# Chapter 3: Optical Fibre¶

## Example 3.1,Page number 3-19¶

In :
import math

#Given Data:
n1=1.61                 #Core index

#Calculations:
NA=math.sqrt(n1**2-n2**2)      #Formula

print"Numerical Aperture of Fibre is = ",NA

Numerical Aperture of Fibre is =  0.435430821142


## Example 3.2,Page number 3-19¶

In :
import math

#Given Data:
n1=1.65                 #Core index

#Calculations:
NA=math.sqrt(n1**2-n2**2)      #Formula

print"Numerical Aperture of Fibre is =",NA

Numerical Aperture of Fibre is = 0.617737808459


## Example 3.3,Page number 3-19¶

In :
import math

#Given Data:
n1=1.48                 #R.I. of Core

#Calculations:
NA=math.sqrt(n1**2-n2**2)      #Formula to find NA
phi=math.asin(NA)*180/(3.1472)    #Acceptance angle

print"Numerical Aperture of Fibre is =",NA
print"Acceptance angle of Fibre is =",phi,"degrees"

Numerical Aperture of Fibre is = 0.508232230383
Acceptance angle of Fibre is = 30.491727191 degrees


## Example 3.4,Page number 3-20¶

In :
import math

#given data:
u1=3.6        #Refractive Index of the Substance at 850 nm
u2=3.4        #Refractive Index of the Substance at 1300 nm
Vv=3*10**8     #Velocity of light in free space

#Calculations:
# i)Finding wavelength at 850 nm
Vs1=Vv/u1             #Velocity of light in substance at 850 nm
print"Velocity of light in substance at 850 nm =" ,Vs1,"m/sec"

lam1=850*10**-9/u1    #Wavelength of light in substance at 850nm
print" Wavelength of light in substance at 850nm =",lam1,"m"

#ii)Finding wavelength at 1300 nm
Vs2=Vv/u2            #Velocity of light in substance at 1300 nm
print"Velocity of light in substance at 1300 nm =",Vs2," m/sec"

lam2=1300*10**-9/u2   #Wavelength of light in substance at 1300nm
print"Wavelength of light in substance at 1300nm =" ,lam2,"m "

Velocity of light in substance at 850 nm = 83333333.3333 m/sec
Wavelength of light in substance at 850nm = 2.36111111111e-07 m
Velocity of light in substance at 1300 nm = 88235294.1176  m/sec
Wavelength of light in substance at 1300nm = 3.82352941176e-07 m


## Example 3.5,Page number 3-20¶

In :
import math

#Given Data:
u1=1.5                  #R.I. of Core
delta= (u1-u2)/u1          #Fractional Refractive index

#Calculations:
NA=u1*sqrt(2*delta)             #Formula to find NA
theta0=math.asin(NA)*180/(3.1472)       #Acceptance angle
thetac=math.asin(u2/u1)*180/(3.1472)   #Critical angle

print"Numerical Aperture of Fibre is =",NA
print"Acceptance angle of Fibre is =",theta0,"degrees"
print" Critical angle of Fibre is =",thetac,"degrees"

Numerical Aperture of Fibre is = 0.387298334621
Acceptance angle of Fibre is = 22.7458994397 degrees
Critical angle of Fibre is = 75.0309676099 degrees


## Example 3.6,Page number 3-20¶

In :
import math

#Given Data:
NA=0.22              # Numerical Aperture of Fibre
delta=0.012          #Fractional index

#Calculations:
#Delta=(u1-u2)/u1
u1=NA/math.sqrt(2*delta)  #Formula
u2=u1-(u1*delta)     #Formula

print"Refractive Index of core of fibre is =",u1
print"Refractive Index of cladding of fibre is =",u2

Refractive Index of core of fibre is = 1.42009389361
Refractive Index of cladding of fibre is = 1.40305276689


## Example 3.7,Page number 3-21¶

In :
import math

#Given Data:
u1=1.466                 #R.I. of Core
V=2.4                    #Cut off parameter
lamda=0.8*10**-6            #wavelength in meter

#Calculations:
NA=math.sqrt(u1**2-u2**2)       #Formula to find Numerical Aperture
print"Numerical Aperture of Fibre is =",NA
#(printing mistake in book)printed answer is 1.13 but correct answer is 0.13
print"(printing mistake in book)"

# V = 2*3.142*a*NA / lamda
print"Core radius of Fibre is (a) =",a,"m"

#w/a= 1.1
w=1.1*a                  #Spot size
print"Spot size of Fibre is =",w,"m"

theta=2*lamda*180/3.142/(3.142*w)       #Divergence angle
print"Divergence angle of Fibre is =",theta,"degrees"

w10=lamda*10/(3.142*w)     #Spot size at 10 m
print"Spot size at 10 m of Fibre is =",w10,"m"

Numerical Aperture of Fibre is = 0.1324990566
(printing mistake in book)
Core radius of Fibre is (a) = 2.30596263706e-06 m
Spot size of Fibre is = 2.53655890076e-06 m
Divergence angle of Fibre is = 11.5009886523 degrees
Spot size at 10 m of Fibre is = 1.00378073182 m


## Example 3.8,Page number 3-21¶

In :
import math

#Given Data:
w=98                   #Spot size in meter
d=50*10**-6             #Core diameter in meter
u1=1.47                #R.I. of Core
lamda=0.85*10**-6         #Wavlength in meter
NA=math.sqrt(u1**2-u2**2)     #Formula to find NA

#Calculations:
V=2*3.142*a*NA/lamda        #cut off parameter
N=(V**2)/2               #Number of modes

print"Cut off parameter of Fibre is =",V
print"Number of modes of Fibre is =",N

Cut off parameter of Fibre is = 44.6646240577
Number of modes of Fibre is = 997.464321107


## Example 3.9,Page number 3-21¶

In :
import math

#Given Data:
u1=1.47                 #R.I. of Core
lamda=1.3*10**-6           #wavelength in meter

#Calculations:
NA=math.sqrt(u1**2-u2**2)      #Formula to find Numerical Aperture

#The condition for single mode is V<2.405
#2*3.142*a*NA/lamda < 2.405

print"Maximum radius of Fibre is =",a,"meter"

Maximum radius of Fibre is = 2.90662126448e-06 meter


## Example 3.10,Page number 3-22¶

In :
import math

#Given Data:
u1=1.465                  #R.I. of Core
lamda=1.25*10**-6            #operating wavelength

#Calculations:
delta=(u1-u2)/u1            #Fractional Refractive index
print"Fractional Refractive index of Fibre is =",delta

#For single mode propagation codition is
# a/lamda < 1.4/(3.142*sqrt(u1(u1-u2)))

u=u1-(math.sqrt(2*delta)/(2*3.142*(a/lamda)))     #effective refractive index
print"Effective Refractive index for lowest mode propagation is =",u

Fractional Refractive index of Fibre is = 0.00341296928328
Effective Refractive index for lowest mode propagation is = 1.46247461864


## Example 3.11,Page number 3-22¶

In :
import math

#Given Data:
u1=1.54                  #R.I. of Core
lamda=1.3*10**-6            #wavelength in meter

#Calculations:
NA=math.sqrt(u1**2-u2**2)      #Formula to find Numerical Aperture

V=2*3.142*a*NA/lamda         #cut off parameter
print"Cut off parameter of Fibre is =",V

N=(V**2)/2                #Number of modes
print" Number of modes of Fibre is =",N

Cut off parameter of Fibre is = 42.1404937865
Number of modes of Fibre is = 887.910608284


## Example 3.11.1,Page number 3-25¶

In :
import math

#Given Data:
u1=1.52                  #R.I. of Core
lamda=1.3*10**-6            #wavelength in meter
d=29*10**-6               #core diameter in meter
a=d/2

#Calculations:
NA=math.sqrt(u1**2-u2**2)     #Formula to find Numerical Aperture
V=2*3.142*a*NA/lamda        #Normalised frequency
Nm=(V**2)/2               #Number of modes

print"Normalised frequency of Fibre is (V)=",V
print"The Maximum Number of modes the Fibre will support is (Nm) =",Nm

Normalised frequency of Fibre is (V)= 4.05242861605
The Maximum Number of modes the Fibre will support is (Nm) = 8.2110888441


## Example 3.12,Page number 3-22¶

In :
import math

#Given Data:
u1=1.5                  #R.I. of Core
d=10*10**-6              #diameter of core
lamda=1.3*10**-6           #wavelength
V=2.405                 #cut off parameter for single mode

#Calculations:

#We know, V=2*3.142*a*NA/lamda
NA=V*lamda/(2*3.142*a)       #Numerical Aperture

theta=math.asin(NA)*180/3.142   #Acceptance angle
print"Acceptance angle of Fibre is =",theta,"Degrees"

#Also, NA=u1*sqrt(2*delta)
delta=(NA/u1)**2/2           #Fractional index
print"Maximum Fractional Refractive index of Fibre is =",delta

#delta=(u1-u2)/u1
print"Refractive index of cladding of Fibre is =",u2

Acceptance angle of Fibre is = 5.71002346964 Degrees
Maximum Fractional Refractive index of Fibre is = 0.00220035113094
Refractive index of cladding of Fibre is = 1.4966994733


## Example 3.13,Page number 3-23¶

In :
import math

#Given Data:
n1=1.5               #R.I. of core
delta=0.0005         #Fractional index difference

#Calculations:
#(a):
#Delta=(u1-u2)/u1
print"(a)Refractive Index of cladding of fibre is =",n2

#(b):
phi=math.asin(n2/n1)*180/3.142        #Critical internal reflection angle
print"(b)Critical internal reflection angle of Fibre is =",phi,"degrees"

#(c):
theta0=math.asin(math.sqrt(n1**2-n2**2))*180/3.142    #External critical Acceptance angle
print"(c)External critical Acceptance angle of Fibre is =",theta0,"degrees"

#(d):
NA=n1*math.sqrt(2*delta)          #Formula to find Numerical Aperture
print"(d)Numerical Aperture of Fibre is =",NA

(a)Refractive Index of cladding of fibre is = 1.49925
(b)Critical internal reflection angle of Fibre is = 88.1766396681 degrees
(c)External critical Acceptance angle of Fibre is = 2.71810509125 degrees
(d)Numerical Aperture of Fibre is = 0.0474341649025


## Example 3.14,Page number 3-24¶

In :
import math

#Given Data:
NA1=0.20              #Numerical Aperture of Fibre

#Calculations:
#NA=sqrt(n1**2-n2**2)
#In air, n0=1
n1=math.sqrt(NA1**2+n2**2)     #R.I.of core

#Now, in water
n0=1.33
NA2=math.sqrt(n1**2-n2**2)/n0      #Numerical Aperture in water
theta0=math.asin(NA2)*180/3.142    #Acceptance angle of fibre in water
print"Acceptance angle of Fibre in water is =",theta0,"degrees"

Acceptance angle of Fibre in water is = 8.6475921767 degrees


## Example 3.15,Page number 3-24¶

In :
import math

#Given Data:
n1=1.45             #R.I.of core

#Calculations:
NA=math.sqrt(n1**2-n2**2)   #Numerical Aperture
print"Numerical Aperture of Fibre is =",NA

theta0=math.asin(NA)*180/3.142    #Acceptance angle of fibre
print"Acceptance angle of Fibre is =",theta0,"degrees"

Numerical Aperture of Fibre is = 0.377491721764
Acceptance angle of Fibre is = 22.1755250876 degrees


## Example 3.16,Page number 3-24¶

In :
import math

#Given Data:
NA=0.16             #Numerical Aperture of Fibre
n1=1.45             #R.I. of core
d=90*10**-6          #Core diameter

#Calculations:
#NA=sqrt(n1**2-n2**2)
print"(a)Refractive Index of cladding of fibre is =",n2

theta0=math.asin(NA)*180/3.142    #Acceptance angle of fibre
print"(b)Acceptance angle of Fibre is =",theta0,"degrees"

(a)Refractive Index of cladding of fibre is = 1.44114537782
(b)Acceptance angle of Fibre is = 9.20570258795 degrees


## Example 3.17,Page number 3-25¶

In :
import math

#Given Data:
n1=1.48               #R.I. of core
delta=0.055           #Realtive R.I.
lamda=1*10**-6        #Wavelength of light

#Calculations:
#Delta=(u1-u2)/u1
NA=n1*math.sqrt(2*delta)  #Formula to find Numerical Aperture
print"Numerical Aperture of Fibre is =",NA

theta0=math.asin(NA)*180/3.142     #Acceptance angle of fibre
print"Acceptance angle of Fibre is =",theta0,"degrees"

V=2*3.142*a*NA/lamda            #V number
N=(V**2)/2                   #Number of guided modes

#In book,instead of NA , value of delta is taken into calculation.
#Thus there is calculation mistake in values of V and N.

print"V number of Fibre is =",V
print"Number of guided mode of Fibre is =",N
print"(Calculation mistake in book)"

Numerical Aperture of Fibre is = 0.490860468973
Acceptance angle of Fibre is = 29.3933421943 degrees
V number of Fibre is = 154.228359351
Number of guided mode of Fibre is = 11893.1934141
(Calculation mistake in book)

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