Force in Ampere per meter\n", + "M=2300 #Magnetization in Ampere per meter\n", + "u0=4*math.pi*10**-7 #Permeability in vacuum\n", + "\n", + "#Calculations:\n", + "#H=(B/u0)-M\n", + "B=u0*(H+M) #Flux Density\n", + "ur=B/(u0*H) #Relative Permeability\n", + "\n", + "print\"Corresponding Flux Density is =\",B,\"Wb/m^2\"\n", + "print\"Relative Permeability is =\",ur\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Corresponding Flux Density is = 0.00313907937947 Wb/m^2\n", + "Relative Permeability is = 12.6161616162\n" + ] + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 6.2,Page number 2-26" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Given Data:\n", + "x=3.7*10**-3 #Susceptibility at T=300 K\n", + "T=300 #Temperature in kelvin\n", + "T1=250 #Temperature in kelvin\n", + "T2=600 #Temperature in kelvin\n", + "\n", + "#Calculations:\n", + "C=x*T #Curie's law\n", + "ur1=C/T1 #Relative permeability at 250 K\n", + "ur2=C/T2 #Relative permeability at 600 K\n", + "\n", + "print\"Relative Permeability at 250 K is =\",ur1\n", + "print\"Relative Permeability at 600 K is =\",ur2\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Relative Permeability at 250 K is = 0.00444\n", + "Relative Permeability at 600 K is = 0.00185\n" + ] + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 6.3,Page number 2-27" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Given Data:\n", + "u=0.8*10**-23 #Magnetic dipole moment of an atom in paramagnetic gas in J/T\n", + "B=0.8 #Magnetic field in tesla\n", + "K=1.38*10**-23 #Boltzmann constant\n", + "\n", + "#To find Temperature at which Average thermal energy is equal to Magnetic energy \n", + "#i.e. uB=3KT/2\n", + "T=2*u*B/(3*K) #Required temperature\n", + "\n", + "print\"Required temperature is =\",T,\"Kelvin\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Required temperature is = 0.309178743961 Kelvin\n" + ] + } + ], + "prompt_number": 3 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 6.4,Page number 2-27" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Given Data:\n", + "T=27+273 #Temperature in kelvin\n", + "B=0.5 #Magnetic field in tesla\n", + "C=2*10**-3 #Curie's Constant\n", + "u0=4*math.pi*10**-7 #Permeability in vacuum\n", + "\n", + "# C=u0*M*T/B (Curie's law)\n", + "M=C*B/(u0*T) #Magnetization of material at 300 K\n", + "\n", + "print\"Magnetization of material at 300 K is =\",M,\"A/m\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Magnetization of material at 300 K is = 2.65258238486 A/m\n" + ] + } + ], + "prompt_number": 5 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 6.5,Page number 2-27" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Given Data:\n", + "B=10.9*10**-5 #Horizontal component of B in wb/m^2\n", + "u0=4*math.pi*10**-7 #Permeability in free space\n", + "\n", + "H=B/u0 #Horizontal component of magnetic field\n", + "print\"Horizontal component of magnetic field is =\",H,\"Ampere/meter\"\n", + "print\"(Print mistake in unit in book)\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Horizontal component of magnetic field is = 86.7394439851 Ampere/meter\n", + "(Print mistake in unit in book)\n" + ] + } + ], + "prompt_number": 7 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 6.6,Page number 2-28" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Given Data:\n", + "u0=4*math.pi*10**-7 #Permeability in vacuum\n", + "ur=900 #Relative permeability of medium\n", + "l=2 #length in meter\n", + "A=60*10**-4 #Crosss sectional area of ring in m^2\n", + "phi=5.9*10**-3 #flux in weber\n", + "n=700 #Number of turns\n", + "\n", + "#Calculations:\n", + "#We know, phi=B*A\n", + "B=phi/A #Flux density\n", + "#But, B=u*H\n", + "H=B/(u0*ur) #Magnetic field strength\n", + "\n", + "I=H*l/n #Required current\n", + "print\"Current required to produce given flux is =\",I,\"Ampere\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Current required to produce given flux is = 2.48416445567 Ampere\n" + ] + } + ], + "prompt_number": 8 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 6.7,Page number 2-28" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Given Data:\n", + "\n", + "u0=4*math.pi*10**-7 #Permeability in vacuum\n", + "ur=900 #Relative permeability of medium\n", + "r=25*10**-2 #radius of ring\n", + "A=25*10**-4 #Crosss sectional area of ring in m^2\n", + "Ag=1*10**-3 #Air gap\n", + "phi=2.7*10**-3 #flux in weber\n", + "N=400 #Number of turns\n", + "\n", + "#Calculations:\n", + "#We know, phi=B*A\n", + "B=phi/A #Flux density\n", + "#But, B=u*H\n", + "H=B/(u0*ur) #Magnetic field strength\n", + "L=H*2*math.pi*r+(B*Ag/u0) #Total amp turns required (iron+air)\n", + "I=L/N #Required current\n", + "\n", + "print\"Current required to produce given flux is =\",I,\"Ampere\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Current required to produce given flux is = 5.89859173174 Ampere\n" + ] + } + ], + "prompt_number": 10 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 6.8,Page number 2-29" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Given Data:\n", + "\n", + "u0=4*math.pi*10**-7 #Permeability in vacuum\n", + "A=0.2*10**-4 #Crosss sectional area of iron bar in m^2\n", + "H=1600 #magnetising field in A/m\n", + "phi=2.4*10**-5 #Magnetic flux in weber\n", + "\n", + "\n", + "#Calculations:\n", + "#We know, phi=B*A\n", + "B=phi/A #Flux density\n", + "u=B/H #magnetic permeability\n", + "ur=u/u0 #relative permeability\n", + "xm=ur-1 #susceptibility of the iron bar\n", + "\n", + "print\"magnetic permeability of iron bar is =\",u,\"N/(A^2)\"\n", + "print\"susceptibility of the iron bar is =\",xm\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "magnetic permeability of iron bar is = 0.00075 N/(A^2)\n", + "susceptibility of the iron bar is = 595.831036595\n" + ] + } + ], + "prompt_number": 11 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 6.9,Page number 2-29" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Given Data:\n", + "u0=4*math.pi*10**-7 #Permeability in vacuum\n", + "xm=948*10**-11 #susceptibility of the iron bar\n", + "\n", + "#Calculations:\n", + "ur=1+xm #relative permeability\n", + "u=u0*ur #permeability of medium\n", + "\n", + "print\"Relative Permeability of medium is =\",ur\n", + "print\"Permeability of medium is =\",u,\"H/m\"\n", + "\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Relative Permeability of medium is = 1.00000000948\n", + "Permeability of medium is = 1.25663707335e-06 H/m\n" + ] + } + ], + "prompt_number": 13 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 6.10,Page number 2-30" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Given Data:\n", + "B=2.5 #Magnetic field in tesla\n", + "u0=4*math.pi*10**-7 #Permeability in free space\n", + "i0=0.7 #current in the core\n", + "ri=11*10**-2 #inner radii of core\n", + "ro=12*10**-2 #outer radii of core\n", + "\n", + "#Calculations:\n", + "r=(ri+ro)/2 #Average radii of core\n", + "n=3000/(2*math.pi*r) #Number of turns\n", + "\n", + "#We know, B=u0*ur*n*i0 .Thus,\n", + "ur=B/(u0*n*i0)\n", + "\n", + "print\"Relative Permeability of medium is =\",ur\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Relative Permeability of medium is = 684.523809524\n" + ] + } + ], + "prompt_number": 14 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 6.11,Page number 2-31" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Given Data:\n", + "B=1.0 #Flux density in tesla\n", + "u0=4*math.pi*10**-7 #Permeability in free space\n", + "i=2.0 #current in the core\n", + "n=10*100 #n=N/l i.e. turns per meter\n", + "\n", + "#Calculations:\n", + "H=n*i #Magnetising force produced in wire\n", + "print\"Magnetising force produced in wire is =\",H,\"Amp-turn/meter\"\n", + "\n", + "#We know that, B=u0(H+I).Thus,\n", + "I=B/u0-H #Magnetisation of material\n", + "print\"Magnetisation of material is =\",I,\"Amp-turn/meter\"\n", + "\n", + "#u=B/H, i.e. ur*u0=B/H.\n", + "ur=B/(u0*H) #Relative permeability of core\n", + "print\" Relative Permeability of core is =\",ur\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Magnetising force produced in wire is = 2000.0 Amp-turn/meter\n", + "Magnetisation of material is = 793774.715459 Amp-turn/meter\n", + " Relative Permeability of core is = 397.88735773\n" + ] + } + ], + "prompt_number": 15 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 6.12,Page number 2-31" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Given Data:\n", + "M=40 #Mass of an iron core\n", + "D=7.5*10**3 #Density of iron\n", + "f=100 #Frequency\n", + "A=3800*10**-1 #Loss due to Area of hysterisis loop in J/m^3\n", + "\n", + "#Calculations:\n", + "V=M/D #Volume of iron core\n", + "L1=A*V #Loss of energy in core per cycle\n", + "print\"Loss of energy in core per cycles is =\",L1,\"joules\"\n", + "\n", + "N=f*60 #Number of cycles per minute\n", + "L=L1*N #Loss of energy per minute\n", + "\n", + "print\"Loss of energy per minute is =\",L,\"joules\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Loss of energy in core per cycles is = 2.02666666667 joules\n", + "Loss of energy per minute is = 12160.0 joules\n" + ] + } + ], + "prompt_number": 16 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 6.13,Page number 2-32" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Given Data:\n", + "u0=4*math.pi*10**-7 #Permeability in vacuum\n", + "l=30*10**-2 #length in meter\n", + "A=1*10**-4 #Crosss sectional area of ring in m^2\n", + "phi=2*10**-6 #flux in weber\n", + "N=300 #Number of turns\n", + "I=0.032 #Current in winding\n", + "\n", + "#Calculations:\n", + "#(i):\n", + "B=phi/A #Flux density\n", + "print\"(i)Flux Density in the ring is =\",B,\"Wb/m^2\"\n", + "\n", + "#(ii):\n", + "H=N*I/l #Magnetic intensity\n", + "print\"(ii)Magnetic intensity is =\",H,\"Amp-turn/meter\"\n", + "\n", + "#(iii):\n", + "u=B/H #Permeability of ring\n", + "print\"(iii)Permeability of ring is =\",u,\" Wb/A-m\"\n", + "ur=u/u0 #Relative permeability of ring\n", + "print\"Relative Permeability of ring is =\",ur\n", + "\n", + "#(iv):\n", + "xm=ur-1 #susceptibility of the ring\n", + "print\"(iv)Magnetic susceptibility of the ring is =\",xm\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "(i)Flux Density in the ring is = 0.02 Wb/m^2\n", + "(ii)Magnetic intensity is = 32.0 Amp-turn/meter\n", + "(iii)Permeability of ring is = 0.000625 Wb/A-m\n", + "Relative Permeability of ring is = 497.359197162\n", + "(iv)Magnetic susceptibility of the ring is = 496.359197162\n" + ] + } + ], + "prompt_number": 17 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 6.14,Page number 2-32" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Given Dta:\n", + "M=12*10**3 #Mass of an iron core in grams\n", + "D=7.5 #Density of iron in gm/cc\n", + "f=50 #Frequency\n", + "A=3000 #loss due to Area of hysterisis loop in ergs/cm^3\n", + "\n", + "#Calculations:\n", + "V=M/D #Volume of iron core\n", + "L1=A*V #Loss of energy in core per cycle\n", + "\n", + "L=L1*f*3600 #Loss of energy per hour\n", + "\n", + "print\"Loss of energy per hour is =\",L,\"Erg\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Loss of energy per hour is = 8.64e+11 Erg\n" + ] + } + ], + "prompt_number": 18 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 6.15,Page number 2-33" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Given Data:\n", + "A=0.5*10**3 #Area of B-H loop in Joules per m^3\n", + "V=10**-3 #Volume of specimen in m^3\n", + "n=50 #Frequency of a.c.\n", + "\n", + "#Calculations:\n", + "H=n*V*A #Hysteresis power loss\n", + "\n", + "print\"Hysteresis power loss is =\",H,\"Watt\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Hysteresis power loss is = 25.0 Watt\n" + ] + } + ], + "prompt_number": 19 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 6.16,Page number 2-33" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Given Data:\n", + "u0=4*math.pi*10**-7 #Permeability in vacuum\n", + "ur=1000 #Relative permeability of medium\n", + "V=10**-4 #Volume of iron rod in m^3\n", + "n=500 #Number of turns per meter\n", + "i=0.5 #Current in windings of solenoid in Amperes\n", + "\n", + "#Calculations:\n", + "#We know I=(ur-1)H\n", + "#and H=ni , hence\n", + "I=(ur-1)*n*i #Intensity of magnetisation\n", + "M=I*V #Magnetic moment\n", + "\n", + "print\"Magnetic moment of the rod is =\",M,\"A-m^2\"\n", + "\n", + "\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Magnetic moment of the rod is = 24.975 A-m^2\n" + ] + } + ], + "prompt_number": 22 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 6.17,Page number 2-34" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Given Data:\n", + "u0=4*math.pi*10**-7 #Permeability in vacuum\n", + "ur=600 #Relative permeability of iron\n", + "d=12*10**-2 #mean diameter of ring in m\n", + "N=500 #Number of turns\n", + "i=0.3 #Current in windings of solenoid in Amperes\n", + "\n", + "#Calculations:\n", + "r=d/2 #Radius of ring\n", + "\n", + "B=u0*ur*N*i/(2*math.pi*r) #Flux densityin the core\n", + "print\"Flux densityin the core is =\",B,\"Wb/m^2\"\n", + "\n", + "H=B/(u0*ur) #Magnetic intensity\n", + "print\"Magnetic intensity is =\",H,\"Amp-turns/m\"\n", + "\n", + "#We know that, B=u0(H+I)\n", + "I1=(B-u0*H)/u0 #magnetisation\n", + "I2=u0*I1 #Electronic current loop\n", + "\n", + "I=I2/B*100 #Percentage flux density due to electroniuc loop currents\n", + "print\"Percentage flux density due to electroniuc loop currents is =\",I,\"percent\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Flux densityin the core is = 0.3 Wb/m^2\n", + "Magnetic intensity is = 397.88735773 Amp-turns/m\n", + "Percentage flux density due to electroniuc loop currents is = 99.8333333333 percent\n" + ] + } + ], + "prompt_number": 23 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 6.18,Page number 2-35" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Given Data:\n", + "\n", + "u0=4*math.pi*10**-7 #Permeability in vacuum\n", + "ur=900 #Relative permeability of iron ring\n", + "d=40*10**-2 #diameter of ring\n", + "l=5*10**-3 #air gap in the ring\n", + "A=5.8*10**-4 #Crosss sectional area of ring in m^2\n", + "phi=1.5*10**-4 #flux in weber\n", + "N=600 #Number of turns\n", + "\n", + "#Calculations:\n", + "r=d/2 #Radius of ring\n", + "\n", + "#We know, phi=B*A\n", + "B=phi/A #Flux density\n", + "\n", + "#But, B=u*H\n", + "H=B/(u0*ur) #Magnetic field strength\n", + "\n", + "m1=H*ur*l #amp-turns in air gap\n", + "m2=H*2*math.pi*r #amp-turns by ring\n", + "m=m1+m2 #total mmf(amp-turns) required\n", + "\n", + "I=m/N #Required current\n", + "print\"Current required to produce given flux is =\",I,\"Amperes\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Current required to produce given flux is = 2.19395891742 Amperes\n" + ] + } + ], + "prompt_number": 24 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 6.18.1,Page number 2-38" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Given Data:\n", + "u0=4*math.pi*10**-7 #Permeability in vacuum\n", + "X=-0.5*10**-5 #Magnetic susceptibility of silicon\n", + "H=9.9*10**4 #Magnetic field intensity\n", + "\n", + "#Calculations:\n", + "\n", + "#As, X=I/H. thus,\n", + "I=X*H #intensity of magnetisation\n", + "print\"Intensity of magnetisation is =\",I\n", + "\n", + "B=u0*(H+I) #Magnetic flux density\n", + "print\"Magnetic flux density is =\",B,\"Wb/ m^2\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Intensity of magnetisation is = -0.495\n", + "Magnetic flux density is = 0.124406447047 Wb/ m^2\n" + ] + } + ], + "prompt_number": 25 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 6.18.2,Page number 2-38" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Given Data:\n", + "u0=4*math.pi*10**-7 #Permeability in vacuum\n", + "ur=380 #Relative permeability\n", + "d=20*10**-2 #diameter of solenoid in m\n", + "r=d/2 #radius of ring in m\n", + "A=5*10**-4 #Crosss sectional area of ring in m^2\n", + "phi=2*10**-3 #flux in weber\n", + "N=200 #Number of turns\n", + "\n", + "#Calculations:\n", + "l=math.pi*d #air gap in the ring\n", + "S=(l/(u0*ur*A)) #Reluctance of iron ring\n", + "print\"Reluctance of iron ring is =\",S,\"Amp-turn/ Wb \"\n", + "\n", + "#ohm's law for magnetic circuit is phi=N*I/S. thus,\n", + "I=S*phi/N #required current\n", + "print\"Current required to obtain given magnetic flux is =\",I,\"Amperes\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Reluctance of iron ring is = 2631578.94737 Amp-turn/ Wb \n", + "Current required to obtain given magnetic flux is = 26.3157894737 Amperes\n" + ] + } + ], + "prompt_number": 27 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 6.18.3,Page number 2-39" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Given Values:\n", + "u0=4*math.pi*10**-7 #Permeability in vacuum\n", + "ur=1 #Relative permeability of air\n", + "r=15*10**-2 #radius of ring in m\n", + "A=6*10**-4 #Crosss sectional area of ring in m^2\n", + "I=4 #Coil current in amp\n", + "N=500 #Number of turns\n", + "\n", + "#Calculations:\n", + "m=N*I #MMF of coil\n", + "print\"MMF of coil is =\",m,\"Ampere-turn\"\n", + "\n", + "l=2*math.pi*r #air gap\n", + "R=(l/(u0*ur*A)) #Reluctance of iron ring\n", + "print\"Reluctance of iron ring is =\",R,\"Ampere-turn/Wb\"\n", + "\n", + "phi=m/R #Magnetic flux\n", + "print\"Magnetic flux is =\",phi,\"Weber\"\n", + "\n", + "B=phi/A #Magnetic Flux density\n", + "print\"Magnetic flux density is =\",B,\"Weber/m^2\"\n", + "\n", + "H=B/(u0*ur) #Magnetic field intensity\n", + "print\"Magnetic field intensity is =\",H,\"Amperes/m\"\n", + "\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "MMF of coil is = 2000 Ampere-turn\n", + "Reluctance of iron ring is = 1250000000.0 Ampere-turn/Wb\n", + "Magnetic flux is = 1.6e-06 Weber\n", + "Magnetic flux density is = 0.00266666666667 Weber/m^2\n", + "Magnetic field intensity is = 2122.06590789 Amperes/m\n" + ] + } + ], + "prompt_number": 28 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 6.19,Page number 2-36" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Given Data:\n", + "u0=4*math.pi*10**-7 #Permeability in vacuum\n", + "ur=6*10**-3 #Relative permeability of iron\n", + "r=0.5 #radius of ring in m\n", + "l=1*10**-2 #air gap in the ring\n", + "A=5*10**-4 #Crosss sectional area of ring in m^2\n", + "i=5 #current in ampere\n", + "N=900 #Number of turns\n", + "\n", + "#Calculations:\n", + "S=(l/(u0*A))+((2*math.pi*r-l)/ur*A) #Reluctance of iron\n", + "print\"Reluctance of iron is =\",S,\"Ampere-turn/Wb\"\n", + "\n", + "m=N*i #mmf produced\n", + "print\"mmf produced is =\",m,\"Ampere-turn\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Reluctance of iron is = 15915494.5702 Ampere-turn/Wb\n", + "mmf produced is = 4500 Ampere-turn\n" + ] + } + ], + "prompt_number": 29 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 6.20,Page number 2-36" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Given Data:\n", + "H=5*10**3 #coercivity of bar magnet in amp/m\n", + "l=10*10**-2 #length of solenoid in m\n", + "N=50 #No of turns\n", + "\n", + "#Calculations:\n", + "\n", + "#We know that, H=NI/l ,hence\n", + "I=l*H/N #current through solenoid\n", + "\n", + "print\"Current through solenoid is =\",I,\"Amperes\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Current through solenoid is = 10.0 Amperes\n" + ] + } + ], + "prompt_number": 30 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 6.21,Page number 2-36" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Given Data:\n", + "ur=1200 #Relative permeability of medium\n", + "V=10**-3 #volume of iron rod\n", + "N=5*10**2 #no of turns per m\n", + "i=0.5 #current through solenoid in amp\n", + "\n", + "#Calculations:\n", + "x=ur-1 #susceptibility of the ring\n", + "H=N*i #Magnetisisng field\n", + "\n", + "#We know, x=I/H\n", + "I=x*H #magnetisation\n", + "\n", + "#Also, I=M/V , thus\n", + "M=I*V #magnetic moment\n", + "print\"Magnetic moment is =\",M,\"Ampere-turn-m^2\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Magnetic moment is = 299.75 Ampere-turn-m^2\n" + ] + } + ], + "prompt_number": 31 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 6.22,Page number 2-37" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Given Dta:\n", + "ur=100 #Relative permeability of medium\n", + "l=0.2 #length of iron rod\n", + "d=10*10**-3 #diameter of solenoid in m\n", + "N=300 #no of turns per m\n", + "i=0.5 #current through solenoid in amp\n", + "r=d/2 #radius of solenoid\n", + "\n", + "#Calculations:\n", + "x=ur-1 #susceptibility of the ring\n", + "H=N*i #Magnetisisng field\n", + "\n", + "#We know, x=I/H\n", + "I=x*H #magnetisation\n", + "\n", + "V=math.pi*(r**2)*l #volume of iron rod\n", + "\n", + "#Also, I=M/V , thus\n", + "M=I*V #magnetic moment\n", + "print\"Magnetic moment is =\",M,\"Ampere-turn-m^2\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Magnetic moment is = 0.233263254529 Ampere-turn-m^2\n" + ] + } + ], + "prompt_number": 33 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 6.23,Page number 2-38" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Given Data:\n", + "l=1.2 #length of circuit in meter\n", + "u=7.3*10**-3 #permeability of silicon sheet\n", + "A=100 #cross sectional area in cm^2\n", + "N=150 #No of turns\n", + "B=0.3 #magmetic field in Wb/m^2\n", + "\n", + "#Calculations:\n", + "\n", + "#We know, B=u*H\n", + "H=B/u #Magnetic field strength\n", + "\n", + "m=H*l #amp-turns in air gap\n", + "\n", + "I1=m/N #Required current\n", + "print\"Current required to obtain given magnetic field is =\",I1,\"Amperes\"\n", + "\n", + "I=I1/A #Required current per unit area\n", + "print\"Current required per unit area to obtain given magnetic field is =\",I,\"Amperes\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Current required to obtain given magnetic field is = 0.328767123288 Amperes\n", + "Current required per unit area to obtain given magnetic field is = 0.00328767123288 Amperes\n" + ] + } + ], + "prompt_number": 34 + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file diff --git a/Applied_Physics-II/screenshots/Ex2_2.png b/Applied_Physics-II/screenshots/Ex2_2.png Binary files differnew file mode 100755 index 00000000..1d450952 --- /dev/null +++ b/Applied_Physics-II/screenshots/Ex2_2.png diff --git a/Applied_Physics-II/screenshots/Ex2_2_1.png b/Applied_Physics-II/screenshots/Ex2_2_1.png Binary files differnew file mode 100755 index 00000000..1d450952 --- /dev/null +++ b/Applied_Physics-II/screenshots/Ex2_2_1.png diff --git a/Applied_Physics-II/screenshots/Ex2_2_2.png b/Applied_Physics-II/screenshots/Ex2_2_2.png Binary files differnew file mode 100755 index 00000000..1d450952 --- /dev/null +++ b/Applied_Physics-II/screenshots/Ex2_2_2.png |
