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 "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": [
      " \n",
      "#importing modules\n",
      "from __future__ import division\n",
      "import math\n",
      "\n",
      "#Variable declaration\n",
      "beta = 0.51;    #Fringe width(mm)\n",
      "d = 2.2;        #Distance between the slits(mm)\n",
      "D = 2;      #Distance between the slits and the screen(m)\n",
      "\n",
      "#Calculation\n",
      "beta = beta*10**-1;     #Fringe width(cm)\n",
      "d = d*10**-1;    #Distance between the slits(cm)\n",
      "D=D*10**2;    #Distance between the slits and the screen(cm)\n",
      "lamda = beta*d/D;    #Wavelength of light(cm)\n",
      "lamda = lamda*10**8;     #Wavelength of light(A)\n",
      "\n",
      "#Result\n",
      "print \"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": [
      " \n",
      "#importing modules\n",
      "from __future__ import division\n",
      "import math\n",
      "\n",
      "#Variable declaration\n",
      "lambda1 = 4250;    #First wavelength emitted by source of light(A)\n",
      "lambda2 = 5050;    #Second wavelength emitted by source of light(A)\n",
      "D = 1.5;    #Distance between the source and the screen(m)\n",
      "d = 0.025;       #Distance between the slits(mm)\n",
      "n = 3;    #Number of fringe from the centre\n",
      "\n",
      "#Calculation\n",
      "lambda1 = lambda1*10**-10;     #First wavelength emitted(m)\n",
      "lambda2 = lambda2*10**-10;     #Second wavelength emitted(m)\n",
      "d = d*10**-3;     #Distance between the slits(m)\n",
      "x3 = n*lambda1*D/d;    #Position of third bright fringe due to lambda1(m)\n",
      "x3_prime = n*lambda2*D/d;    #Position of third bright fringe due to lambda2(m)\n",
      "x = x3_prime-x3;      #separation between the third bright fringe(m)\n",
      "x = x*10**2;    #separation between the third bright fringe(cm)\n",
      "\n",
      "#Result\n",
      "print \"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": [
      " \n",
      "    \n",
      "#importing modules\n",
      "import math\n",
      "\n",
      "#Variable declaration\n",
      "lamda = 5.5*10**-5;    #Wavelength emitted by source of light(cm)\n",
      "n = 4;    #Number of fringes shifted\n",
      "t = 3.9*10**-4;    #Thickness of the thin glass sheet(cm)\n",
      "\n",
      "#Calculation\n",
      "mew = (n*lamda/t)+1;    #Refractive index of the sheet of glass\n",
      "mew = math.ceil(mew*10**4)/10**4;     #rounding off the value of v to 4 decimals\n",
      "\n",
      "#Result\n",
      "print \"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": [
      " \n",
      "#importing modules\n",
      "import math\n",
      "\n",
      "#Variable declaration\n",
      "lamda = 5893;    #Wavelength of monochromatic lihgt used(A)\n",
      "n = 1;    #Number of fringe for the least thickness of the film\n",
      "cosr = 1;    #for normal incidence\n",
      "mew = 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\n",
      "t = (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\n",
      "t1 = n*lamda/(2*mew*cosr);    #Thickness of the film that appears bright(A)\n",
      "\n",
      "#Result\n",
      "print \"The thickness of the film that appears bright is\",t, \"angstrom\"\n",
      "print \"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\n",
        "The 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": [
      " \n",
      "#importing modules\n",
      "import math\n",
      "\n",
      "#Variable declaration\n",
      "lamda = 5893;    #Wavelength of monochromatic lihgt used(A)\n",
      "n = 10;    #Number of fringe that are found \n",
      "d = 1;     #Distance of 10 fringes(cm)\n",
      "\n",
      "#Calculation\n",
      "beta = d/n;    #Fringe width(cm)\n",
      "lamda = lamda*10**-8;    #Wavelength of monochromatic lihgt used(cm)\n",
      "theta = lamda/(2*beta);    #Angle of the wedge(rad)\n",
      "theta = theta*10**4;\n",
      "theta = math.ceil(theta*10**4)/10**4;     #rounding off the value of theta to 4 decimals\n",
      "\n",
      "#Result\n",
      "print \"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": [
      " \n",
      "import math\n",
      "from __future__ import division\n",
      "\n",
      "#Variable declaration\n",
      "lamda = 5900;    #Wavelength of monochromatic lihgt used(A)\n",
      "t = 0.010;    #Spacer thickness(mm)\n",
      "l = 10;    #Wedge length(cm)\n",
      "\n",
      "#Calculation\n",
      "t = t*10**-1;    #Spacer thickness(cm)\n",
      "theta = t/l;    #Angle of the wedge(rad)\n",
      "lamda = lamda*10**-8;    #Wavelength of monochromatic lihgt used(cm)\n",
      "beta = lamda/(2*theta);    #Fringe width(cm)\n",
      "\n",
      "#Result\n",
      "print \"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": [
      " \n",
      "import math\n",
      "\n",
      "#Variable declaration\n",
      "D4 = 0.4;    #Diameter of 4th dark ring(cm)\n",
      "D12 = 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\n",
      "D20 = math.sqrt((2*D12**2)-D4**2);    #Diameter of 20th dark ring(cm)\n",
      "D20 = math.ceil(D20*10**4)/10**4;     #rounding off the value of D20 to 4 decimals\n",
      "\n",
      "#Result\n",
      "print \"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": [
      " \n",
      "#importing modules\n",
      "import math\n",
      "from __future__ import division\n",
      "\n",
      "#Variable declaration\n",
      "Dn = 0.30;    #Diameter of nth dark ring with air film(cm)\n",
      "dn = 0.25;    #Diameter of nth dark ring with liquid film(cm)\n",
      "\n",
      "#Calculation\n",
      "mew = (Dn/dn)**2;    #Refractive index of the liquid\n",
      "\n",
      "#Result\n",
      "print \"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": [
      " \n",
      "#importing modules\n",
      "import math\n",
      "\n",
      "#Variable declaration\n",
      "x = 0.002945;    #Distance through which movable mirror is shifted(cm)\n",
      "N = 100;    #Number of fringes shifted\n",
      "\n",
      "#Calculation\n",
      "x = x*10**-2;    #Distance through which movable mirror is shifted(m)\n",
      "lamda = 2*x/N;   #Wavelength of light(m)\n",
      "lamda = lamda*10**10;    #Wavelength of light(A)\n",
      "\n",
      "#Result\n",
      "print \"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": [
      " \n",
      "import math\n",
      "\n",
      "#Variable declaration\n",
      "lambda1 = 5896;    #Wavelength of D1 line of sodium(A)\n",
      "lambda2 = 5890;    #Wavelength of D2 line of sodium(A)\n",
      "\n",
      "#Calculation\n",
      "lamda = (lambda1+lambda2)/2;\n",
      "x = (lamda**2)/(2*(lambda1-lambda2));    #Shift in movable mirror of Michelson Interferometer(A)\n",
      "x = x*10**-7;           #Shift in movable mirror of Michelson Interferometer(mm)\n",
      "x = math.ceil(x*10**4)/10**4;     #rounding off the value of D20 to 4 decimals\n",
      "\n",
      "#Result\n",
      "print \"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": {}
  }
 ]
}