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+{
+ "metadata": {
+  "celltoolbar": "Raw Cell Format",
+  "name": "",
+  "signature": "sha256:fad8e22fb99cc3e157ab8315172e2ff6ddae35bb6f49be764a49d2d1d3f70fcc"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 1: Interference"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 1.1, Page number 1-16"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "#Given Data:\n",
+      "i=45*math.pi/180         #angle of incidence\n",
+      "u=1.33                   #Refractive index of a soap film\n",
+      "lamda=5.896*10**-7       #wavelength of required yellow light\n",
+      "\n",
+      "#Calculations:\n",
+      "#u=sin i/sin r           #Snell's law   .So,\n",
+      "r=math.asin(math.sin(i)/u)     #angle of reflection\n",
+      "\n",
+      "#Now, condition for bright fringe is\n",
+      "#2ut*cos r=(2n-1)lamda/2\n",
+      "#Here n=1\n",
+      "t=lamda/(2*2*u*math.cos(r))      #minimum thickness of film at which light will appear bright yellow\n",
+      "print\"Minimum thickness of film at which light will appear bright yellow of required wavelength is =\",t,\"m\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Minimum thickness of film at which light will appear bright yellow of required wavelength is = 1.30853030399e-07 m\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 1.2, Page number 1-16"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "#Given Data:\n",
+      "theta=40./3600*math.pi/180     #angle of wedge in radians\n",
+      "B=0.12*10**-2                 #fringe spacing\n",
+      "\n",
+      "#Calculations:\n",
+      "#We know, B=lam/(2*u*theta). Here u=1\n",
+      "lamda=2*B*theta               #wavelength of light used\n",
+      "print\"Wavelength of light used is =\",lamda,\"m\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Wavelength of light used is = 4.65421133865e-07 m\n"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 1.3, Page number 1-17"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "#Given Data:\n",
+      "i=30*math.pi/180        #angle of incidence\n",
+      "u=1.46                  #Refractive index of a oil\n",
+      "lamda=5.890*10**-7       #wavelength of required yellow light\n",
+      "n=8                     #eighth dark band\n",
+      "\n",
+      "#Calculations:\n",
+      "#u=sin i/sin r      #Snell's law   .So,\n",
+      "r=math.asin(math.sin(i)/u)     #angle of reflection\n",
+      "\n",
+      "#Now, condition for dark fringe is\n",
+      "#2ut*cos r=n*lamda\n",
+      "t=n*lamda/(2*u*math.cos(r))     #thickness of film\n",
+      "print\"Thickness of the film is =\",t,\"m\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Thickness of the film is = 1.71755887917e-06 m\n"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 1.4, Page number 1-17"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "#Given Data:\n",
+      "\n",
+      "B=0.1*10**-2               #fringe spacing\n",
+      "lamda=5.893*10**-7         #Wavelength of light\n",
+      "u=1.52                     #Refractive index of wedge\n",
+      "\n",
+      "#Calculations:\n",
+      "#We know, B=lamda/(2*u*theta). Here u=1\n",
+      "theta1=lamda/(2*u*B)            #angle of wedge in radians\n",
+      "theta=theta1*3600*180/math.pi   #angle of wedge in seconds\n",
+      "print\"Angle of wedge is =\",theta,\"seconds of an arc\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Angle of wedge is = 39.9841612899 seconds of an arc\n"
+       ]
+      }
+     ],
+     "prompt_number": 7
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 1.5, Page number 1-18"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "#Given Data:\n",
+      "t=0.2/(100)**2*10**-2     #thickness of film in meter\n",
+      "lamda=5.5*10**-7         #wavelength of light in meter\n",
+      "r=0                      #normal incidence\n",
+      "n=1                      #first band\n",
+      "\n",
+      "#Calculations:\n",
+      "\n",
+      "#Condition for dark fringe is\n",
+      "#2ut*cos r =n*lamda\n",
+      "u=n*lamda/(2*t*math.cos(r))   #Refractive index of a oil\n",
+      "print\"Refractive index of a oil is =\",u\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Refractive index of a oil is = 1.375\n"
+       ]
+      }
+     ],
+     "prompt_number": 9
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 1.6, Page number 1-18"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "#Given Data:\n",
+      "lamda=5.893*10**-7    #Wavelength of light\n",
+      "u=1.42                #Refractive index of a soap film\n",
+      "r=0                   #normal incidence\n",
+      "n=1                   #first band\n",
+      "\n",
+      "#Calculations:\n",
+      "\n",
+      "#(i)\n",
+      "#Condition for dark fringe is\n",
+      "#2ut*cos r=n*lamda\n",
+      "t1=n*lamda/(2*u*cos(r))      #thickness of film for dark black fringe\n",
+      "print\"Thickness of the film for dark black fringe is =\",t1,\"m\"\n",
+      "\n",
+      "#(ii)\n",
+      "#Now, condition for bright fringe is\n",
+      "#2ut*cos r=(2n-1)lamda/2\n",
+      "t2=lamda/(2*2*u*math.cos(r))      #Thickness of film for bright fringe\n",
+      "print\"Thickness of film for bright fringe is =\",t2,\"m\"\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Thickness of the film for dark black fringe is = 2.075e-07 m\n",
+        "Thickness of film for bright fringe is = 1.0375e-07 m\n"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 1.7, Page number 1-19"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "#Given Data:\n",
+      "i=30*math.pi/180         #angle of incidence\n",
+      "u=1.43               #Refractive index of a soap film\n",
+      "lamda=6*10**-7          #wavelength of light\n",
+      "n=1                  #For minimum thickness\n",
+      "\n",
+      "#Calculations:\n",
+      "#u=sin i/sin r      #Snell's law   .So,\n",
+      "r=(math.asin(math.sin(i)/u))    #angle of reflection\n",
+      "\n",
+      "#Now, condition of minima in transmitted system is\n",
+      "#2ut*cos(r)=(2n-1)lamda/2\n",
+      "t=lamda/(2*2*u*math.cos(r))      #minimum thickness of film\n",
+      "print\"Minimum thickness of film is \",t,\"m\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Minimum thickness of film is  1.11962124395e-07 m\n"
+       ]
+      }
+     ],
+     "prompt_number": 49
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 1.8, Page number 1-19"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "#Given Data:\n",
+      "\n",
+      "lamda = 5893*10**-10      #Wavelength of light\n",
+      "theta = 1               #assuming value of theta\n",
+      "\n",
+      "#We know, B=lamda/(2*u*theta). Here u=1\n",
+      "B = lamda/(2*theta)       #fringe spacing\n",
+      "n=20                  #interference fringes\n",
+      "\n",
+      "#Calculations:\n",
+      "#t=n*B*tan(theta)\n",
+      "t = 20*B*theta          #Thickness of wire\n",
+      "print\"Thickness of wire is =\",t,\"m\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Thickness of wire is = 5.893e-06 m\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 1.9, Page number 1-20"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "#Given Data:\n",
+      "u1=1.3                #Refractive index of oil\n",
+      "u2=1.5                #Refractive index of glass\n",
+      "lamda1=7*10**-7       #Wavelength of light\n",
+      "lamda2=5*10**-7       #Wavelength of light\n",
+      "\n",
+      "#Calculations:\n",
+      "\n",
+      "#for finding value of n, solve:\n",
+      "#(2n+1)*lamda1/2=(2(n+1)+1)*lamda2/2\n",
+      "#We get,n=2\n",
+      "n=2\n",
+      "\n",
+      "toil=(2*n+1)*lamda1/(2*u1*2)     #thickness of oil layer\n",
+      "print\"Thickness of oil layer is =\",toil,\"m\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Thickness of oil layer is = 6.73076923077e-07 m\n"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "Example 1.10, Page number 1-21"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": []
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "#Given Data:\n",
+      "u1=1.2               #Refractive index of drop of oil\n",
+      "u2=1.33              #Refractive index of water\n",
+      "lamda=4.8*10**-7     #wavelength of light\n",
+      "n=3                  #order\n",
+      "r=0                  #normal incidence,so r=0\n",
+      "\n",
+      "#Calculations:\n",
+      "t=n*lamda/(2*u1)      #Thickness of oil drop\n",
+      "print\"Thickness of oil drop is =\",t,\"m\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Thickness of oil drop is = 6e-07 m\n"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 1.11, Page number 1-22"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "#Given Data:\n",
+      "i=math.asin(4/5)          #angle of incidence\n",
+      "u=4/3                    #Refractive index of a soap film\n",
+      "lamda1=6.1*10**-7        #wavelength of light\n",
+      "lamda2=6*10**-7          #wavelength of light\n",
+      "\n",
+      "#Calculations:\n",
+      "#u=sin i/sin r      #Snell's law   .So,\n",
+      "r=math.asin(math.sin(i)/u)     #angle of reflection\n",
+      "\n",
+      "#Now, condition for dark band is\n",
+      "#2ut*cos r=n*lamda\n",
+      "#for consecutive bands, n=lamda2/(lamda1-lamda2). hence\n",
+      "\n",
+      "t=lamda2*lamda1/((lamda1-lamda2)*2*u*math.sqrt(1-(math.sin(i)/u)**2))     #thickness of film\n",
+      "print\"Thickness of the film is =\",t,\"m\"\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Thickness of the film is = 1.83e-05 m\n"
+       ]
+      }
+     ],
+     "prompt_number": 55
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 1.12, Page number 1-40"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "#Given Data:\n",
+      "n=10                #10th dark ring\n",
+      "Dn=0.5*10**-2       #Diameter of ring\n",
+      "lamda=6*10**-7      #wavelength of light\n",
+      "\n",
+      "#Calculations:\n",
+      "#As Dn^2=4*n*R*lamda\n",
+      "R=Dn**2/(4*n*lamda)    #Radius of curvature of the lens\n",
+      "print\"Radius of curvature of the lens is =\",R,\"m\"\n",
+      "\n",
+      "t=Dn**2/(8*R)        #thickness of air  film\n",
+      "print\"Thickness of air  film is =\",t,\"m\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Radius of curvature of the lens is = 1.04166666667 m\n",
+        "Thickness of air  film is = 3e-06 m\n"
+       ]
+      }
+     ],
+     "prompt_number": 9
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 1.13, Page number 1-41"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "#Given Data:\n",
+      "\n",
+      "B=0.25*10**-2             #fringe spacing\n",
+      "lamda=5.5*10**-7          #Wavelength of light\n",
+      "u=1.4                     #Refractive index of wedge\n",
+      "\n",
+      "#Calculations:\n",
+      "#We know, B=lamda/(2*u*theta).\n",
+      "theta1=lamda/(2*u*B)            #angle of wedge in radians\n",
+      "theta=theta1*3600*180/math.pi   #angle of wedge in seconds\n",
+      "print\"Angle of wedge is =\",theta,\"seconds\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Angle of wedge is = 16.2065204908 seconds\n"
+       ]
+      }
+     ],
+     "prompt_number": 13
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 1.14, Page number 1-41"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "#Given Data:\n",
+      "n=4              #4th dark ring\n",
+      "m=12             #m=n+p\n",
+      "D4=0.4*10**-2    #Diameter of 4th ring\n",
+      "D12=0.7*10**-2   #Diameter of 12th ring\n",
+      "\n",
+      "#Calculations:\n",
+      "\n",
+      "#(Dn+p)^2-Dn^2=4*p*lamda*R\n",
+      "#Solving, (D12^2-D4^2)/(D20^2-D4^2)\n",
+      "#We get above value =1/2. Hence\n",
+      "D20=math.sqrt(2*D12**2-D4**2)      #Diameter of 20th ring\n",
+      "print\"Diameter of 20th ring is =\",D20,\"m\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Diameter of 20th ring is = 0.00905538513814 m\n"
+       ]
+      }
+     ],
+     "prompt_number": 14
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 1.15, Page number 1-42"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "#Given Data:\n",
+      "n=6                  #6th bright ring\n",
+      "D6=0.31*10**-2       #Diameter of 6th ring\n",
+      "lamda=6*10**-7       #wavelength of light\n",
+      "R=1                  #Radius of curvature\n",
+      "\n",
+      "#Calculations:\n",
+      "\n",
+      "#Diameter of nth bright ring is \n",
+      "#Dn^2=2(2n-1)*lamda*R/u. Hence\n",
+      "u=2*(2*n-1)*lamda*R/(D6)**2    #Refractive index of liquid\n",
+      "print\"Refractive index of liquid is =\",u\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Refractive index of liquid is = 1.37356919875\n"
+       ]
+      }
+     ],
+     "prompt_number": 15
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 1.16, Page number 1-42"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "#Given Data:\n",
+      "lamda=6*10**-7         #wavelength of light\n",
+      "k=0.125*10**-4         #k=D(n+1)^2-Dn^2.\n",
+      "u=1                    #Refractive index of medium between lens and plate\n",
+      "#Calculations:\n",
+      "\n",
+      "#(i)\n",
+      "lamda1=4.5*10**-7      #new wavelength of light\n",
+      "#Difference between squres of diameters of successive rings is directly proportional to wavelength.So,\n",
+      "k1=lamda1/lamda*k      #new Difference between squres of diameters of successive rings after changing wavelength\n",
+      "print\"New Difference between squres of diameters of successive rings after changing wavelength is =\",k1,\"m^2\"\n",
+      "\n",
+      "#(ii)\n",
+      "u2=1.33          #Refractive index of liquid introduced between lens and plate\n",
+      "#Difference between squres of diameters of successive rings is inversely proportional to Refractive index.so,\n",
+      "k2=u/u2*k        #new Difference between squres of diameters of successive rings after changing refractive index\n",
+      "print\"New Difference between squres of diameters of successive rings after changing refrective index is =\",k2,\"m^2\"\n",
+      "\n",
+      "#(iii)\n",
+      "#Difference between squres of diameters of successive rings is directly proportional to Radius of curvature.So,\n",
+      "#after doubling radius of curvature,\n",
+      "k3=2*k           #new Difference between squres of diameters of successive rings after doubling radius of curvature\n",
+      "print\"New Difference between squres of diameters of successive rings after doubling radius of curvature is =\",k3,\"m^2\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "New Difference between squres of diameters of successive rings after changing wavelength is = 9.375e-06 m^2\n",
+        "New Difference between squres of diameters of successive rings after changing refrective index is = 9.3984962406e-06 m^2\n",
+        "New Difference between squres of diameters of successive rings after doubling radius of curvature is = 2.5e-05 m^2\n"
+       ]
+      }
+     ],
+     "prompt_number": 16
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 1.17, Page number 1-43"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "#Given Data:\n",
+      "Dn=0.225*10**-2         #Diameter of nth ring\n",
+      "Dm=0.45*10**-2          #Diameter of (n+9)th ring\n",
+      "lamda=6*10**-7          #wavelength of light\n",
+      "R=0.9                   #Radius of curvature\n",
+      "p=9\n",
+      "\n",
+      "#Calculations:\n",
+      "#(Dn+p)^2-Dn^2=4*p*lamda*R/u\n",
+      "u=4*p*lamda*R/((Dm)**2-Dn**2)     #Refractive index of liquid\n",
+      "print\"Refractive index of liquid is =\",u\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Refractive index of liquid is = 1.28\n"
+       ]
+      }
+     ],
+     "prompt_number": 17
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 1.18, Page number 1-44"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "#Given Data:\n",
+      "D10=0.5*10**-2          #Diameter of 10th ring\n",
+      "lamda=5.5*10**-7        #wavelength of light\n",
+      "u=1.25                  #Refractive index of liquid\n",
+      "\n",
+      "\n",
+      "#Calculations:\n",
+      "#As Dn^2=4*n*R*lamda/u\n",
+      "#Dn^2 is inversely proportional to refractive index.\n",
+      "D10n=D10/math.sqrt(u)        #new diameter of 10th ring after changing medium between lens and plate\n",
+      "print\"new diameter of 10th ring after changing medium between lens and plate is =\",D10n,\"m\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "new diameter of 10th ring after changing medium between lens and plate is = 0.004472135955 m\n"
+       ]
+      }
+     ],
+     "prompt_number": 18
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 1.19, Page number 1-45"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "#Given Data:\n",
+      "D5=0.336*10**-2       #Diameter of 5th ring\n",
+      "D15=0.59*10**-2       #Diameter of 15th ring\n",
+      "lamda=5.89*10**-7       #wavelength of light\n",
+      "p=10                  #n=5,n+p=15\n",
+      "\n",
+      "#Calculations:\n",
+      "#(Dn+p)^2-Dn^2=4*p*lamdaR/u\n",
+      "R=((D15)**2-D5**2)/(4*p*lamda)      #Radius of curvature of the lens\n",
+      "print\"Radius of curvature of the lens is =\",R,\"m\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Radius of curvature of the lens is = 0.998319185059 m\n"
+       ]
+      }
+     ],
+     "prompt_number": 20
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 1.20, Page number 1-45"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "#Given Data:\n",
+      "n=10               #10th dark ring\n",
+      "D10=0.6*10**-2     #Diameter of ring\n",
+      "lamda=6*10**-7     #wavelength of light\n",
+      "u=4./3              #Refractive index of water\n",
+      "\n",
+      "\n",
+      "#Calculations:\n",
+      "#As Dn^2=4*n*R*lamda/u\n",
+      "R=(D10**2)*u/(4*n*lamda)    #Radius of curvature of the lens\n",
+      "print\"Radius of curvature of the lens is =\",R,\"m\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Radius of curvature of the lens is = 2.0 m\n"
+       ]
+      }
+     ],
+     "prompt_number": 27
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 1.20.1, Page number 1-52"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "#Given Data:\n",
+      "i=45*math.pi/180        #angle of incidence\n",
+      "u=1.2                   #Refractive index of a film\n",
+      "t=4*10**-7              #thickness of film\n",
+      "\n",
+      "#Calculations:\n",
+      "#u=sin i/sin r       #Snell's law   .So,\n",
+      "r=math.asin(math.sin(i)/u)     #angle of reflection\n",
+      "\n",
+      "#Now, condition for dark fringe is\n",
+      "#2ut*cos r=n*lamda\n",
+      "lamda1=2*u*t*math.cos(r)/1          #n=1\n",
+      "print\"For n=1 wavelength is =\",lamda1,\"m\"\n",
+      "print\"This is in the visible spectrum and it will remain absent.\" \n",
+      "\n",
+      "lamda2=2*u*t*math.cos(r)/2          #n=2\n",
+      "print\"For n=2 wavelength is =\",lamda2,\"m\"\n",
+      "print\"This is not in the visible spectrum\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "For n=1 wavelength is = 7.75628777187e-07 m\n",
+        "This is in the visible spectrum and it will remain absent.\n",
+        "For n=2 wavelength is = 3.87814388593e-07 m\n",
+        "This is not in the visible spectrum\n"
+       ]
+      }
+     ],
+     "prompt_number": 30
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 1.20.2, Page number 1-53"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "#Given Data:\n",
+      "r=45*math.pi/180        #angle of refraction\n",
+      "u=1.45                  #Refractive index of a medium\n",
+      "lamda=5.5*10**-7        #wavelength of required yellow light\n",
+      "n=1\n",
+      "\n",
+      "#Calculations:\n",
+      "\n",
+      "#Now, condition for dark fringe is\n",
+      "#2ut*cos r=n*lamda\n",
+      "t=n*lamda/(2*u*math.cos(r))      #thickness of thin medium\n",
+      "print\"Thickness of the thin medium is =\",t,\"m\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Thickness of the thin medium is = 2.68212917002e-07 m\n"
+       ]
+      }
+     ],
+     "prompt_number": 31
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 1.21, Page number 1-45"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "#Given Data:\n",
+      "u=1.33                 #Refractive index of a soap film\n",
+      "r=0                    #normal incidence\n",
+      "t=5*10**-7             #thickness of film\n",
+      "\n",
+      "#Calculations:\n",
+      "\n",
+      "#Now, condition for maxima is\n",
+      "#2ut*cos r=(2n-1)lamda/2\n",
+      "lamda1=4*u*t*math.cos(r)/(2*1-1)          #n=1\n",
+      "print\"For n=1 wavelength is =\",lamda1,\"m\"\n",
+      "lamda2=4*u*t*math.cos(r)/(2*2-1)          #n=2\n",
+      "print\"For n=2 wavelength is =\",lamda2,\"m\"\n",
+      "lamda3=4*u*t*math.cos(r)/(2*3-1)          #n=3\n",
+      "print\"For n=3 wavelength is =\",lamda3,\"m\"\n",
+      "lamda4=4*u*t*math.cos(r)/(2*4-1)          #n=4\n",
+      "print\"For n=4 wavelength is =\",lamda4,\"m\"\n",
+      "\n",
+      "print\"Out of these wavelengths wavelength for n=3 lies in the visible spectrum.\"\n",
+      "print\"Hence, wavelength for n=3 is the most reflected wavelength.\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "For n=1 wavelength is = 2.66e-06 m\n",
+        "For n=2 wavelength is = 8.86666666667e-07 m\n",
+        "For n=3 wavelength is = 5.32e-07 m\n",
+        "For n=4 wavelength is = 3.8e-07 m\n",
+        "Out of these wavelengths wavelength for n=3 lies in the visible spectrum.\n",
+        "Hence, wavelength for n=3 is the most reflected wavelength.\n"
+       ]
+      }
+     ],
+     "prompt_number": 32
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 1.22, Page number 1-46"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "#Given Data:\n",
+      "u=1.5                #Refractive index of a oil\n",
+      "lamda=5.88*10**-7    #wavelength of required yellow light\n",
+      "n=1                  #for smallest thickness\n",
+      "r=60*math.pi/180     #angle of reflection\n",
+      "\n",
+      "#Calculations:\n",
+      "\n",
+      "#Now, condition for dark fringe is\n",
+      "#2ut*cos r=n*lamda\n",
+      "t=n*lamda/(2*u*math.cos(r))      #thickness of film\n",
+      "print\"Thickness of the film is =\",t,\"m\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Thickness of the film is = 3.92e-07 m\n"
+       ]
+      }
+     ],
+     "prompt_number": 33
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 1.23, Page number 1-46"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "#Given Data:\n",
+      "theta=20./3600*math.pi/180     #angle of wedge in radians\n",
+      "B=0.25*10**-2                 #fringe spacing\n",
+      "u=1.4                         #Refractive index of film\n",
+      "\n",
+      "#Calculations:\n",
+      "#We know, B=lamda/(2*u*theta).\n",
+      "lamda=2*B*theta*u             #wavelength of light\n",
+      "print\"Wavelength of light is =\",lamda,\"m\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Wavelength of light is = 6.78739153553e-07 m\n"
+       ]
+      }
+     ],
+     "prompt_number": 35
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 1.24, Page number 1-47"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "#Given Data:\n",
+      "#Dn=2*D40\n",
+      "\n",
+      "#Calculations:\n",
+      "#As Dn^2 = 4*n*R*lamda/u  and   Dn^2 = 4*D40^2\n",
+      "#i.e. 4*n*R*lamda/u = 4*4*40*R*lamda/u .hence,\n",
+      "n=4*40          #order of the required ring\n",
+      "print\"Order of the dark ring which will have double the diameter of that of 40th ring is =\",n\n",
+      " \n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Order of the dark ring which will have double the diameter of that of 40th ring is = 160\n"
+       ]
+      }
+     ],
+     "prompt_number": 36
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 1.25, Page number 1-47"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "#Given Data:\n",
+      "lamda1=6*10**-7       #wavelength of light\n",
+      "lamda2=4.5*10**-7    #wavelength of light\n",
+      "R=0.9                 #Radius of curvature\n",
+      "\n",
+      "#Calculations:\n",
+      "#As Dn^2=4*n*R*lamda.\n",
+      "#Dn^2=D(n+1)^2 for different wavelengths.we get,\n",
+      "n=lamda2/(lamda1-lamda2)       #nth dark ring due to lam1 which coincides with (n+1)th dark ring due lamda2\n",
+      "D3=math.sqrt(4*n*R*lamda1)     #diameter of 3rd dark ring for lamda1\n",
+      "print\"Diameter of 3rd dark ring for lam1 is =\",D3,\"m\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Diameter of 3rd dark ring for lam1 is = 0.00254558441227 m\n"
+       ]
+      }
+     ],
+     "prompt_number": 38
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 1.26, Page number 1-48"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "#Given Data:\n",
+      "i=45*math.pi/180        #angle of incidence\n",
+      "u=4./3                  #Refractive index of soap film\n",
+      "lamda=5*10**-7          #wavelength of light\n",
+      "t=1.5*10**-6            #thickness of film\n",
+      "\n",
+      "#Calculations:\n",
+      "#u=sin i/sin r        #Snell's law   .So,\n",
+      "r=math.asin(math.sin(i)/u)     #angle of reflection\n",
+      "\n",
+      "#Now, condition for dark band is\n",
+      "#2ut*cos r=n*lamda\n",
+      "n=2*u*t*math.cos(r)/lamda    #order of band\n",
+      "print\"order of dark band is =\",n\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "order of dark band is = 6.78232998313\n"
+       ]
+      }
+     ],
+     "prompt_number": 39
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 1.27, Page number 1-49"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "#Given Data:\n",
+      "D5=0.336*10**-2         #Diameter of 5th ring\n",
+      "D15=0.59*10**-2         #Diameter of 15th ring\n",
+      "lamda=5.89*10**-7       #wavelength of light\n",
+      "p=10                    #n=5,n+p=15\n",
+      "\n",
+      "#Calculations:\n",
+      "#(Dn+p)^2-Dn^2=4*p*lamda*R/u\n",
+      "R=((D15)**2-D5**2)/(4*p*lamda)    #Radius of curvature of the lens\n",
+      "print\"Radius of curvature of the lens is =\",R,\"m\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Radius of curvature of the lens is = 0.998319185059 m\n"
+       ]
+      }
+     ],
+     "prompt_number": 40
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 1.29, Page number 1-50"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "#As Dn^2=4*n*R*lamda.\n",
+      "#thus, Dn is directly proportional to sqaure root of n\n",
+      "D5=math.sqrt(5)     #D5 is directly proportional to sqaure root of 5\n",
+      "D4=math.sqrt(4)     #D4 is directly proportional to sqaure root of 4\n",
+      "k1=D5-D4\n",
+      "print\"Separation between D5 and D4 is directly proportional to =\",k1\n",
+      "\n",
+      "D80=math.sqrt(80)     #D80 is directly proportional to sqaure root of 80\n",
+      "D79=math.sqrt(79)     #D79 is directly proportional to sqaure root of 79\n",
+      "k2=D80-D79\n",
+      "print\"Separation between D80 and D79 is directly proportional to =\",k2\n",
+      "\n",
+      "print\"Thus, (D80-D79) < (D5-D4). Hence proved.\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Separation between D5 and D4 is directly proportional to = 0.2360679775\n",
+        "Separation between D80 and D79 is directly proportional to = 0.0560774926836\n",
+        "Thus, (D80-D79) < (D5-D4). Hence proved.\n"
+       ]
+      }
+     ],
+     "prompt_number": 41
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 1.30, Page number 1-51"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "#Given Data:\n",
+      "D5=0.336*10**-2       #Diameter of 5th ring\n",
+      "D15=0.59*10**-2       #Diameter of 15th ring\n",
+      "p=10                  #n=5,n+p=15\n",
+      "R=1                   #Radius of curvature\n",
+      "\n",
+      "#Calculations:\n",
+      "#(Dn+p)^2-Dn^2=4*p*lamda*R/u\n",
+      "lamda=((D15)**2-D5**2)/(4*p*R)    #Wavelength of light\n",
+      "print\"Wavelength of light is =\",lamda,\"m\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Wavelength of light is = 5.8801e-07 m\n"
+       ]
+      }
+     ],
+     "prompt_number": 42
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 1.31, Page number 1-51"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "\n",
+      "#Condition for bright band is\n",
+      "#2ut*cos r = (2n-1)*lamda1\n",
+      "\n",
+      "#for consecutive bands, 2n=(lamda1+lamda2)/(lamda1-lamda).\n",
+      "#thus, 2ut*cos r = lamda2*lamda1/(lamda1-lamda2)\n",
+      "\n",
+      "#And, thicknessof film \n",
+      "#t= lamda2*lamda1/((2*u*cosr)(lamda1-lamda2))\n",
+      "print\"Hence expression for thickness of film is obtained.\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Hence expression for thickness of film is obtained.\n"
+       ]
+      }
+     ],
+     "prompt_number": 56
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [],
+     "language": "python",
+     "metadata": {},
+     "outputs": []
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file