{ "metadata": { "celltoolbar": "Raw Cell Format", "name": "", "signature": "sha256:5f637d4fa0aa8bb3bc54a28134ac85a937c06f0d3a591f536de2719824a41f5c" }, "nbformat": 3, "nbformat_minor": 0, "worksheets": [ { "cells": [ { "cell_type": "heading", "level": 1, "metadata": {}, "source": [ "Chapter 3: Optical Fibre" ] }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 3.1,Page number 3-19" ] }, { "cell_type": "code", "collapsed": false, "input": [ "import math\n", "\n", "#Given Data:\n", "n1=1.61 #Core index\n", "n2=1.55 #Cladding index\n", "\n", "#Calculations:\n", "NA=math.sqrt(n1**2-n2**2) #Formula\n", "\n", "print\"Numerical Aperture of Fibre is = \",NA\n", "\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "Numerical Aperture of Fibre is = 0.435430821142\n" ] } ], "prompt_number": 2 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 3.2,Page number 3-19" ] }, { "cell_type": "code", "collapsed": false, "input": [ "import math\n", "\n", "#Given Data:\n", "n1=1.65 #Core index\n", "n2=1.53 #Cladding index\n", "\n", "#Calculations:\n", "NA=math.sqrt(n1**2-n2**2) #Formula\n", "\n", "print\"Numerical Aperture of Fibre is =\",NA\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "Numerical Aperture of Fibre is = 0.617737808459\n" ] } ], "prompt_number": 3 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 3.3,Page number 3-19" ] }, { "cell_type": "code", "collapsed": false, "input": [ "import math\n", "\n", "#Given Data:\n", "n1=1.48 #R.I. of Core\n", "n2=1.39 #R.I. of Cladding\n", "\n", "#Calculations:\n", "NA=math.sqrt(n1**2-n2**2) #Formula to find NA\n", "phi=math.asin(NA)*180/(3.1472) #Acceptance angle\n", "\n", "print\"Numerical Aperture of Fibre is =\",NA\n", "print\"Acceptance angle of Fibre is =\",phi,\"degrees\"" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "Numerical Aperture of Fibre is = 0.508232230383\n", "Acceptance angle of Fibre is = 30.491727191 degrees\n" ] } ], "prompt_number": 8 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 3.4,Page number 3-20" ] }, { "cell_type": "code", "collapsed": false, "input": [ "import math\n", "\n", "#given data:\n", "u1=3.6 #Refractive Index of the Substance at 850 nm\n", "u2=3.4 #Refractive Index of the Substance at 1300 nm\n", "Vv=3*10**8 #Velocity of light in free space\n", "\n", "#Calculations:\n", "# i)Finding wavelength at 850 nm\n", "Vs1=Vv/u1 #Velocity of light in substance at 850 nm\n", "print\"Velocity of light in substance at 850 nm =\" ,Vs1,\"m/sec\"\n", "\n", "lam1=850*10**-9/u1 #Wavelength of light in substance at 850nm\n", "print\" Wavelength of light in substance at 850nm =\",lam1,\"m\"\n", "\n", "\n", "#ii)Finding wavelength at 1300 nm\n", "Vs2=Vv/u2 #Velocity of light in substance at 1300 nm\n", "print\"Velocity of light in substance at 1300 nm =\",Vs2,\" m/sec\"\n", "\n", "lam2=1300*10**-9/u2 #Wavelength of light in substance at 1300nm\n", "print\"Wavelength of light in substance at 1300nm =\" ,lam2,\"m \"" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "Velocity of light in substance at 850 nm = 83333333.3333 m/sec\n", " Wavelength of light in substance at 850nm = 2.36111111111e-07 m\n", "Velocity of light in substance at 1300 nm = 88235294.1176 m/sec\n", "Wavelength of light in substance at 1300nm = 3.82352941176e-07 m \n" ] } ], "prompt_number": 2 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 3.5,Page number 3-20" ] }, { "cell_type": "code", "collapsed": false, "input": [ "import math\n", "\n", "#Given Data:\n", "u1=1.5 #R.I. of Core\n", "u2=1.45 #R.I.of Cladding\n", "delta= (u1-u2)/u1 #Fractional Refractive index\n", "\n", "#Calculations:\n", "NA=u1*sqrt(2*delta) #Formula to find NA\n", "theta0=math.asin(NA)*180/(3.1472) #Acceptance angle\n", "thetac=math.asin(u2/u1)*180/(3.1472) #Critical angle\n", "\n", "print\"Numerical Aperture of Fibre is =\",NA\n", "print\"Acceptance angle of Fibre is =\",theta0,\"degrees\" \n", "print\" Critical angle of Fibre is =\",thetac,\"degrees\" \n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "Numerical Aperture of Fibre is = 0.387298334621\n", "Acceptance angle of Fibre is = 22.7458994397 degrees\n", " Critical angle of Fibre is = 75.0309676099 degrees\n" ] } ], "prompt_number": 7 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 3.6,Page number 3-20" ] }, { "cell_type": "code", "collapsed": false, "input": [ "import math\n", "\n", "#Given Data:\n", "NA=0.22 # Numerical Aperture of Fibre\n", "delta=0.012 #Fractional index\n", "\n", "#Calculations:\n", "#Delta=(u1-u2)/u1\n", "u1=NA/math.sqrt(2*delta) #Formula\n", "u2=u1-(u1*delta) #Formula\n", "\n", "print\"Refractive Index of core of fibre is =\",u1\n", "print\"Refractive Index of cladding of fibre is =\",u2\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "Refractive Index of core of fibre is = 1.42009389361\n", "Refractive Index of cladding of fibre is = 1.40305276689\n" ] } ], "prompt_number": 1 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 3.7,Page number 3-21" ] }, { "cell_type": "code", "collapsed": false, "input": [ "import math\n", "\n", "#Given Data:\n", "u1=1.466 #R.I. of Core\n", "u2=1.46 #R.I.of Cladding\n", "V=2.4 #Cut off parameter\n", "lamda=0.8*10**-6 #wavelength in meter\n", "\n", "#Calculations:\n", "NA=math.sqrt(u1**2-u2**2) #Formula to find Numerical Aperture\n", "print\"Numerical Aperture of Fibre is =\",NA\n", "#(printing mistake in book)printed answer is 1.13 but correct answer is 0.13\n", "print\"(printing mistake in book)\"\n", "\n", "# V = 2*3.142*a*NA / lamda\n", "a=V*lamda/(2*3.142*NA) #core radius\n", "print\"Core radius of Fibre is (a) =\",a,\"m\"\n", "\n", "#w/a= 1.1\n", "w=1.1*a #Spot size\n", "print\"Spot size of Fibre is =\",w,\"m\"\n", "\n", "theta=2*lamda*180/3.142/(3.142*w) #Divergence angle\n", "print\"Divergence angle of Fibre is =\",theta,\"degrees\"\n", "\n", "w10=lamda*10/(3.142*w) #Spot size at 10 m\n", "print\"Spot size at 10 m of Fibre is =\",w10,\"m\"\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "Numerical Aperture of Fibre is = 0.1324990566\n", "(printing mistake in book)\n", "Core radius of Fibre is (a) = 2.30596263706e-06 m\n", "Spot size of Fibre is = 2.53655890076e-06 m\n", "Divergence angle of Fibre is = 11.5009886523 degrees\n", "Spot size at 10 m of Fibre is = 1.00378073182 m\n" ] } ], "prompt_number": 5 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 3.8,Page number 3-21" ] }, { "cell_type": "code", "collapsed": false, "input": [ "import math\n", "\n", "#Given Data:\n", "w=98 #Spot size in meter\n", "d=50*10**-6 #Core diameter in meter\n", "a=d/2 #core radius\n", "u1=1.47 #R.I. of Core\n", "u2=1.45 #R.I.of Cladding\n", "lamda=0.85*10**-6 #Wavlength in meter\n", "NA=math.sqrt(u1**2-u2**2) #Formula to find NA\n", "\n", "#Calculations:\n", "V=2*3.142*a*NA/lamda #cut off parameter\n", "N=(V**2)/2 #Number of modes\n", "\n", "print\"Cut off parameter of Fibre is =\",V\n", "print\"Number of modes of Fibre is =\",N\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "Cut off parameter of Fibre is = 44.6646240577\n", "Number of modes of Fibre is = 997.464321107\n" ] } ], "prompt_number": 6 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 3.9,Page number 3-21" ] }, { "cell_type": "code", "collapsed": false, "input": [ "import math\n", "\n", "#Given Data:\n", "u1=1.47 #R.I. of Core\n", "u2=1.46 #R.I.of Cladding\n", "lamda=1.3*10**-6 #wavelength in meter\n", "\n", "#Calculations:\n", "NA=math.sqrt(u1**2-u2**2) #Formula to find Numerical Aperture\n", "\n", "#The condition for single mode is V<2.405\n", "#2*3.142*a*NA/lamda < 2.405\n", "\n", "a=2.405*lamda/(2*3.142*NA) #Maximum radius of fibre\n", "\n", "print\"Maximum radius of Fibre is =\",a,\"meter\"\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "Maximum radius of Fibre is = 2.90662126448e-06 meter\n" ] } ], "prompt_number": 7 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 3.10,Page number 3-22" ] }, { "cell_type": "code", "collapsed": false, "input": [ "import math\n", "\n", "#Given Data:\n", "u1=1.465 #R.I. of Core\n", "u2=1.46 #R.I.of Cladding\n", "lamda=1.25*10**-6 #operating wavelength\n", "\n", "#Calculations:\n", "delta=(u1-u2)/u1 #Fractional Refractive index\n", "print\"Fractional Refractive index of Fibre is =\",delta\n", "\n", "#For single mode propagation codition is \n", "# a/lamda < 1.4/(3.142*sqrt(u1(u1-u2)))\n", "\n", "a=lamda*1.4/(3.142*u1*math.sqrt(delta)) #core radius\n", "\n", "u=u1-(math.sqrt(2*delta)/(2*3.142*(a/lamda))) #effective refractive index\n", "print\"Effective Refractive index for lowest mode propagation is =\",u\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "Fractional Refractive index of Fibre is = 0.00341296928328\n", "Effective Refractive index for lowest mode propagation is = 1.46247461864\n" ] } ], "prompt_number": 8 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 3.11,Page number 3-22" ] }, { "cell_type": "code", "collapsed": false, "input": [ "import math\n", "\n", "#Given Data:\n", "u1=1.54 #R.I. of Core\n", "u2=1.5 #R.I.of Cladding\n", "lamda=1.3*10**-6 #wavelength in meter\n", "a=25*10**-6 #core radius in meter\n", "\n", "#Calculations:\n", "NA=math.sqrt(u1**2-u2**2) #Formula to find Numerical Aperture\n", "\n", "V=2*3.142*a*NA/lamda #cut off parameter\n", "print\"Cut off parameter of Fibre is =\",V\n", "\n", "N=(V**2)/2 #Number of modes\n", "print\" Number of modes of Fibre is =\",N\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "Cut off parameter of Fibre is = 42.1404937865\n", " Number of modes of Fibre is = 887.910608284\n" ] } ], "prompt_number": 9 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 3.11.1,Page number 3-25" ] }, { "cell_type": "code", "collapsed": false, "input": [ "import math\n", "\n", "#Given Data:\n", "u1=1.52 #R.I. of Core\n", "u2=1.5189 #R.I.of Cladding\n", "lamda=1.3*10**-6 #wavelength in meter\n", "d=29*10**-6 #core diameter in meter\n", "a=d/2\n", "\n", "#Calculations:\n", "NA=math.sqrt(u1**2-u2**2) #Formula to find Numerical Aperture\n", "V=2*3.142*a*NA/lamda #Normalised frequency\n", "Nm=(V**2)/2 #Number of modes\n", "\n", "print\"Normalised frequency of Fibre is (V)=\",V\n", "print\"The Maximum Number of modes the Fibre will support is (Nm) =\",Nm\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "Normalised frequency of Fibre is (V)= 4.05242861605\n", "The Maximum Number of modes the Fibre will support is (Nm) = 8.2110888441\n" ] } ], "prompt_number": 11 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 3.12,Page number 3-22" ] }, { "cell_type": "code", "collapsed": false, "input": [ "import math\n", "\n", "#Given Data:\n", "u1=1.5 #R.I. of Core\n", "d=10*10**-6 #diameter of core\n", "a=d/2 #core radius\n", "lamda=1.3*10**-6 #wavelength\n", "V=2.405 #cut off parameter for single mode\n", "\n", "#Calculations:\n", "\n", "#We know, V=2*3.142*a*NA/lamda\n", "NA=V*lamda/(2*3.142*a) #Numerical Aperture\n", "\n", "theta=math.asin(NA)*180/3.142 #Acceptance angle\n", "print\"Acceptance angle of Fibre is =\",theta,\"Degrees\"\n", "\n", "#Also, NA=u1*sqrt(2*delta)\n", "delta=(NA/u1)**2/2 #Fractional index\n", "print\"Maximum Fractional Refractive index of Fibre is =\",delta\n", "\n", "#delta=(u1-u2)/u1\n", "u2=u1*(1-delta) #R.I.of cladding\n", "print\"Refractive index of cladding of Fibre is =\",u2\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "Acceptance angle of Fibre is = 5.71002346964 Degrees\n", "Maximum Fractional Refractive index of Fibre is = 0.00220035113094\n", "Refractive index of cladding of Fibre is = 1.4966994733\n" ] } ], "prompt_number": 13 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 3.13,Page number 3-23" ] }, { "cell_type": "code", "collapsed": false, "input": [ "import math\n", "\n", "#Given Data:\n", "n1=1.5 #R.I. of core\n", "delta=0.0005 #Fractional index difference\n", "\n", "#Calculations:\n", "#(a):\n", "#Delta=(u1-u2)/u1\n", "n2=n1-(n1*delta) #R.I. of cladding\n", "print\"(a)Refractive Index of cladding of fibre is =\",n2\n", "\n", "#(b):\n", "phi=math.asin(n2/n1)*180/3.142 #Critical internal reflection angle\n", "print\"(b)Critical internal reflection angle of Fibre is =\",phi,\"degrees\"\n", "\n", "#(c):\n", "theta0=math.asin(math.sqrt(n1**2-n2**2))*180/3.142 #External critical Acceptance angle\n", "print\"(c)External critical Acceptance angle of Fibre is =\",theta0,\"degrees\"\n", "\n", "#(d):\n", "NA=n1*math.sqrt(2*delta) #Formula to find Numerical Aperture\n", "print\"(d)Numerical Aperture of Fibre is =\",NA\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "(a)Refractive Index of cladding of fibre is = 1.49925\n", "(b)Critical internal reflection angle of Fibre is = 88.1766396681 degrees\n", "(c)External critical Acceptance angle of Fibre is = 2.71810509125 degrees\n", "(d)Numerical Aperture of Fibre is = 0.0474341649025\n" ] } ], "prompt_number": 14 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 3.14,Page number 3-24" ] }, { "cell_type": "code", "collapsed": false, "input": [ "import math\n", "\n", "#Given Data:\n", "NA1=0.20 #Numerical Aperture of Fibre\n", "n2=1.59 #R.I. of cladding\n", "\n", "#Calculations:\n", "#NA=sqrt(n1**2-n2**2)\n", "#In air, n0=1\n", "n1=math.sqrt(NA1**2+n2**2) #R.I.of core\n", "\n", "#Now, in water \n", "n0=1.33\n", "NA2=math.sqrt(n1**2-n2**2)/n0 #Numerical Aperture in water\n", "theta0=math.asin(NA2)*180/3.142 #Acceptance angle of fibre in water\n", "print\"Acceptance angle of Fibre in water is =\",theta0,\"degrees\"\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "Acceptance angle of Fibre in water is = 8.6475921767 degrees\n" ] } ], "prompt_number": 15 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 3.15,Page number 3-24" ] }, { "cell_type": "code", "collapsed": false, "input": [ "import math\n", "\n", "#Given Data:\n", "n1=1.45 #R.I.of core\n", "n2=1.40 #R.I. of cladding\n", "\n", "#Calculations:\n", "NA=math.sqrt(n1**2-n2**2) #Numerical Aperture\n", "print\"Numerical Aperture of Fibre is =\",NA\n", "\n", "theta0=math.asin(NA)*180/3.142 #Acceptance angle of fibre\n", "print\"Acceptance angle of Fibre is =\",theta0,\"degrees\"\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "Numerical Aperture of Fibre is = 0.377491721764\n", "Acceptance angle of Fibre is = 22.1755250876 degrees\n" ] } ], "prompt_number": 16 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 3.16,Page number 3-24" ] }, { "cell_type": "code", "collapsed": false, "input": [ "import math\n", "\n", "#Given Data:\n", "NA=0.16 #Numerical Aperture of Fibre\n", "n1=1.45 #R.I. of core\n", "d=90*10**-6 #Core diameter\n", "\n", "#Calculations:\n", "#NA=sqrt(n1**2-n2**2)\n", "n2=math.sqrt(n1**2-NA**2) #R.I.of cladding\n", "print\"(a)Refractive Index of cladding of fibre is =\",n2\n", "\n", "theta0=math.asin(NA)*180/3.142 #Acceptance angle of fibre\n", "print\"(b)Acceptance angle of Fibre is =\",theta0,\"degrees\"\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "(a)Refractive Index of cladding of fibre is = 1.44114537782\n", "(b)Acceptance angle of Fibre is = 9.20570258795 degrees\n" ] } ], "prompt_number": 18 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 3.17,Page number 3-25" ] }, { "cell_type": "code", "collapsed": false, "input": [ "import math\n", "\n", "#Given Data:\n", "n1=1.48 #R.I. of core\n", "delta=0.055 #Realtive R.I.\n", "lamda=1*10**-6 #Wavelength of light\n", "a=50*10**-6 #core radius\n", "\n", "#Calculations:\n", "#Delta=(u1-u2)/u1\n", "n2=n1-(n1*delta) #R.I. of cladding\n", "NA=n1*math.sqrt(2*delta) #Formula to find Numerical Aperture\n", "print\"Numerical Aperture of Fibre is =\",NA\n", "\n", "\n", "theta0=math.asin(NA)*180/3.142 #Acceptance angle of fibre\n", "print\"Acceptance angle of Fibre is =\",theta0,\"degrees\"\n", "\n", "V=2*3.142*a*NA/lamda #V number\n", "N=(V**2)/2 #Number of guided modes\n", "\n", "#In book,instead of NA , value of delta is taken into calculation.\n", "#Thus there is calculation mistake in values of V and N.\n", "\n", "print\"V number of Fibre is =\",V\n", "print\"Number of guided mode of Fibre is =\",N\n", "print\"(Calculation mistake in book)\"\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "Numerical Aperture of Fibre is = 0.490860468973\n", "Acceptance angle of Fibre is = 29.3933421943 degrees\n", "V number of Fibre is = 154.228359351\n", "Number of guided mode of Fibre is = 11893.1934141\n", "(Calculation mistake in book)\n" ] } ], "prompt_number": 19 }, { "cell_type": "code", "collapsed": false, "input": [], "language": "python", "metadata": {}, "outputs": [] } ], "metadata": {} } ] }