{ "metadata": { "name": "", "signature": "sha256:2b54094513f6b50d68fe3b6c91cd7d7cafc36254b75abea2a0d6c88b266fd509" }, "nbformat": 3, "nbformat_minor": 0, "worksheets": [ { "cells": [ { "cell_type": "heading", "level": 1, "metadata": {}, "source": [ "Chapter2:INTERFERENCE" ] }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Ex2.2:pg-41" ] }, { "cell_type": "code", "collapsed": false, "input": [ "import math\n", "#to calculate ratio of intensity\n", "#I1/I2=1/25\n", "#formula is a1/a2=math.sqrt(I1/I2)=1/5\n", "a2=5 #a2=5*a1\n", "a1=1\n", "I=((1+5)**2)/((1-5)**2)\n", "print \"ratio of intensity at the maxima and minima in the interference pattern is Imax/Imin=((a1+a2)**2)/((a1-a2)**2)=\",((a1+a2)**2)/((a1-a2)**2),\"unitless\"\n", "#answer is given in terms of ratio\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "ratio of intensity at the maxima and minima in the interference pattern is Imax/Imin=((a1+a2)**2)/((a1-a2)**2)= 2 unitless\n" ] } ], "prompt_number": 1 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Ex2.3:pg-42" ] }, { "cell_type": "code", "collapsed": false, "input": [ "import math\n", "#to calculate ratio of intensity at this point to that at the centre of a bright fringe\n", "#the intensity at any pont is I=a1**2+a2**2+2*a1*a2*cos del\n", "#let a1=a2=a\n", "#phase difference del is 0\n", "#then I0=a**2+a**2+2*a*a*cos 0\n", "#we get I0=4a**2\n", "I0=4 #intensity\n", "#path difference is lemda/8\n", "#phase difference =2*math.pi/lemda*path difference=math.pi/4\n", "#I1=a**2+a**2+2a*a*cos math.pi/4\n", "#I1=3.414a**2\n", "I1=3.414\n", "intensity=I1/I0\n", "print \" ratio of intensity =\",intensity,\"unitless\"\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ " ratio of intensity = 0.8535 unitless\n" ] } ], "prompt_number": 2 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Ex2.4:pg-42" ] }, { "cell_type": "code", "collapsed": false, "input": [ "import math\n", "#to calculate ratio of maximum intensity to minimum intensity\n", "#formula is I1/I2=a1**2/a2**2=100/1\n", "#a1/a2=10/1\n", "a1=10 #a1=10*a2\n", "a2=1\n", "print \"the ratio of maximum intensity to minmum intensity in the interference pattern Imax/Imin=((a1+a2)**2)/((a1-a2)**2)=\",((a1+a2)**2)/((a1-a2)**2),\"unitless\" \n", "#answer is given in terms of ratio in the book\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "the ratio of maximum intensity to minmum intensity in the interference pattern Imax/Imin=((a1+a2)**2)/((a1-a2)**2)= 1 unitless\n" ] } ], "prompt_number": 3 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Ex2.5:pg-43" ] }, { "cell_type": "code", "collapsed": false, "input": [ "import math\n", "#to calculate relative intensities\n", "#Imax/Imin=(a1+a2)**2/(a1-a2)**2+105/95\n", "#(a1+a2)/(a1-a2)=1.051\n", "#we get a1/a2=40 \n", "a1=40 #a1=40*a2\n", "a2=1\n", "print \"the ratio of the intensities of interfering sources is I1/I2=a1**2/a2**2=\",a1**2/a2**2,\"unitless\"\n", "#answer is given in terms of ratio in the book\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "the ratio of the intensities of interfering sources is I1/I2=a1**2/a2**2= 1600 unitless\n" ] } ], "prompt_number": 1 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Ex2.7:pg-52" ] }, { "cell_type": "code", "collapsed": false, "input": [ "import math\n", "#to calculate distance between the two coherent sources\n", "lamda=5890*10**-10 #wavelength in m\n", "omega=9.424*10**-4 #width of the fringes in m\n", "D=.80 #distance in m\n", "twod=D*lamda/omega\n", "print \"the distance between two coherent sources is twod=\",twod,\"m\"\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "the distance between two coherent sources is twod= 0.0005 m\n" ] } ], "prompt_number": 5 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Ex2.8:pg-53" ] }, { "cell_type": "code", "collapsed": false, "input": [ "import math\n", "#to calculate fringe width \n", "mu=1.5 #refractive index (unitless)\n", "alpha=math.pi/180 #refracting angle in radian\n", "Y1=20*10**-2 #distance between the source and the biprism in m\n", "Y2=80*10**-2 #distance in m\n", "D=Y1+Y2 # distance in m\n", "lamda=6900*10**-10 #wavelength in m\n", "twod=2*(mu-1)*alpha*Y1 \n", "omega=D*lamda/twod\n", "print \"the fringe width is omega=\",round(omega,7),\"m\"\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "the fringe width is omega= 0.0001977 m\n" ] } ], "prompt_number": 2 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Ex2.9:pg-53" ] }, { "cell_type": "code", "collapsed": false, "input": [ "import math\n", "#to calculate wavelength of light\n", "omega=(1.888*10**-2)/20 #in (m)\n", "D=1.20 #distance of eye piece from the source in m\n", "twod=0.00075 #distance between two virtual sources in m\n", "lamda=(omega*twod/D)/(10e-11)\n", "print \"the wavelength of light is lamda=\",lamda,\"Angstrom\"\n", "\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "the wavelength of light is lamda= 5900.0 Angstrom\n" ] } ], "prompt_number": 44 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Ex2.10:pg-54" ] }, { "cell_type": "code", "collapsed": false, "input": [ "import math\n", "#to calculate thickness of glass plate\n", "n=3\n", "mu=1.5 #refractive index (unitless)\n", "lamda=5450*10**-10 #wavelength in m\n", "t=n*lamda/(mu-1)\n", "print \"the thickness of glass plate is t=\",t,\"m\"\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "the thickness of glass plate is t= 3.27e-06 m\n" ] } ], "prompt_number": 4 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Ex2.11:pg-54" ] }, { "cell_type": "code", "collapsed": false, "input": [ "import math\n", "#to calculate refractive index of the sheet\n", "t=6.3*10**-6 #thickness of thin sheet of transparent material in m\n", "lamda=5460*10**-10 #wavelength in m\n", "n=6\n", "mu=(n*lamda/t)+1\n", "print \"the refractive index of the sheet is mu=\",mu,\"unitless\"\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "the refractive index of the sheet is mu= 1.52 unitless\n" ] } ], "prompt_number": 9 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Ex2.12:pg-54" ] }, { "cell_type": "code", "collapsed": false, "input": [ "import math\n", "#to calculate refractive index of mica\n", "t=1.2*10**-8 #thickness of thin sheet of mica in m\n", "n=1 \n", "lamda=6*10**-7 #wavelength in m\n", "mu=(n*lamda/t)+1\n", "print \"the refractive index of mica is mu=\",mu,\"unitless\"\n", "#answer is given wrong in the book=1.50\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "the refractive index of mica is mu= 51.0 unitless\n" ] } ], "prompt_number": 11 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Ex2.13:pg-54" ] }, { "cell_type": "code", "collapsed": false, "input": [ "import math\n", "#to calculate intensity\n", "mu=1.5 #refractive index(unitless)\n", "t=1.5*10**-6 #thickness of thin glass plate in m\n", "pathdifference=(mu-1)*t # in m\n", "lamda=5*10**-7 #wavelength in m\n", "#del=2*math.pi*pathdifference/lamda\n", "#del=3*math.pi\n", "a1=1\n", " #where a1=a2=a\n", "a2=1\n", "#formula is I=a1**2+a2**2+2*a1*a2*cos del\n", "# where cos 3math.pi=-1\n", "I=a1**2+a2**2+2*a1*a2*(-1) \n", "print \"the intensity at the centre of the screen is I=\",I,\"unitless\" \n", "#to calculate lateral shift\n", "D=1 #distance in m\n", "twod=5*10**-4 #distance between two slits in m\n", "mu=1.5 #refractive index (unitless)\n", "t=1.5*10**-6 #thickness of thin glass plate in m\n", "x0=D*(mu-1)*t/twod\n", "print \"the lateral shift of the central maximum is x0=\",x0,\"m\"\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "the intensity at the centre of the screen is I= 0 unitless\n", "the lateral shift of the central maximum is x0= 0.0015 m\n" ] } ], "prompt_number": 8 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Ex2.14:pg-55" ] }, { "cell_type": "code", "collapsed": false, "input": [ "import math\n", "#to calculate spacing between the slits\n", "lamda=6*10**-5 #wavelength in cm\n", "omegatheta=0.1*math.pi/180 #angular width of a fringe in radians\n", "twod=lamda/omegatheta\n", "print \"the spacing between the slits is twod=\",\"{:.2e}\".format(twod),\"cm\"\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "the spacing between the slits is twod= 3.44e-02 cm\n" ] } ], "prompt_number": 9 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Ex2.15:pg-55" ] }, { "cell_type": "code", "collapsed": false, "input": [ "import math\n", "#to calculate distance of the third bright fringe on the screen from the central maximum \n", "lamda=6.5*10**-5 #wavelength in cm\n", "twod=0.2 #distance between the slits in cm\n", "D=120 #distance between the plane of the slits and the screen in cm\n", "n=3 \n", "X3=D*n*lamda/twod\n", "print \"the distance of the third bright fringe from the central maximum is X3=\",X3,\"cm\"\n", "#to calculate the least distance from the central maximum \n", "lamda1=6.5*10**-5 #wavelength in cm\n", "lamda2=5.2*10**-5 #wavelength in cm\n", "#Xn=Dnlamda1/2d=D(n+1)lamda2/2d\n", "#we get,\n", "n=lamda2/(lamda1-lamda2)\n", "print \"n=\",n,\"unitless\"\n", "Xn=D*n*lamda1/twod\n", "print \"the distance from the central maximum when the bright fringes due to both wavelengths coincide is Xn=\",Xn,\"cm\"\n", " \n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "the distance of the third bright fringe from the central maximum is X3= 0.117 cm\n", "n= 4.0 unitless\n", "the distance from the central maximum when the bright fringes due to both wavelengths coincide is Xn= 0.156 cm\n" ] } ], "prompt_number": 17 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Ex2.16:pg-56" ] }, { "cell_type": "code", "collapsed": false, "input": [ "import math\n", "#to calculate refractive index \n", "D=10 #distance in cm\n", "twod=0.2 #distance detween the slits in cm\n", "t=0.05 #thickness of transparent plate in cm\n", "deltaX=0.5 #in cm\n", "mu=(deltaX*twod/(D*t))+1\n", "print \"the refractive index of the transparent plate is mu=\",mu,\"unitless\"\n", "#to calculate order\n", "n=10 \n", "lamda=7000*10**-8 #wavelength in cm\n", "#path difference =n*lamda\n", "n1=n*lamda/(5000*10**-8) \n", "print \"the order will be visible is n1=\",n1,\"unitless\"\n", "#to calculate distance between the two coherent sources\n", "D=100 #distance in m \n", "lamda=6000*10**-8 #wavelength in cm\n", "omega=0.05 #distance between two consecutive bright fringes on the screen in cm\n", "twod=D*lamda/omega\n", "print \"the distance between the coherent sources is twod=\",twod,\"cm\"\n", "#to calculate wavelength\n", "Xn=1 #distance of fourth bright fringe from the central fringe in cm\n", "twod=0.02 #distance between the two coherent sources in cm\n", "n=4 \n", "D=100 #distance in cm \n", "lamda=Xn*twod/(n*D)\n", "print \"the wavelength of light is lamda=\",lamda,\"cm\"\n", "#to calculate wavelength\n", "#position of nth bright fringe from the centre of the central fringe is Xn=D*n*lamda/2d----eq(1)\n", "#fringe width umega=D*lamda/2d---------------------eq(2)\n", "#from eq(1) and eq(2) we get, Xn=n*omega\n", "#for 11th bright fringe X11=11*omega\n", "#position for nth dark fringe Xn'=(2n+1)D*lamda/4d\n", "#X4'=(7/2)*omega\n", "#distance between 11th and 4th dark fringe =0.8835 cm\n", "#we get \n", "omega=0.1178 #in cm\n", "twod=0.05 #distance between slis in cm\n", "D=100 # distance in cm\n", "lamda=omega*twod/D \n", "print \"the wavelength of light is lamda=\",lamda,\"cm\"\n", "#to calculate changed fringe width\n", "#X10-X0=10*omega\n", "#given that X10-X0=14.73-12.34=2.39mm \n", "omega=0.239 #in mm\n", "lamda=6000 #wavelength in angstrom\n", "lamda1=5000 #lamda'=5000 angstrom\n", "omega1=omega*lamda1/lamda\n", "print \"the changed fringe width is omega1=\",omega1,\"mm\"\n", "#to calculate thickness of mica sheet\n", "n=3\n", "mu=1.6 #refractive index(unitless)\n", "lamda=5.89*10**-5 #wavelength in cm\n", "t=n*lamda/(mu-1)\n", "print \"the thickness of mica sheet is t=\",t,\"cm\"\n", "#answer of thickness is given wrong in the book =0.002945 cm\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "the refractive index of the transparent plate is mu= 1.2 unitless\n", "the order will be visible is n1= 14.0 unitless\n", "the distance between the coherent sources is twod= 0.12 cm\n", "the wavelength of light is lamda= 5e-05 cm\n", "the wavelength of light is lamda= 5.89e-05 cm\n", "the changed fringe width is omega1= 0.199166666667 mm\n", "the thickness of mica sheet is t= 0.0002945 cm\n" ] } ], "prompt_number": 18 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Ex2.17:pg-75" ] }, { "cell_type": "code", "collapsed": false, "input": [ "import math\n", "#to calculate the smallest thickness of the plate\n", "mu=1.5 #refractive index(unitless)\n", "r=60*math.pi/180 #angle of refraction in radians\n", "lamda=5890*10**-10 #wavelength in m\n", "n=1\n", "#formula is t=n*lamda/(2*mu*cosr) where cosr=0.5\n", "t=n*lamda/(2*mu*0.5)\n", "print \"the smallest thickness of the plate which will appear dark by reflection is t=\",\"{:.3e}\".format(t),\"m\"\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "the smallest thickness of the plate which will appear dark by reflection is t= 3.927e-07 m\n" ] } ], "prompt_number": 10 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Ex2.18:pg-75" ] }, { "cell_type": "code", "collapsed": false, "input": [ "import math\n", "#to calculate least thickness of the film\n", "lamda=5893*10**-10#wavelength in m\n", "r=0 #in degree\n", "mu=1.42 #refractive index\n", "n=1\n", "#the formula is t=n*lamda/(2*mu*cosr), where cos0=1\n", "t=n*lamda/(2*mu*1)\n", "print \"the least thickness of the film that will appear black is t=\",t,\"m\"\n", "t=(2*n-1)*lamda/(2*mu*1*2)\n", "print \"the least thickness of the film that will appear bright is t=\",t,\"m\"\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "the least thickness of the film that will appear black is t= 2.075e-07 m\n", "the least thickness of the film that will appear bright is t= 1.0375e-07 m\n" ] } ], "prompt_number": 20 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Ex2.19:pg-76" ] }, { "cell_type": "code", "collapsed": false, "input": [ "import math\n", "#to calculate thickness of the film\n", "lamda1=6.1*10**-7 #wavelength in m\n", "lamda2=6*10**-7 # wavelength in m\n", "#the two dark consecutive fringes are overlapping for the wavelength lamda1 and lamda2 respectively\n", "#then, n*lamda1=(n+1)*lamda2\n", "#we get,\n", "n=lamda2/(lamda1-lamda2)\n", "sini=4.0/5\n", "mu=4.0/3\n", "#formula is mu=sini/sinr\n", "sinr=0.6\n", "cosr=math.sqrt(1-(sinr)**2)\n", "t=n*lamda1/(2*mu*cosr)\n", "print \"the thickness of the film is t=\",\"{:.2e}\".format(t),\"m\"\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "the thickness of the film is t= 1.72e-05 m\n" ] } ], "prompt_number": 11 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Ex2.20:pg-77" ] }, { "cell_type": "code", "collapsed": false, "input": [ "import math\n", "#to calculate thickness of the film\n", "mu=1.33 #refractive index of soap film (unitless)\n", "i=45*math.pi/180\n", "#the formula is mu=sini/sinr \n", "sinr=0.5317 \n", "cosr=math.sqrt(1-(sinr)**2)\n", "#for destructive interference\n", "lamda=5890*10**-10 #wavelength in m\n", "n=1\n", "t=n*lamda/(2*mu*cosr)\n", "print \"the thickness of the film is t=\",\"{:.3e}\".format(t),\"m\"\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "the thickness of the film is t= 2.614e-07 m\n" ] } ], "prompt_number": 12 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Ex2.21:pg-77" ] }, { "cell_type": "code", "collapsed": false, "input": [ "import math\n", "#to calculate angle of the wedge\n", "lamda=6000*10**-10 #wavelength in m\n", "mu=1.4 #refractive index in unitless\n", "omega=2*10.0**-3 #distance in m\n", "theta=lamda/(2*mu*omega)\n", "print \"the angle of the wedge is theta =\",\"{:.2e}\".format(theta),\"radians\"\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "the angle of the wedge is theta = 1.07e-04 radians\n" ] } ], "prompt_number": 13 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Ex2.22:pg-77" ] }, { "cell_type": "code", "collapsed": false, "input": [ "import math\n", "#to calculate wavelength of light \n", "theta=10*22.0/(7*60*60*180.0) #angle of wedge in radians\n", "omega=5*10.0**-2 #distance between the successive fringes in m\n", "mu=1.4 #refractive index\n", "lamda=2.0*mu*theta*omega/(10e-10)\n", "print \"the wavelength of light is lamda=\",round(lamda),\"Angstrom\"\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "the wavelength of light is lamda= 6790.0 Angstrom\n" ] } ], "prompt_number": 23 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Ex2.23:pg-78" ] }, { "cell_type": "code", "collapsed": false, "input": [ "import math\n", "#to calculate wavelength of the light\n", "D15=0.590*10**-2 #diamater of 15th ring in m\n", "D5=0.336*10**-2 #diameter of 5th ring in m\n", "p=1 # in m\n", "R=1 #radius of plano convex lens in m\n", "#formula is lamda=Dn+p**2-Dn**2/4pR\n", "lamda=((D15**2)-(D5**2))/(4*p*R)/(10e-10)\n", "print \"the wavelength of the monochromatic light is lamda=\",int(lamda),\"Angstrom\"\n", "\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "the wavelength of the monochromatic light is lamda= 5880 Angstrom\n" ] } ], "prompt_number": 14 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Ex2.24:pg-78" ] }, { "cell_type": "code", "collapsed": false, "input": [ "import math\n", "#to calculate refractive index of the liquid\n", "n=6\n", "lamda=6000*10**-10 #wavelength in m\n", "R=1 #radius of curvature of the curved surface in m\n", "Dn=3.1*10**-3 #diameter of 6th bright ring in m\n", "mu=2*(2*n-1)*lamda*R/Dn**2\n", "print \"the refractive index of the liquid is mu=\",round(mu,3),\"unitless\"\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "the refractive index of the liquid is mu= 1.374 unitless\n" ] } ], "prompt_number": 28 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Ex2.25:pg-78" ] }, { "cell_type": "code", "collapsed": false, "input": [ "import math\n", "#to calculate radius of curvature\n", "lamda=5900*10**-10 #wavelength in m\n", "n=10\n", "Dn=5*10**-3 # diameter of 10th dark ring in m\n", "R=Dn**2/(4*n*lamda)\n", "print \"the radius of curvature of the lens is R=\",round(R,3),\"m\"\n", "#to calculate thichness\n", "t=n*lamda/2\n", "print \"the thickness of the air film is t=\",t,\"m\"\n", "\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "the radius of curvature of the lens is R= 1.059 m\n", "the thickness of the air film is t= 2.95e-06 m\n" ] } ], "prompt_number": 15 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Ex2.26:pg-79" ] }, { "cell_type": "code", "collapsed": false, "input": [ "import math\n", "#to calculate the distance from the apex of the wedge at which the maximum due to the two wavelengths first coincide\n", "#condition for maxima for normal incidence air film is 2t=(2n+1)lamda/2\n", "#let nth order maximum due to lamda1 coincides with (n+1)th order maximum due to lamda2 \n", "#we get , n=(3lamda2-lamda1)/2(lamda1-lamda2)\n", "# we also get, 2t=lamda1*lamda2/(lamda1-lamda2)\n", "#t=X*theta\n", "lamda1=5896.0*10**-8 #wavelength in cm\n", "lamda2=5890.0*10**-8 #wavelength in cm\n", "theta=0.3*math.pi/180 #angle of wedge\n", "X=lamda1*lamda2/(2*(lamda1-lamda2)*theta)\n", "print \"the distance from the apex of the wedge is X=\",round(X,3),\"cm\"\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "the distance from the apex of the wedge is X= 5.527 cm\n" ] } ], "prompt_number": 17 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Ex2.27:pg-80" ] }, { "cell_type": "code", "collapsed": false, "input": [ "import math\n", "#to calculate radius of curvature \n", "n=10.0\n", "Dn=0.50 #diameter of 10th ring in cm\n", "lamda=6000*10**-8 #wavelength in cm\n", "R=Dn**2/(4*n*lamda)\n", "print \"the radius of curvature of the lens is R=\",R,\"cm\"\n", "#answer is given wrong in the book =106 cm\n", "#to calculate thickness of the film\n", "t=Dn**2/(8*R)\n", "print \"the thickness of the film is t=\",\"{:.1e}\".format(t),\"cm\"\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "the radius of curvature of the lens is R= 104.166666667 cm\n", "the thickness of the film is t= 3.0e-04 cm\n" ] } ], "prompt_number": 19 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Ex2.28:pg-80" ] }, { "cell_type": "code", "collapsed": false, "input": [ "import math\n", "#to calculate diameter\n", "#the difference of (n+p)th and nth dark ring is Dn+p**2-Dn**2=4nRlamda\n", "N=12 #where N=n+p\n", "n=4\n", "D12=0.7 #diameter of 12th dark ring in cm\n", "D4=0.4 #diameter of 4th dark ring in cm\n", "#D12**2-D4**2=4pRlamda where p=8 ----eq(1)\n", "#D20**2-D4**2=4pRlamda where p=16 -----eq(2)\n", "#divide eq(2) by eq(1) ,we get\n", "D20=math.sqrt((2*D12**2)-D4**2)\n", "print \"the diameter of 20th dark ring is D20=\",round(D20,2),\"cm\"\n", "\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "the diameter of 20th dark ring is D20= 0.91 cm\n" ] } ], "prompt_number": 21 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Ex2.29:pg-80" ] }, { "cell_type": "code", "collapsed": false, "input": [ "import math\n", "#to calculate diameter \n", "lamda1=6*10**-5 #wavelength in cm\n", "lamda2=4.5*10**-5 #wavelength in cm\n", "R=90 #radius of curvature of the curved surface in cm\n", "#Dn**2=4nRlamda1 -------eq(1)\n", "#Dn+1**2=4(n+1)Rlamda2-------eq(2)\n", "#the nth dark ring due to lamda1 coincides with (n+1)th dark ring due to lamda2\n", "#from eq(1) and eq(2)-4nRlamda1=4(n+1)Rlamda2\n", "# we get,\n", "n=lamda2/(lamda1-lamda2)\n", "Dn=math.sqrt(4*n*R*lamda1)\n", "print \"the diameter of nth dark ring for lamda1 is Dn=\",round(Dn,3),\"cm\"\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "the diameter of nth dark ring for lamda1 is Dn= 0.255 cm\n" ] } ], "prompt_number": 22 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Ex2.30:pg-81" ] }, { "cell_type": "code", "collapsed": false, "input": [ "import math\n", "#to calculate the difference of square of diameters for nth and (n+p)th ring when light of wavelength lamda is changed to lamda'\n", "lamda=6*10**-5 #wavelength in cm\n", "lamda1=4.5*10**-5 #wavelength in cm\n", "#Let D=(D**2-Dn**2)=0.125 cm**2 \n", "D=0.125\n", "#formula is D'(n+p)**2-D'n**2=lamda'*(D(n+p)**2-Dn**2)/lamda\n", "print \"the difference of square of diameters is D1(n+p)**2-D1n**2=(lamda1*D)/lamda=\",(lamda1*D)/lamda,\"cm**2\"\n", "#to calculate difference of square of diamaters when liqquid of refractive index mu' is introduced\n", "mu=1 #refractive index (unitless)\n", "mu1=1.33 # mu'=1.33\n", "#formula is D'(n+p)**2-D'n**2=(mu/mu')*(D(n+p)**2-Dn**2)\n", "print \"the difference of square of diameters is D1(n+p)**2-D1n**2=(mu*D)/mu1=\",round((mu*D)/mu1,3),\"cm**2\"\n", "#to calculate difference of square of diameters when radius of curvature of convex surface of the plano convex lens is doubled\n", "R1=2 #radius of curvature is R'=2R\n", "R=1\n", "#formula is D'(n+p)**2-D'n**2=(R'/R)*(D(n+p)**2-Dn**2)\n", "print \"the difference of square of diameters is D1(n+p)**2-D1n**2=(R1*D)/R=\",round((R1*D)/R,2),\"cm**2\"\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "the difference of square of diameters is D1(n+p)**2-D1n**2=(lamda1*D)/lamda= 0.09375 cm**2\n", "the difference of square of diameters is D1(n+p)**2-D1n**2=(mu*D)/mu1= 0.094 cm**2\n", "the difference of square of diameters is D1(n+p)**2-D1n**2=(R1*D)/R= 0.25 cm**2\n" ] } ], "prompt_number": 27 } ], "metadata": {} } ] }