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diff --git a/Applied_Physics_II/Chapter_1.ipynb b/Applied_Physics_II/Chapter_1.ipynb deleted file mode 100755 index c6e00369..00000000 --- a/Applied_Physics_II/Chapter_1.ipynb +++ /dev/null @@ -1,1108 +0,0 @@ -{ - "metadata": { - "name": "" - }, - "nbformat": 3, - "nbformat_minor": 0, - "worksheets": [ - { - "cells": [ - { - "cell_type": "heading", - "level": 1, - "metadata": {}, - "source": [ - "Chapter 1: Interference of Light" - ] - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 1.2.1, Page number 1-11" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math\n", - "\n", - "#Variable declaration\n", - "i = 45 #angle of incidence(degrees)\n", - "t = 4*10**-5 #thickness of film(cm)\n", - "u = 1.2\n", - "\n", - "#Calculations & Result\n", - "r = math.degrees(math.asin(math.sin(i*math.pi/180)/u))\n", - "for n in range(1,4):\n", - " lamda = (2*u*t*math.cos(r*math.pi/180))/n\n", - " print \"For n = %d, wavelength = %.2f A\"%(n,lamda/10**-8)\n", - "print \"Since the visible range of wavelengths lie between 4000 to 7000 A, none of the obtained wavelengths lie in this range\"" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "For n = 1, wavelength = 7756.29 A\n", - "For n = 2, wavelength = 3878.14 A\n", - "For n = 3, wavelength = 2585.43 A\n", - "Since the visible range of wavelengths lie between 4000 to 7000 A, none of the obtained wavelengths lie in this range\n" - ] - } - ], - "prompt_number": 1 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 1.2.2, Page number 1-12" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math\n", - "\n", - "#Variable declaration\n", - "r = 90 #angle of refraction(degrees)\n", - "t = 5*10**-5 #thickness of film(cm)\n", - "u = 1.33\n", - "\n", - "#Calculations & Result\n", - "for n in range(1,4):\n", - " lamda = (4*u*t*int(math.cos(math.radians(90))))/((2*n)-1)\n", - " print \"For n = %d, wavelength = %.2f A\"%(n,lamda)\n", - "print \"Since the visible range of wavelengths lie between 4000 to 7000 A, none of the obtained wavelengths do not lie in this range\"\n", - "\n", - "print \"\\nPlease note: Since r=90, cos(r)=0\\nHence, the answers given in the textbook are incorrect\"" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "For n = 1, wavelength = 0.00 A\n", - "For n = 2, wavelength = 0.00 A\n", - "For n = 3, wavelength = 0.00 A\n", - "Since the visible range of wavelengths lie between 4000 to 7000 A, none of the obtained wavelengths do not lie in this range\n", - "\n", - "Please note: Since r=90, cos(r)=0\n", - "Hence, the answers given in the textbook are incorrect\n" - ] - } - ], - "prompt_number": 33 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 1.2.3, Page number 1-12" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math\n", - "\n", - "#Variable declaration\n", - "i = 45 #angle of incidence(degrees)\n", - "t = 1.5*10**-4 #thickness of film(cm)\n", - "lamda = 5*10**-5 #wavelength(cm)\n", - "u = 4./3. #refractive index\n", - "\n", - "#Calculations\n", - "r = math.degrees(math.asin(math.sin(i*math.pi/180)/u))\n", - "n = (2*u*t*math.cos(r*math.pi/180))/lamda\n", - "\n", - "#Result\n", - "print \"The order of interfernce is %.2f, close to 7\"%n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "The order of interfernce is 6.78, close to 7\n" - ] - } - ], - "prompt_number": 3 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 12.2.4, Pae number 1-13" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math\n", - "\n", - "#Variable declaration\n", - "i = 45 #angle of incidence(degrees)\n", - "lamda = 5896*10**-8 #wavelength(cm)\n", - "u = 1.33 #refractive index\n", - "n = 1\n", - "\n", - "#Calculations\n", - "r = math.degrees(math.asin(math.sin(i*math.pi/180)/u))\n", - "t = ((2*n-1)*lamda)/(2*u*math.cos(r*math.pi/180)*2)\n", - "\n", - "#Result\n", - "print \"The required thickness is\",round((t/1E-5),2),\"*10^-5 cm\"\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "The required thickness is 1.31 *10^-5 cm\n" - ] - } - ], - "prompt_number": 4 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 1.2.5, Page number 1-14" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "\n", - "lamda1 = 7000 #wavelength(A)\n", - "lamda2 = 5000 #wavelength(A)\n", - "u = 1.3 #R.I. of oil\n", - "\n", - "#Calculations\n", - "'''\n", - "2utcosr = (2n-1)7000/2 ----(1)\n", - "2utcosr = (2n+1)5000/2 ----(2)\n", - "Divinding (1) by (2), we get the following expression\n", - "1 = (2n+1)5000\n", - " -----------\n", - " (2n-1)7000\n", - "Solving the above expression, we get,\n", - "'''\n", - "n = 12000/4000\n", - "t = ((2*n-1)*lamda)/(2*u*math.cos(r*math.pi/180)*2)\n", - "\n", - "#Result\n", - "print \"The required thickness is\",round((t/1E-5),4),\"*10^-5 cm\"" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "The required thickness is 6.6936 *10^-5 cm\n" - ] - } - ], - "prompt_number": 5 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 1.2.6, Page number 1-15" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "\n", - "import math\n", - "\n", - "#Variable declaration\n", - "i = 30 #angle of incidence(degrees)\n", - "lamda = 5890*10**-8 #wavelength(cm)\n", - "u = 1.46 #refractive index\n", - "n = 8\n", - "\n", - "#Calculations\n", - "r = math.degrees(math.asin(math.sin(i*math.pi/180)/u))\n", - "t = (n*lamda)/(2*u*math.cos(r*math.pi/180))\n", - "\n", - "#Result\n", - "print \"The required thickness is\",round((t/1E-4),3),\"*10^-4 cm\"\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "The required thickness is 1.718 *10^-4 cm\n" - ] - } - ], - "prompt_number": 6 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 1.2.7, Page number 1-15" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "\n", - "import math\n", - "\n", - "#Variable declaration\n", - "r = 60 #angle of refraction(degrees)\n", - "lamda = 5890*10**-8 #wavelength(cm)\n", - "u = 1.5 #refractive index\n", - "n = 1 #for minimumm thickness\n", - "\n", - "#Calculations\n", - "#For r = 60\n", - "t1 = (n*lamda)/(2*u*math.cos(r*math.pi/180))\n", - "\n", - "#For normal incidence \n", - "r = 0\n", - "t2 = (n*lamda)/(2*u*math.cos(r*math.pi/180))\n", - "\n", - "#Result\n", - "print \"For r = 60, the required thickness is\",round((t1/1E-5),2),\"*10^-5 cm\"\n", - "print \"For r = 0, the required thickness is\",round((t2/1E-5),2),\"*10^-5 cm\"\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "For r = 60, the required thickness is 3.93 *10^-5 cm\n", - "For r = 0, the required thickness is 1.96 *10^-5 cm\n" - ] - } - ], - "prompt_number": 7 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 1.2.8, Page number 1-16" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "\n", - "import math\n", - "\n", - "#Variable declaration\n", - "r = 0 #for normal incidence(degrees)\n", - "lamda = 5.5*10**-5 #wavelength(cm)\n", - "n = 1 #for minimumm thickness\n", - "A = 10**4 #area(cm^2)\n", - "V = 0.2 #volume(cc)\n", - "\n", - "#Calculations\n", - "t = V/A\n", - "#for nth dark band,\n", - "u = (n*lamda)/(2*t*math.cos(r*math.pi/180))\n", - "\n", - "#Result\n", - "print \"Refractive index =\",u" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Refractive index = 1.375\n" - ] - } - ], - "prompt_number": 2 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 1.2.9, Page number 1-17" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "\n", - "import math\n", - "\n", - "#Variable declaration\n", - "r = 60 #angle of incidence(degrees)\n", - "t = 2*10**-7 #thickness of film(cm)\n", - "u = 1.2\n", - "\n", - "#Calculations & Result\n", - "for n in range(1,4):\n", - " lamda = (4*u*t*math.cos(r*math.pi/180))/(2*n-1)\n", - " print \"For n = %d, wavelength = %.2f A\"%(n,lamda/10**-10)\n", - "print \"Since the visible range of wavelengths lie between 4000 to 7000 A, the wavelength in the visible spectrum is 4800 A\"" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "For n = 1, wavelength = 4800.00 A\n", - "For n = 2, wavelength = 1600.00 A\n", - "For n = 3, wavelength = 960.00 A\n", - "Since the visible range of wavelengths lie between 4000 to 7000 A, the wavelength in the visible spectrum is 4800 A\n" - ] - } - ], - "prompt_number": 5 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 1.3.1, Page number 1-21" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "\n", - "#Variable declaration \n", - "a = 40. #angle(sec)\n", - "lamda = 1.2 #distance between fringes(cm)\n", - "alpha = 10 #no. of fringes\n", - "\n", - "#Calculations\n", - "Bair = lamda/alpha #cm\n", - "alpha = (a*math.pi)/(3600*180) #radians\n", - "lamda = 2*alpha*Bair\n", - "\n", - "#Result\n", - "print \"Wavelength of monochromatic light =\",round((lamda/1E-8),1),\"A\"" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Wavelength of monochromatic light = 4654.2 A\n" - ] - } - ], - "prompt_number": 14 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 1.3.2, Page number 1-22" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "\n", - "#Variable declaration \n", - "lamda = 5893*10**-8 #wavelength(cm)\n", - "u = 1.52 #refractive index\n", - "B = 0.1 #fringe spacing(cm)\n", - "\n", - "#Calculations\n", - "alpha = (lamda/(2*u*B))*180*3600/math.pi #seconds\n", - "\n", - "#Result\n", - "print \"Angle of wedge =\",round(alpha,2),\"secs\"\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Angle of wedge = 39.98 secs\n" - ] - } - ], - "prompt_number": 16 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 1.3.3, Page number 1-22" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "\n", - "#Variable declaration \n", - "u = 1.4 #refractive index\n", - "B = 0.25 #fringe spacing(cm)\n", - "a = 20 #angle(secs)\n", - "\n", - "#Calculations\n", - "alpha = (a*math.pi)/(3600*180) #radians\n", - "lamda = 2*u*alpha*B\n", - "\n", - "#Result\n", - "print \"Wavelength of light =\",round((lamda/1E-8),2),\"A\"" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Wavelength of light = 6787.39 A\n" - ] - } - ], - "prompt_number": 18 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 1.3.4, Page number 1-23" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "\n", - "#Variable declaration \n", - "u = 1.5 #refractive index\n", - "lamda = 5.82*10**-5 #wavelength(cm)\n", - "a = 20 #angle(secs)\n", - "\n", - "#Calculations\n", - "alpha = (a*math.pi)/(3600*180) #radians\n", - "B = lamda/(2*u*alpha)\n", - "N = 1/B\n", - "\n", - "#Result\n", - "print \"Number of interfernce fronges pr cm is\",round(N)" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Number of interfernce fronges pr cm is 5.0\n" - ] - } - ], - "prompt_number": 20 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 1.3.5, Page number 1-24" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "\n", - "#Variable declaration \n", - "u = 1 #refractive index for air film\n", - "lamda = 6*10**-5 #wavelength(cm)\n", - "B = 1./10 #distance between fringes(cm)\n", - "\n", - "#Calculations\n", - "alpha = lamda/(2*u*B) #radians\n", - "d = alpha*10\n", - "\n", - "#Result\n", - "print \"Daimeter of wire =\",d,\"cm\"" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Daimeter of wire = 0.003 cm\n" - ] - } - ], - "prompt_number": 23 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 1.3.6, Page number 1-24" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "\n", - "alpha = 0.01*10**-1/10 #angle(radians)\n", - "u = 1 #refractive index for air film\n", - "lamda = 5900*10**-10 #wavlength(m)\n", - "\n", - "#Calculation\n", - "B = lamda/(2*u*alpha)\n", - "\n", - "#Result\n", - "print \"Seperation between fringes is\",B/10**-3,\"mm\"" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Seperation between fringes is 2.95 mm\n" - ] - } - ], - "prompt_number": 26 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 1.4.1, Page number 1-32" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "\n", - "#Variable declaration \n", - "n = 40\n", - "\n", - "#Calculation\n", - "#Equating the equation 4*R*n*lamda=4*4R*n*lamda, we get\n", - "\n", - "N = (4*4*n)/4\n", - "\n", - "#result\n", - "print \"Ring number =\",N" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Ring number = 160\n" - ] - } - ], - "prompt_number": 27 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 1.4.2, Page number 1-32" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "\n", - "#Variable declaration \n", - "n = 10\n", - "Dn = 0.5 #diameter of dark ring(cm)\n", - "lamda = 5*10**-5 #waelength(cm)\n", - "\n", - "#Calculations\n", - "R = Dn**2/(4*n*lamda)\n", - "\n", - "#Result\n", - "print \"Radius of curvature =\",R,\"cm\"" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Radius of curvature = 125.0 cm\n" - ] - } - ], - "prompt_number": 28 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 1.4.3, Page number 1-33" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "\n", - "#Variable declaration \n", - "n = 5\n", - "p = 10\n", - "D5 = 0.336 #diameter of 5th ring(cm)\n", - "lamda = 5890*10**-8 #waelength(cm)\n", - "D15 = 0.59 #diameter of 15th ringcm\n", - "\n", - "#Calculations\n", - "R = (D15**2-D5**2)/(4*p*lamda)\n", - "\n", - "#Result\n", - "print \"Radius of curvature =\",round(R,2),\"cm\"" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Radius of curvature = 99.83 cm\n" - ] - } - ], - "prompt_number": 34 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 1.4.4, Page number 1-33" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "\n", - "#Varaible declaration\n", - "Dn = 0.42 #diameter of dark ring(cm)\n", - "p = 8 \n", - "R = 200 #radius of curvature(cm)\n", - "Dn8 = 0.7 #diameter of (n+8)th ring(cm)\n", - "\n", - "#Calculations\n", - "lamda = (Dn8**2-Dn**2)/(4*R*p)\n", - "\n", - "#Result\n", - "print \"Wavelength =\",lamda/1E-8,\"A\"" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Wavelength = 4900.0 A\n" - ] - } - ], - "prompt_number": 36 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 1.4.5, Page number 1-34" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "\n", - "#Varaible declaration\n", - "Dn = 0.218 #cm\n", - "Dn10 = 0.451 #cm\n", - "lamda = 5893*10**-8 #wavelength(cm)\n", - "R = 90 #radius of curvature(cm)\n", - "p = 10\n", - "\n", - "#Calculation\n", - "u = (4*p*lamda*R)/(Dn10**2-Dn**2)\n", - "\n", - "#Result\n", - "print \"Refractive index =\",round(u,3)" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Refractive index = 1.361\n" - ] - } - ], - "prompt_number": 37 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 1.4.6, Page number 1-34" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "#Varaible declaration\n", - "D5 = 0.42 #diameter of dark ring(cm)\n", - "\n", - "#Calculations\n", - "'''\n", - "For 5th dark ring,\n", - "D5^2 = 20*R*lamda -----1\n", - "\n", - "For 10th dark ring,\n", - "D10^2 = 40*R*lamda -----2\n", - "\n", - "Substituting 1 in 2,\n", - "'''\n", - "\n", - "D10 = math.sqrt((40*D5**2)/20)\n", - "\n", - "#Result\n", - "print \"Diameter of the 10th dark ring =\",round(D10,3),\"cm\"" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Diameter of the 10th dark ring = 0.594 cm\n" - ] - } - ], - "prompt_number": 35 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 1.4.7, Page number 1-35" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "\n", - "#Varaible declaration\n", - "lamda_n = 6000 #wavelength of nth ring(A)\n", - "lamda_n1 = 5000 #wavelength for (n+1)th ring(cm)\n", - "\n", - "#Calculations\n", - "'''\n", - "Dn^2 = 4*R*n*lamda_n ---1\n", - "\n", - "Dn+1^2 = 4*R(n+1)*lamda_n1 ---2\n", - "\n", - "Equating 1 and 2, we get,\n", - "'''\n", - "\n", - "n = 5\n", - "R = 2\n", - "Dn = math.sqrt(4*R*n*lamda_n*10**-8)\n", - "\n", - "#Result\n", - "print \"Diameter =\",round(Dn,3),\"cm\"" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Diameter = 0.049 cm\n" - ] - } - ], - "prompt_number": 42 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 1.4.8, Page number 1-35" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "\n", - "#Varaible declaration\n", - "Dair = 2.3 #diameter of ring in air(cm)\n", - "Dliq = 2 #diameter of ring in liquid(cm)\n", - "\n", - "#Calculations\n", - "u = Dair**2/Dliq**2\n", - "\n", - "#Result\n", - "print \"Refractive index =\",u" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Refractive index = 1.3225\n" - ] - } - ], - "prompt_number": 43 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 1.4.11, Page number 1-37" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "\n", - "D4 = 0.4 #diameter of 4th ring(cm)\n", - "D12 = 0.7 #diameter of 12th ring(cm)\n", - "\n", - "#Calculations\n", - "'''\n", - "For D4, \n", - "D4 = math.sqrt(4R*4*lamda)\n", - "'''\n", - "rt_Rl = 0.1\n", - "R = 80 \n", - "\n", - "#For D20,\n", - "D20 = math.sqrt(R)*rt_Rl\n", - "\n", - "#Result\n", - "print \"Diameter of 20th ring =\",round(D20,3),\"cm\"\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Diameter of 20th ring = 0.894 cm\n" - ] - } - ], - "prompt_number": 46 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 1.4.12, Page number 1-36" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "\n", - "#Variable declaration \n", - "n = 5\n", - "p = 10\n", - "D5 = 0.336 #diameter of 5th ring(cm)\n", - "D15 = 0.590 #diameter of 15th ring(cm)\n", - "R = 100 #radius of curvature(cm)\n", - "\n", - "#Calculations\n", - "lamda = (D15**2-D5**2)/(4*R*p)\n", - "\n", - "#Result\n", - "print \"Wavlength =\",lamda/1E-8,\"A\"" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Wavlength = 5880.1 A\n" - ] - } - ], - "prompt_number": 50 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 1.7.1, Page number 1-42" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "\n", - "#Variable declaration\n", - "lamda = 560 #wavelength(nm)\n", - "u = 2 #refractive index\n", - "\n", - "#Calculations\n", - "lamda_dash = lamda/u\n", - "t = lamda_dash/4\n", - "\n", - "#Result\n", - "print \"Thickness of film =\",t,\"nm\"" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Thickness of film = 70 nm\n" - ] - } - ], - "prompt_number": 55 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 1.7.2, Page number 1-42" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "\n", - "#Variable declaration\n", - "lamda = 6000 #wavelength(E)\n", - "u = 1.2 #refractive index\n", - "\n", - "#Calculations\n", - "lamda_dash = lamda/u\n", - "t = lamda_dash/4\n", - "\n", - "#Result\n", - "print \"Thickness of film =\",t,\"A\"" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Thickness of film = 1250.0 A\n" - ] - } - ], - "prompt_number": 56 - } - ], - "metadata": {} - } - ] -}
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