{
"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": []
}
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}