path: root/Engineering_Physics/chapter2.ipynb

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  {

   "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": {}

  }

 ]

}