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diff --git a/Engineering_Physics/Chapter3_1.ipynb b/Engineering_Physics/Chapter3_1.ipynb new file mode 100644 index 00000000..e20ce99f --- /dev/null +++ b/Engineering_Physics/Chapter3_1.ipynb @@ -0,0 +1,238 @@ +{ + "metadata": { + "name": "Chapter3" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": "3: Interference" + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Example number 3.1, Page number 71" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "#To calculate the velocity of light\n\n#importing modules\nfrom __future__ import division\nimport math\n\n#Variable declaration\nbeta = 0.51; #Fringe width(mm)\nd = 2.2; #Distance between the slits(mm)\nD = 2; #Distance between the slits and the screen(m)\n\n#Calculation\nbeta = beta*10**-1; #Fringe width(cm)\nd = d*10**-1; #Distance between the slits(cm)\nD=D*10**2; #Distance between the slits and the screen(cm)\nlamda = beta*d/D; #Wavelength of light(cm)\nlamda = lamda*10**8; #Wavelength of light(A)\n\n#Result\nprint \"The wavelength of light is\",lamda, \"angstrom\"", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "The wavelength of light is 5610.0 angstrom\n" + } + ], + "prompt_number": 4 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Example number 3.2, Page number 71" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "#To calculate the seperation between the third fringe\n\n#importing modules\nfrom __future__ import division\nimport math\n\n#Variable declaration\nlambda1 = 4250; #First wavelength emitted by source of light(A)\nlambda2 = 5050; #Second wavelength emitted by source of light(A)\nD = 1.5; #Distance between the source and the screen(m)\nd = 0.025; #Distance between the slits(mm)\nn = 3; #Number of fringe from the centre\n\n#Calculation\nlambda1 = lambda1*10**-10; #First wavelength emitted(m)\nlambda2 = lambda2*10**-10; #Second wavelength emitted(m)\nd = d*10**-3; #Distance between the slits(m)\nx3 = n*lambda1*D/d; #Position of third bright fringe due to lambda1(m)\nx3_prime = n*lambda2*D/d; #Position of third bright fringe due to lambda2(m)\nx = x3_prime-x3; #separation between the third bright fringe(m)\nx = x*10**2; #separation between the third bright fringe(cm)\n\n#Result\nprint \"The separation between the third bright fringe due to the two wavelengths is\",x, \"cm\"\n", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "The separation between the third bright fringe due to the two wavelengths is 1.44 cm\n" + } + ], + "prompt_number": 5 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Example number 3.3, Page number 71" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "#To calculate the refractive index of the sheet of glass\n\n#importing modules\nimport math\n\n#Variable declaration\nlamda = 5.5*10**-5; #Wavelength emitted by source of light(cm)\nn = 4; #Number of fringes shifted\nt = 3.9*10**-4; #Thickness of the thin glass sheet(cm)\n\n#Calculation\nmew = (n*lamda/t)+1; #Refractive index of the sheet of glass\nmew = math.ceil(mew*10**4)/10**4; #rounding off the value of v to 4 decimals\n\n#Result\nprint \"The refractive index of the sheet of glass is\",mew", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "The refractive index of the sheet of glass is 1.5642\n" + } + ], + "prompt_number": 6 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Example number 3.4, Page number 72" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "#To calculate the least thickness of the film that appears bright and dark\n\n#importing modules\nimport math\n\n#Variable declaration\nlamda = 5893; #Wavelength of monochromatic lihgt used(A)\nn = 1; #Number of fringe for the least thickness of the film\ncosr = 1; #for normal incidence\nmew = 1.42; #refractive index of the soap film\n\n#Calculation\n#As for constructive interference, \n#2*mew*t*cos(r) = (2*n-1)*lambda/2, solving for t\nt = (2*n-1)*lamda/(4*mew*cosr); #Thickness of the film that appears bright(A)\n#As for destructive interference, \n#2*mu*t*cos(r) = n*lambda, solving for t\nt1 = n*lamda/(2*mew*cosr); #Thickness of the film that appears bright(A)\n\n#Result\nprint \"The thickness of the film that appears bright is\",t, \"angstrom\"\nprint \"The thickness of the film that appears dark is\",t1, \"angstrom\"", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "The thickness of the film that appears bright is 1037.5 angstrom\nThe thickness of the film that appears dark is 2075.0 angstrom\n" + } + ], + "prompt_number": 7 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Example number 3.5, Page number 72" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "#To calculate the angle of the wedge\n\n#importing modules\nimport math\n\n#Variable declaration\nlamda = 5893; #Wavelength of monochromatic lihgt used(A)\nn = 10; #Number of fringe that are found \nd = 1; #Distance of 10 fringes(cm)\n\n#Calculation\nbeta = d/n; #Fringe width(cm)\nlamda = lamda*10**-8; #Wavelength of monochromatic lihgt used(cm)\ntheta = lamda/(2*beta); #Angle of the wedge(rad)\ntheta = theta*10**4;\ntheta = math.ceil(theta*10**4)/10**4; #rounding off the value of theta to 4 decimals\n\n#Result\nprint \"The angle of the wedge is\",theta,\"*10**-4 rad\"", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "The angle of the wedge is 2.9465 *10**-4 rad\n" + } + ], + "prompt_number": 12 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Example number 3.6, Page number 72" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "#To calculate the separation between consecutive bright fringes\n\n#importing modules\nimport math\nfrom __future__ import division\n\n#Variable declaration\nlamda = 5900; #Wavelength of monochromatic lihgt used(A)\nt = 0.010; #Spacer thickness(mm)\nl = 10; #Wedge length(cm)\n\n#Calculation\nt = t*10**-1; #Spacer thickness(cm)\ntheta = t/l; #Angle of the wedge(rad)\nlamda = lamda*10**-8; #Wavelength of monochromatic lihgt used(cm)\nbeta = lamda/(2*theta); #Fringe width(cm)\n\n#Result\nprint \"The separation between consecutive bright fringes is\",beta, \"cm\"", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "The separation between consecutive bright fringes is 0.295 cm\n" + } + ], + "prompt_number": 13 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Example number 3.7, Page number 72" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "#To calculate the diameter of 20th dark ring\n\n#importing modules\nimport math\n\n#Variable declaration\nD4 = 0.4; #Diameter of 4th dark ring(cm)\nD12 = 0.7; #Diameter of 12th dark ring(cm)\n\n#Calculation\n#We have (dn_plus_k**2)-Dn**2 = 4*k*R*lamda\n#D12**2-D4**2 = 32*R*lamda and D20**2-D12**2 = 32*R*lamda for k = 8\n#since RHS are equal, by equating the LHS we get D12**2-D4**2 = D20**2-D12**2\nD20 = math.sqrt((2*D12**2)-D4**2); #Diameter of 20th dark ring(cm)\nD20 = math.ceil(D20*10**4)/10**4; #rounding off the value of D20 to 4 decimals\n\n#Result\nprint \"The diameter of 20th dark ring is\",D20, \"cm\"", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "The diameter of 20th dark ring is 0.9056 cm\n" + } + ], + "prompt_number": 14 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Example number 3.8, Page number 73" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "#To calculate the refractive index of the liquid\n\n#importing modules\nimport math\nfrom __future__ import division\n\n#Variable declaration\nDn = 0.30; #Diameter of nth dark ring with air film(cm)\ndn = 0.25; #Diameter of nth dark ring with liquid film(cm)\n\n#Calculation\nmew = (Dn/dn)**2; #Refractive index of the liquid\n\n#Result\nprint \"The refractive index of the liquid is\", mew\n", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "The refractive index of the liquid is 1.44\n" + } + ], + "prompt_number": 15 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Example number 3.9, Page number 73" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "#To calculate the wavelength of light\n\n#importing modules\nimport math\n\n#Variable declaration\nx = 0.002945; #Distance through which movable mirror is shifted(cm)\nN = 100; #Number of fringes shifted\n\n#Calculation\nx = x*10**-2; #Distance through which movable mirror is shifted(m)\nlamda = 2*x/N; #Wavelength of light(m)\nlamda = lamda*10**10; #Wavelength of light(A)\n\n#Result\nprint \"The wavelength of light is\",lamda, \"angstrom\"", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "The wavelength of light is 5890.0 angstrom\n" + } + ], + "prompt_number": 16 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Example number 3.10, Page number 73" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "#To calculate the shift in movable mirror\n\n#importing modules\nimport math\n\n#Variable declaration\nlambda1 = 5896; #Wavelength of D1 line of sodium(A)\nlambda2 = 5890; #Wavelength of D2 line of sodium(A)\n\n#Calculation\nlamda = (lambda1+lambda2)/2;\nx = (lamda**2)/(2*(lambda1-lambda2)); #Shift in movable mirror of Michelson Interferometer(A)\nx = x*10**-7; #Shift in movable mirror of Michelson Interferometer(mm)\nx = math.ceil(x*10**4)/10**4; #rounding off the value of D20 to 4 decimals\n\n#Result\nprint \"The shift in movable mirror is\",x, \"mm\"", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "The shift in movable mirror is 0.2894 mm\n" + } + ], + "prompt_number": 17 + }, + { + "cell_type": "code", + "collapsed": false, + "input": "", + "language": "python", + "metadata": {}, + "outputs": [] + } + ], + "metadata": {} + } + ] +}
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