{ "metadata": { "name": "", "signature": "sha256:742c44cb267900361b88ee7b473acd2b1b40f32a4db7e15db6918955b1159df0" }, "nbformat": 3, "nbformat_minor": 0, "worksheets": [ { "cells": [ { "cell_type": "heading", "level": 1, "metadata": {}, "source": [ "Chapter01:Fiber Optics Communications System" ] }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "\n", "Ex1.7.1:Pg-1.14" ] }, { "cell_type": "code", "collapsed": false, "input": [ "#Given\n", "import math\n", "n1= 1.5 # for glass\n", "n2= 1.33 # for water\n", "phi1= (math.pi/6) # phi1 is the angel of incidence\n", " # According to Snell's law...\n", " # n1*sin(phi1)= n2*sin(phi2) \n", "sinphi2= (n1/n2)*math.sin(phi1) # phi2 is the angle of refraction..\n", "phi2 = math.asin(sinphi2)\n", "temp= math.degrees(phi2)\n", "print \" The angel of refraction in degrees =\",round(temp,2) \n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ " The angel of refraction in degrees = 34.33\n" ] } ], "prompt_number": 11 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Ex1.7.2:Pg-1.14" ] }, { "cell_type": "code", "collapsed": false, "input": [ "# Given\n", "import math\n", "n1= 1.50 # RI of glass..\n", "n2 = 1.0 # RI of air...\n", " # According to Snell's law...\n", " # n1*sin(phi1)= n2*sin(phi2) \n", " # From definition of critical angel phi2 = 90 degrees and phi1 will be critical angel\n", "t1=(n2/n1)*math.sin(math.pi/2)\n", "phiC=math.asin(t1)\n", "temp= math.degrees(phiC)\n", "print \" The Critical angel in degrees =\",round(temp,2) \n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ " The Critical angel in degrees = 41.81\n" ] } ], "prompt_number": 26 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Ex1.7.3:Pg-1.15" ] }, { "cell_type": "code", "collapsed": false, "input": [ " # Given\n", " # To find RI of glass\n", " # To find the critical angle for glass...\n", " \n", "phi1 = 33 # Angle of incidence..\n", "phi2 = 90 # Angle of refraction..\n", "n2= 1.0 \n", "\n", "n1 = round(sin(math.radians(phi2))/sin(math.radians(phi1)),3) \n", "print \" The Refractive Index is =\",n1 \n", "\n", "#phiC = math.asin((n2/n1)*math.sin(90)) \n", "phiC=math.asin(0.54)\n", "print \" \\n\\nThe Critical angel in degrees =\",round(math.degrees(phiC),2)\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ " The Refractive Index is = 1.836\n", " \n", "\n", "The Critical angel in degrees = 32.68\n" ] } ], "prompt_number": 81 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Ex1.7.4:Pg-1.15" ] }, { "cell_type": "code", "collapsed": false, "input": [ " # Given\n", "import math\n", " \n", "n1= 1.5 # TheRi of medium 1\n", "n2= 1.36 # the RI of medium 2\n", "phi1= 30 # The angle of incidence\n", " # According to Snell's law...\n", " # n1*sin(phi1)= n2*sin(phi2) \n", "phi2 = math.asin((n1/n2)*math.sin(math.radians(phi1))) \n", "print \" The angel of refraction is in degrees from normal = \",round(math.degrees(phi2),2)\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ " The angel of refraction is in degrees from normal = 33.47\n" ] } ], "prompt_number": 91 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Ex1.7.5:Pg-1.16" ] }, { "cell_type": "code", "collapsed": false, "input": [ "# Given\n", "import math\n", " \n", "n1 = 3.6 # RI of GaAs..\n", "n2 = 3.4 # RI of AlGaAs..\n", "phi1 = 80 # Angle of Incidence..\n", " # According to Snell's law...\n", " # n1*sin(phi1)= n2*sin(phi2) \n", " # At critical angle phi2 = 90...\n", "phiC = math.asin((n2/n1)*sin(math.radians(90)) )\n", "print \" The Critical angel in degrees =\",round(math.degrees(phiC),2) \n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ " The Critical angel in degrees = 70.81\n" ] } ], "prompt_number": 97 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Ex1.9.1:Pg-1.22" ] }, { "cell_type": "code", "collapsed": false, "input": [ "# Given\n", "import math \n", "\n", "n1= 1.5 # RI of medium 1\n", "n2 =1.45 # RI of medium 2\n", "\n", "delt= (n1-n2)/n1 \n", "NA = n1*(math.sqrt(2*delt)) \n", "print \" The Numerical aperture =\",round(NA,2)\n", "phiA = math.asin(NA) \n", "print \" \\n\\nThe Acceptance angel in degrees =\",round(math.degrees(phiA),2) \n", "\n", "phiC = math.asin(n2/n1) \n", "print \" \\n\\nThe Critical angel in degrees =\",round(degrees(phiC),2)\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ " The Numerical aperture = 0.39\n", " \n", "\n", "The Acceptance angel in degrees = 22.79\n", " \n", "\n", "The Critical angel in degrees = 75.16\n" ] } ], "prompt_number": 100 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Ex1.9.2:Pg-1.23" ] }, { "cell_type": "code", "collapsed": false, "input": [ "# Given\n", "import math\n", "\n", "n1= 1.5 # RI of core\n", "n2 = 1.48 # RI of cladding..\n", "\n", "NA = math.sqrt((n1**2)-(n2**2)) \n", "print \" The Numerical Aperture =\",round(NA,2) \n", "\n", "phiA = math.asin(NA) \n", "print \" \\n\\nThe Critical angel =\",round(math.degrees(phiA),2) \n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ " The Numerical Aperture = 0.24\n", " \n", "\n", "The Critical angel = 14.13\n" ] } ], "prompt_number": 101 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Ex1.9.3:Pg-1.23" ] }, { "cell_type": "code", "collapsed": false, "input": [ "# Given\n", "import math \n", " \n", "\n", "NA = 0.35 # Numerical Aperture\n", "delt = 0.01 \n", " # NA= n1*(math.sqrt(2*delt) n1 is RI of core\n", "n1 = 0.35/(math.sqrt(2*delt)) \n", "print \"The RI of core =\",round(n1,4) \n", "\n", " # Numerical Aperture is also given by \n", " # NA = math.sqrt(n1**2 - n2**2) # n2 is RI of cladding\n", "n2 = math.sqrt((n1**2-NA**2)) \n", "print \" \\n\\nThe RI of Cladding =\",round(n2,3) \n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "The RI of core = 2.4749\n", " \n", "\n", "The RI of Cladding = 2.45\n" ] } ], "prompt_number": 104 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Ex1.9.4:Pg-1.24" ] }, { "cell_type": "code", "collapsed": false, "input": [ "# Given\n", "import math\n", "\n", "Vc = 2.01*10**8 # velocity of light in core in m/sec...\n", "phiC= 80.0 # Critical angle in degrees...\n", "\n", " # RI of Core (n1) is given by (Velocity of light in air/ velocity of light in air)...\n", "n1= 3*10**8/Vc \n", " # From critical angle and the value of n1 we calculate n2...\n", "n2 = sin(math.radians(phiC))*n1 # RI of cladding...\n", "NA = math.sqrt(n1**2-n2**2) \n", "print \" The Numerical Aperture =\",round(NA,2) \n", "phiA = math.asin(NA) # Acceptance angle...\n", "print \" \\n\\nThe Acceptance angel in degrees =\",round(math.degrees(phiA),2) \n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ " The Numerical Aperture = 0.26\n", " \n", "\n", "The Acceptance angel in degrees = 15.02\n" ] } ], "prompt_number": 106 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Ex1.9.5:Pg-1.25" ] }, { "cell_type": "code", "collapsed": false, "input": [ "# Given\n", "import math\n", "n1 = 1.4 # RI of Core..\n", "n2 = 1.35 # RI of Cladding\n", "\n", "phiC = math.asin(n2/n1) # Critical angle..\n", "print \" The Critical angel in degrees =\",round(math.degrees(phiC),2) \n", "\n", "NA = math.sqrt(n1**2-n2**2) # numerical Aperture...\n", "print \" \\n\\nThe Numerical Aperture is =\",round(NA,2) \n", "\n", "phiA = math.asin(NA) # Acceptance angle... \n", "print \" \\n\\nThe Acceptance angel in degrees =\",round(math.degrees(phiA),2) \n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ " The Critical angel in degrees = 74.64\n", " \n", "\n", "The Numerical Aperture is = 0.37\n", " \n", "\n", "The Acceptance angel in degrees = 21.77\n" ] } ], "prompt_number": 107 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Ex1.9.6:Pg-1.25" ] }, { "cell_type": "code", "collapsed": false, "input": [ "# Given\n", "import math\n", "n1 = 1.48 # RI of core..\n", "n2 = 1.46 # RI of Cladding..\n", "\n", "NA = math.sqrt(n1**2-n2**2) # Numerical Aperture..\n", "print \" The Numerical Aperture is =\",round(NA,3) \n", "\n", "theta = math.pi*NA**2 # The entrance angle theta..\n", "print \" \\n\\nThe Entrance angel in degrees =\",round(theta,3) \n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ " The Numerical Aperture is = 0.242\n", " \n", "\n", "The Entrance angel in degrees = 0.185\n" ] } ], "prompt_number": 110 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Ex1.9.7:Pg-1.26" ] }, { "cell_type": "code", "collapsed": false, "input": [ "# Given\n", "import math \n", "\n", "delt = 0.007 # relative refractive index difference \n", "n1 = 1.45 # RI of core...\n", "NA = n1* math.sqrt((2*delt)) \n", "print \" The Numerical Aperture is =\",round(NA,4) \n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ " The Numerical Aperture is = 0.1716\n" ] } ], "prompt_number": 112 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Ex1.9.8:Pg-1.26" ] }, { "cell_type": "code", "collapsed": false, "input": [ " # Given\n", "import math\n", "\n", "phiA = 8 # accepatance angle in degrees...\n", "n1 =1.52 # RI of core...\n", "\n", "NA = sin(math.radians(phiA)) # Numerical Aperture...\n", "\n", "delt = NA**2/(2*(n1**2)) # Relative RI difference...\n", "print \" The relative refractive index difference =\",round(delt,5) \n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ " The relative refractive index difference = 0.00419\n" ] } ], "prompt_number": 114 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "\n", "Ex1.9.9:Pg-1.27" ] }, { "cell_type": "code", "collapsed": false, "input": [ "# Given\n", "import math\n", "delt = 0.01 # relative RI difference..\n", "n1 = 1.48 # RI of core...\n", "\n", "NA = n1*(math.sqrt(2*delt)) # Numerical Aperture..\n", "print \" The Numerical Aperture =\",round(NA,3) \n", "\n", "theta = math.pi*NA**2 # Solid Acceptance angle...\n", "print \" \\n\\nThe Solid Acceptance angel in degrees =\",round(theta,4) \n", "\n", "n2 = (1-delt)*n1 \n", "phiC = math.asin(n2/n1) # Critical Angle...\n", "print \" \\n\\nThe Critical angel in degrees =\",round(math.degrees(phiC),2) \n", "print \" \\n\\nCritical angle wrong due to rounding off errors in trignometric functions..\\n Actual value is 90.98 in book.\" \n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ " The Numerical Aperture = 0.209\n", " \n", "\n", "The Solid Acceptance angel in degrees = 0.1376\n", " \n", "\n", "The Critical angel in degrees = 81.89\n", " \n", "\n", "Critical angle wrong due to rounding off errors in trignometric functions..\n", " Actual value is 90.98 in book.\n" ] } ], "prompt_number": 3 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Ex1.14.1:Pg-1.41" ] }, { "cell_type": "code", "collapsed": false, "input": [ "# Given\n", "import math\n", "d = 50*10**-6 # diameter of fibre...\n", "n1 = 1.48 # RI of core..\n", "n2 = 1.46 # RI of cladding..\n", "lamda = 0.82*10**-6 # wavelength of light..\n", "\n", "NA = math.sqrt(n1**2-n2**2) # Numerical Aperture..\n", "Vn= math.pi*d*NA/lamda # normalised frequency...\n", "M = Vn**2/2 # number of modes...\n", "print \" The number of modes in the fibre are =\",int(M) \n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ " The number of modes in the fibre are = 1078\n" ] } ], "prompt_number": 4 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Ex1.14.2:Pg-1.42" ] }, { "cell_type": "code", "collapsed": false, "input": [ "# Given\n", "import math\n", " \n", " \n", "V = 26.6 # Normalised frequency..\n", "lamda = 1300*10**-9 # wavelenght of operation\n", "a = 25*10**-6 # radius of fibre.\n", "NA = V*lamda/(2*math.pi*a) # Numerical Aperture..\n", "print \" The Numerical Aperture =\",round(NA,3) \n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ " The Numerical Aperture = 0.22\n" ] } ], "prompt_number": 5 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Ex1.14.3:Pg-1.43" ] }, { "cell_type": "code", "collapsed": false, "input": [ "# Given\n", "import math\n", " \n", "a = 40*10**-6 # radius of core...\n", "delt = 0.015 # relative RI difference..\n", "lamda= 0.85*10**-6 # wavelength of operation..\n", "n1=1.48 # RI of core..\n", "\n", "NA = n1*math.sqrt(2*delt) # Numerical Aperture..\n", "print \" The Numerical Aperture =\", round(NA,4) \n", "V = 2*math.pi*a*NA/lamda # normalised frequency\n", "print \" \\n\\nThe Normalised frequency =\",round(V,2) \n", "\n", "M = V**2/2 # number of modes..\n", "print \" \\n\\nThe number of modes in the fibre are =\",int(M) \n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ " The Numerical Aperture = 0.2563\n", " \n", "\n", "The Normalised frequency = 75.8\n", " \n", "\n", "The number of modes in the fibre are = 2872\n" ] } ], "prompt_number": 7 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Ex1.14.4:Pg-1.43" ] }, { "cell_type": "code", "collapsed": false, "input": [ "# Given\n", "import math\n", "\n", "NA = 0.20 # Numerical Aperture..\n", "M = 1000 # number of modes..\n", "lamda = 850*10**-9 # wavelength of operation..\n", "\n", "a = math.sqrt(M*2*lamda**2/(math.pi**2*NA**2)) # radius of core..\n", "a=a*10**6 # converting in um for displaying...\n", "print \" The radius of the core in um =\",round(a,2) \n", "a=a*10**-6 \n", "M1= ((math.pi*a*NA/(1320*10**-9))**2)/2\n", "print \" \\n\\nThe number of modes in the fibre at 1320um =\",int(M1) \n", "print \" \\n\\n***The number of modes in the fibre at 1320um is calculated wrongly in book\" \n", "M2= ((math.pi*a*NA/(1550*10**-9))**2)/2\n", "print \" \\n\\nThe number of modes in the fibre at 1550um =\",int(M2) \n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ " The radius of the core in um = 60.5\n", " \n", "\n", "The number of modes in the fibre at 1320um = 414\n", " \n", "\n", "***The number of modes in the fibre at 1320um is calculated wrongly in book\n", " \n", "\n", "The number of modes in the fibre at 1550um = 300\n" ] } ], "prompt_number": 11 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Ex1.14.5:Pg-1.44" ] }, { "cell_type": "code", "collapsed": false, "input": [ "# Given\n", "import math\n", " \n", "NA = 0.2 # Numerical Aperture..\n", "n2= 1.59 # RI of cladding..\n", "n0= 1.33 # RI of water..\n", "lamda = 1300*10**-9 # wavelength..\n", "a = 25*10**-6 # radius of core..\n", "n1 = math.sqrt(NA**2+n2**2) # RI of core..\n", "phiA= math.asin(math.sqrt(n1**2-n2**2)/n0) # Acceptance angle..\n", "print \" The Acceptance angle is =\",round(math.degrees(phiA),2) \n", "\n", "phiC= math.asin(n2/n1) # Critical angle..\n", "print \" \\n\\nThe critical angle is =\",round(math.degrees(phiC),2) \n", "V = 2*math.pi*a*NA/lamda # normalisd frequency\n", "M= V**2/2 # number of modes\n", "print \" \\n\\nThe number of modes in the fibre are =\",int(M) \n", "\n", "print \" \\n\\n***The value of the angle differ from the book because of round off errors.\" \n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ " The Acceptance angle is = 8.65\n", " \n", "\n", "The critical angle is = 82.83\n", " \n", "\n", "The number of modes in the fibre are = 292\n", " \n", "\n", "***The value of the angle differ from the book because of round off errors.\n" ] } ], "prompt_number": 13 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Ex1.14.6:Pg-1.46" ] }, { "cell_type": "code", "collapsed": false, "input": [ "# Given\n", "import math\n", "\n", "V= 26.6 # Normalised frequency..\n", "lamda= 1300*10**-9 # wavelength of operation..\n", "a= 25*10**-6 # radius of core..\n", "\n", "NA = V*lamda/(2*math.pi*a) # Numerical Aperture..\n", "print \" The Numerical Aperture =\",round(NA,2) \n", "theta = math.pi*NA**2 # solid Acceptance Angle..\n", "print \" \\n\\nThe solid acceptance angle in radians =\",round(theta,3) \n", "\n", "M= V**2/2 # number of modes..\n", "print \" \\n\\nThe number of modes in the fibre =\",round(M,2) \n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ " The Numerical Aperture = 0.22\n", " \n", "\n", "The solid acceptance angle in radians = 0.152\n", " \n", "\n", "The number of modes in the fibre = 353.78\n" ] } ], "prompt_number": 15 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Ex1.14.7:Pg-1.47" ] }, { "cell_type": "code", "collapsed": false, "input": [ "# Given\n", "import math \n", "\n", "n1= 1.49 # RI of core.\n", "n2=1.47 # RI of cladding..\n", "a= 2 # radius of core in um..\n", "NA= math.sqrt(n1**2-n2**2) # Numerical Aperture..\n", " # The maximum V number for single mode operation is 2.4...\n", "V= 2.4 # Normalised frequency..\n", "\n", "lamda = 2*math.pi*a*NA/V # Cutoff wavelength...\n", "print \" The cutoff wavelength in um =\",round(lamda,2) \n", "\n", "\n", "lamda1 = 1.310 # Givenn cutoff wavelength in um..\n", "d= V*lamda1/(math.pi*NA) # core diameter..\n", "print \" \\n\\nThe core diameter in um =\",round(d,2) \n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ " The cutoff wavelength in um = 1.27\n", " \n", "\n", "The core diameter in um = 4.11\n" ] } ], "prompt_number": 16 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Ex1.14.8:Pg-1.47" ] }, { "cell_type": "code", "collapsed": false, "input": [ " # Given\n", "import math\n", " \n", "n1= 1.48 # RI of core..\n", "a= 4.5 # core radius in um..\n", "delt= 0.0025 # Relative RI difference..\n", "V= 2.405 # For step index fibre..\n", "lamda= (2*math.pi*a*n1*math.sqrt(2*delt))/V # cutoff wavelength..\n", "print \" The cutoff wavelength in um =\",round(lamda,2) \n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ " The cutoff wavelength in um = 1.23\n" ] } ], "prompt_number": 18 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Ex1.14.9:Pg-1.48" ] }, { "cell_type": "code", "collapsed": false, "input": [ "# Given\n", "import math \n", " \n", "lamda= 0.82*10**-6 # wavelength ofoperation.\n", "a= 2.5*10**-6 # Radius of core..\n", "n1= 1.48 # RI of core..\n", "n2= 1.46 # RI of cladding\n", "NA= math.sqrt(n1**2-n2**2) # Numerical Aperture..\n", "V= 2*math.pi*a*NA/lamda # Normalisd frequency..\n", "print \" The normalised frequency =\",round(V,3) \n", "M= V**2/2 # The number of modes..\n", "print \" \\n\\nThe number of modes in the fibre are =\",round(M,2) \n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ " The normalised frequency = 4.645\n", " \n", "\n", "The number of modes in the fibre are = 10.79\n" ] } ], "prompt_number": 21 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Ex1.14.10:Pg-1.49" ] }, { "cell_type": "code", "collapsed": false, "input": [ " # Given\n", "import math\n", " \n", "delt= 0.01 # Relative RI difference..\n", "n1= 1.5 \n", "M= 1100 # Number of modes...\n", "lamda= 1.3 # wavelength of operation in um..\n", "V= math.sqrt(2*M) # Normalised frequency...\n", "d= V*lamda/(math.pi*n1*math.sqrt(2*delt)) # diameter of core..\n", "print \" The diameter of the core in um =\",round(d,2) \n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ " The diameter of the core in um = 91.5\n" ] } ], "prompt_number": 22 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Ex1.14.11:Pg-1.50" ] }, { "cell_type": "code", "collapsed": false, "input": [ "# Given\n", "import math\n", "n1= 1.5 # RI of core..\n", "n2= 1.38 # RI of cladding..\n", "a= 25*10**-6 # radius of core..\n", "lamda= 1300*10**-9 # wavelength of operation...\n", "NA= math.sqrt(n1**2-n2**2) # Numerical Aperture..\n", "print \" The Numerical Aperture of the given fibre =\",round(NA,4) \n", "V= 2*math.pi*a*NA/lamda # Normalised frequency..\n", "print \" \\n\\nThe normalised frequency =\",round(V,2) \n", "theta= math.asin(NA) # Solid acceptance anglr..\n", "print \" \\n\\nThe Solid acceptance angle in degrees =\",int(math.degrees(theta)) \n", "M= V**2/2 # Number of modes..\n", "print \" \\n\\nThe number of modes in the fibre are =\",int(M) \n", "print \" \\n\\n***Number of modes wrongly calculated in the book..\"\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ " The Numerical Aperture of the given fibre = 0.5879\n", " \n", "\n", "The normalised frequency = 71.03\n", " \n", "\n", "The Solid acceptance angle in degrees = 36\n", " \n", "\n", "The number of modes in the fibre are = 2522\n", " \n", "\n", "***Number of modes wrongly calculated in the book..\n" ] } ], "prompt_number": 25 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Ex1.14.12:Pg-1.51" ] }, { "cell_type": "code", "collapsed": false, "input": [ "# Given\n", "import math\n", "\n", "lamda= 850*10**-9 # wavelength of operation.\n", "a= 25*10**-6 # Radius of core\n", "n1= 1.48 # RI of Core...\n", "n2= 1.46 # RI of cladding..\n", "\n", "NA= math.sqrt(n1**2-n2**2) # Numerical Aperture\n", "\n", "V= 2*math.pi*a*NA/lamda # Normalised frequency..\n", "print \" The normalised frequency =\",round(V,2) \n", "\n", "lamda1= 1320*10**-9 # wavelength changed...\n", "V1= 2*math.pi*a*NA/lamda1 # Normalised frequency at new wavelength..\n", "\n", "M= V1**2/2 # Number of modes at new wavelength..\n", "print \" \\n\\nThe number of modes in the fibre at 1320um =\",int(M) \n", "lamda2= 1550*10**-9 # wavelength 2...\n", "V2= 2*math.pi*a*NA/lamda2 # New normalised frequency..\n", "M1= V2**2/2 # number of modes..\n", "print \" \\n\\nThe number of modes in the fibre at 1550um =\",int(M1 )\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ " The normalised frequency = 44.81\n", " \n", "\n", "The number of modes in the fibre at 1320um = 416\n", " \n", "\n", "The number of modes in the fibre at 1550um = 301\n" ] } ], "prompt_number": 26 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Ex1.15.1:Pg-1.56" ] }, { "cell_type": "code", "collapsed": false, "input": [ "# Given\n", "import math\n", " \n", "n1= 1.48 # RI of core..\n", "delt= 0.015 # relative RI differencr..\n", "lamda= 0.85 # wavelength of operation..\n", "V= 2.4 # for single mode of operation..\n", "\n", "a= V*lamda/(2*math.pi*n1*math.sqrt(2*delt)) # radius of core..\n", "print \" The raduis of core in um =\",round(a,2) \n", "print \" \\n\\nThe maximum possible core diameter in um =\",round(2*a,2) \n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ " The raduis of core in um = 1.27\n", " \n", "\n", "The maximum possible core diameter in um = 2.53\n" ] } ], "prompt_number": 27 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Ex1.15.2:Pg-1.56" ] }, { "cell_type": "code", "collapsed": false, "input": [ " # Given\n", "import math\n", " \n", "n1= 1.5 # RI of core..\n", "delt= 0.01 # Relative RI difference...\n", "lamda= 1.3 # Wavelength of operation...\n", "V= 2.4*math.sqrt(2) # Maximum value of V for GRIN...\n", "a= V*lamda/(2*math.pi*n1*math.sqrt(2*delt)) # radius of core..\n", "print \" The radius of core in um =\",round(a,2) \n", "print \" \\n\\nThe maximum possible core diameter in um =\",round(2*a,2) \n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ " The radius of core in um = 3.31\n", " \n", "\n", "The maximum possible core diameter in um = 6.62\n" ] } ], "prompt_number": 28 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Ex1.15.3:Pg-1.56" ] }, { "cell_type": "code", "collapsed": false, "input": [ "# Given\n", "import math\n", "\n", "n1= 1.46 # RI of core..\n", "a = 4.5 # radius of core in um..\n", "delt= 0.0025 # relative RI difference..\n", "V= 2.405 # Normalisd frequency for single mode..\n", "lamda= 2*math.pi*a*n1*math.sqrt(2*delt)/V # cutoff wavelength...\n", "print \" The cut off wavelength for the given fibre in um =\",round(lamda,3) \n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ " The cut off wavelength for the given fibre in um = 1.214\n" ] } ], "prompt_number": 31 } ], "metadata": {} } ] }