{ "metadata": { "name": "", "signature": "sha256:7c369c2593aa95d4aac8153f9063553bb327aa743223d1fff41c98aa45d77ebe" }, "nbformat": 3, "nbformat_minor": 0, "worksheets": [ { "cells": [ { "cell_type": "heading", "level": 1, "metadata": {}, "source": [ "Chapter4:POLARISATION" ] }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Ex4.1:pg-147" ] }, { "cell_type": "code", "collapsed": false, "input": [ "import math\n", "# compare the intensities of ordinary and extraordinary rays\n", "#intensity of ordinary rays is given by Io=a**2 *(sin theta)**2\n", "#where theta=30 degree\n", "#we get Io=a**2/4\n", "Io=1.0/4\n", "#intensity of extraordinary ray is given by IE=(a*cos theta)**2\n", "#we get IE=3*a**2/4\n", "IE=3.0/4\n", "I=IE/Io\n", "print \"the intensities of ordinary and extraordinary rays is I=\",I,\"unitless\"\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "the intensities of ordinary and extraordinary rays is I= 3.0 unitless\n" ] } ], "prompt_number": 1 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Ex4.2:pg-147" ] }, { "cell_type": "code", "collapsed": false, "input": [ "import math\n", "#to calculate angle of refraction\n", "#according to brewster's law mu=tan ip\n", "mu=1.732 #refractive index\n", "ip=math.degrees(math.atan(mu)) #polarising angle in degree\n", "r=90-ip\n", "print \"angle of refraction of ray is r=\",round(r),\"degree\"\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "angle of refraction of ray is r= 30.0 degree\n" ] } ], "prompt_number": 11 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Ex4.3:pg-147" ] }, { "cell_type": "code", "collapsed": false, "input": [ "import math\n", "#to calculate polarising angle and angle of refraction\n", "mu=1.345 #refractive index, mu=1/sinc=1/sin48degree=1/0.7431 \n", "ip=math.degrees(math.atan(mu))\n", "r=90-ip\n", "print \"polarising angle is ip=\",round(ip,3),\"degree\"\n", "print \"angle of refraction is r=\",round(r,3),\"degree\" \n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "polarising angle is ip= 53.369 degree\n", "angle of refraction is r= 36.631 degree\n" ] } ], "prompt_number": 10 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Ex4.4:pg-147" ] }, { "cell_type": "code", "collapsed": false, "input": [ "import math\n", "#to calculate thickness of a half wave plate of quartz\n", "lamda=5*10**-5 #wavelength in cm\n", "mue=1.553 \n", " #refractive index (unitless)\n", "muo=1.544\n", "#for a half plate of positive crystal\n", "t=lamda/(2*(mue-muo))\n", "print \"thickness of a half wave plate of quartz is t=\",\"{:.2e}\".format(t),\"cm\"\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "thickness of a half wave plate of quartz is t= 2.78e-03 cm\n" ] } ], "prompt_number": 9 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Ex4.5:pg-148" ] }, { "cell_type": "code", "collapsed": false, "input": [ "import math\n", "#to calculate thickness of quarter wave plate\n", "lamda=5.890*10**-5 #wavelength of light in cm\n", "mue=1.553\n", " #refractive index\n", "muo=1.544\n", "t=lamda/(4*(mue-muo)) \n", "print \"thickness of quarter wave plate is t=\",\"{:.3e}\".format(t),\"cm\" \n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "thickness of quarter wave plate is t= 1.636e-03 cm\n" ] } ], "prompt_number": 8 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Ex4.6:pg-148" ] }, { "cell_type": "code", "collapsed": false, "input": [ "import math\n", "#to calculate thickness of a doubly refracting plate\n", "lamda=5.890*10**-5 #wavelength in cm\n", "muo=1.53 \n", " #refractive index\n", "mue=1.54\n", "t=lamda/(4*(mue-muo))\n", "print \"thickness of a plate is t=\",\"{:.2e}\".format(t),\"cm\"\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "thickness of a plate is t= 1.47e-03 cm\n" ] } ], "prompt_number": 6 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Ex4.7:pg-152" ] }, { "cell_type": "code", "collapsed": false, "input": [ "import math\n", "#to calculate angle of rotation\n", "alpha=66 #specific rotation of cane sugar in degree\n", "c=15.0/100 #concentration of the solution in gm/cc\n", "l=20 #length of tube in cm\n", "theta=alpha*l*c/10\n", "print \"the angle of rotation of the plane of polarisation is theta=\",theta,\"degree\"\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "the angle of rotation of the plane of polarisation is theta= 19.8 degree\n" ] } ], "prompt_number": 5 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Ex4.8:pg-153" ] }, { "cell_type": "code", "collapsed": false, "input": [ "import math\n", "#to calculate specific rotation \n", "theta=26.4 #in degree\n", "l=20 #length in cm\n", "c=0.2 #gm/cm**3\n", "alpha=10*theta/(l*c)\n", "print \"the specific rotation is alpha=\",alpha,\"degree\"\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "the specific rotation is alpha= 66.0 degree\n" ] } ], "prompt_number": 10 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Ex4.9:pg-153" ] }, { "cell_type": "code", "collapsed": false, "input": [ "import math\n", "#to calculate strength of solution\n", "theta=11 #degree\n", "l=20.0 #length in cm\n", "alpha=66 #specific rotation of sugar in degree\n", "c=10*theta/(l*alpha)\n", "print \"strength of solution is c=\",round(c,4),\"gm/cm**3\"\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "strength of solution is c= 0.0833 gm/cm**3\n" ] } ], "prompt_number": 4 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Ex4.10:pg-153" ] }, { "cell_type": "code", "collapsed": false, "input": [ "import math\n", "#to calculate difference in the refractive indices\n", "#specific rotation is theta/d=29.73 degree/mm\n", "theta=29.73 #where theta=theta/d\n", "lamda=5.086*10**-4 #wavelength in mm\n", "#optical rotation is given by theta=math.pi*d*(mul-mur)/lamda\n", "#where mul and mur are refractive indices for anti-clockwise and clockwise polarised lights\n", "mu=theta*lamda/180 #where mu=mul-mur\n", "print \"difference in refractive indices is mu=\",\"{:.1e}\".format(mu),\"unitless\"\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "difference in refractive indices is mu= 8.4e-05 unitless\n" ] } ], "prompt_number": 3 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Ex4.11:pg-153" ] }, { "cell_type": "code", "collapsed": false, "input": [ "import math\n", "#to calculate optical rotation\n", "#let theta' be the optical rotation by a solution of strength c' in a tube of length l' then\n", "#we get 10*theta'/l'*c'=10*theta/l*c\n", "c=1.0/3 #it is given that solution is 1/3 of its previous concentration i.e. c'/c=1/3,where c=c'/c\n", "l1=30 #where l1=l'\n", " #length in cm \n", "l=20.0\n", "theta=13 #degree\n", "#formula is theta'=l'*c'*theta/(l*c)\n", "theta1=l1*c*theta/l\n", "print \"optical rotation is theta1=\",theta1,\"degree\"\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "optical rotation is theta1= 6.5 degree\n" ] } ], "prompt_number": 5 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Ex4.12:pg-154" ] }, { "cell_type": "code", "collapsed": false, "input": [ "import math\n", "#to calculate specific rotation\n", "theta=52.8 #optical rotation in degree\n", "l=20.0 #length of the solution in cm\n", "c=20/50.0 #concentration of the solution in gm/cc\n", "alpha=10*theta/(l*c)\n", "print \"the specific rotation is alpha=\",alpha,\"degree\"\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "the specific rotation is alpha= 66.0 degree\n" ] } ], "prompt_number": 2 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Ex4.13:pg-154" ] }, { "cell_type": "code", "collapsed": false, "input": [ "import math\n", "#to calculate length \n", "l=40 #length in cm\n", "c=5.0/100 #concentration in percentage\n", "theta1=35 #optical rotation in degree ,where theta1=theta'\n", "c1=10.0/100 #concentration in % ,where c1=c'\n", "theta=20\n", "#formula of specific rotation is alpha=10*theta/l*c\n", "l1=l*c*theta1/(c1*theta)\n", "print \"length is l1=\",l1,\"cm\"\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "length is l1= 35.0 cm\n" ] } ], "prompt_number": 6 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Ex4.14:pg-155" ] }, { "cell_type": "code", "collapsed": false, "input": [ "import math\n", "#to calculate rotation of plane of polarisation of light\n", "mur=1.53914\n", " #refractive index\n", "mul=1.53920\n", "lamda=6.5*10**-5 #wavelength in cm\n", "d=0.02 #distance in cm\n", "thetaR=180*(mul-mur)*d/lamda\n", "print \"rotation of plane of polarisation of light is thetaR=\",round(thetaR,3),\"degree\"\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "rotation of plane of polarisation of light is thetaR= 3.323 degree\n" ] } ], "prompt_number": 12 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Ex4.15:pg-155" ] }, { "cell_type": "code", "collapsed": false, "input": [ "import math\n", "#to calculate % purity of the sugar sample\n", "theta=9.9 #optical rotation in degree\n", "alpha=66 #specific roation of pure sugar solution in dm**-1(gm/cc)**-1\n", "l=20 #length of tube in cm\n", "c=10*theta/(l*alpha) #concentration of solution in gm/c.c\n", "#it is given that 80 gm of impure sugar is dissolved in a litre of water\n", "per=(c*100*10**3)/80 #here c is in gm/litre\n", "print \"percentage of the sugar sample is per=\",per,\"%\"\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "percentage of the sugar sample is per= 93.75 %\n" ] } ], "prompt_number": 17 } ], "metadata": {} } ] }