# Ch-2 Optical Fibers & Fiber Cables¶

## Example 2.1 Page no 31¶

In :
from __future__ import division
from math import sqrt, pi
# Calculation of core diameter
# Given data
n1=1.5#               # Refractive index of core
n2=1.48#              # Refractive index of cladding
N=1000#               # No of modes
lamda=1.3#           # Light wavelength
V=sqrt(2*N)#          # Mode parameter

#core diameter
d=(lamda*V)/(2*pi*sqrt(n1**2-n2**2))#

print"Mode parameter = %0.2f  "%(V)#
print"Core diameter = %0.0f micrometer "%(d)#

# Answer is wrong in the book.

Mode parameter = 44.72
Core diameter = 38 micrometer


## Example 2.2 Page no 35¶

In :
from math import asin, sin, degrees
# Calculation of (a) critical angle (b) acceptance angle and oblique angle (c) numerical aperature
# (d) percentage of light collected by the fiber and (e) diameter of fiber

# Given data
n1=1.5#                           # Refractive index of core
n2=1.45#                          # Refractive index of cladding
V=2.405#                          # Mode parameter
lamda=1.55                       # Wavelength of fiber

# (a) Critical angle of the
theatha=degrees(asin(n2/n1))#

# (b) Oblique angle
oa=90-theatha#
#Acceptance angle
t1=n1*sin(oa*pi/180)#
th1=degrees(asin(t1))#

# (c) Numerical aperature
NA=sqrt(n1**2-n2**2)#

# (d) Percentage of light collected in fiber
P=(NA)**2*100#

#(e) Diameter of fiber
d=V*lamda/pi*(1/sqrt(n1**2-n2**2))#

print"Critical angle = %0.0f degrees "%(theatha)#
print"Oblique angle = %0.0f degrees "%(oa)#
print"Acceptance angle = %0.0f degrees "%(th1)#
print"Numerical aperature = %0.4f  "%(NA)#
print"Percentage of light collected in fiber = %0.1f  "%(P)#
print"Diameter of fiber = %0.2f micrometer "%(d)#

Critical angle = 75 degrees
Oblique angle = 15 degrees
Acceptance angle = 23 degrees
Numerical aperature = 0.3841
Percentage of light collected in fiber = 14.7
Diameter of fiber = 3.09 micrometer


## Example 2.3 Page no 38¶

In :
#  Calculation of (a) critical angle (b) numerical aperature and (c) acceptance angle

# Given data
n1=1.5#                           # Refractive index of core
n2=1.47#                          # Refractive index of cladding)

# (a) Critical angle
theatha=degrees(asin(n2/n1))#
# (b) Numerical aperature
NA=sqrt(n1**2-n2**2)#
# (c) Acceptance angle
theatha1=degrees(asin(NA))#

print"Critical angle = %0.1f degrees "%(theatha)#
print"Numerical aperature = %0.2f  "%(NA)#
print"Acceptance angle = %0.1f degrees "%(theatha1)#

Critical angle = 78.5 degrees
Numerical aperature = 0.30
Acceptance angle = 17.4 degrees


## Example 2.4 Page no 46¶

In :
# Calculation of output power

# Given data
Pi=1#                                       # Input power
A=0.5#                                     # Atteuation

# Output Power
Po=Pi*10**((-A*L)/10)#

print"Output Power = %0.3f mW "%(Po)#

Output Power = 0.178 mW


## Example 2.5 Page no 47¶

In :
from math import log10
# Calculation of maximum transmission distance

# Given data
Pi=1*10**-3#                   # Input power
A=0.5#                        # Atteuation
Po=50*10**-6#                  # Output Power

# Maximum transmission distance
L=(10/A)*log10(Pi/Po)#

print"Maximum transmission distance = %0.0f km "%(L)#

Maximum transmission distance = 26 km


## Example 2.6 Page no 48¶

In :
# Calculation of output power

# Given data
Pin=1*10**-3#                  # Input power
AL1=10#                       # Attenuation 1
AL2=20#                       # Attenuation 2
#Output power 1 and 2
Po1=Pin/10#
Po2=Pin/20#
Po1=Po1*10**3#
Po2=Po2*10**6#

print"Output power = %0.1f mW"%(Po1)#
print"Output power = %0.0f uW"%(Po2)#

Output power = 0.1 mW
Output power = 50 uW


## Example 2.7 Page no 50¶

In :
from math import exp

# Calculation of attenuation and Rayleigh scattering coefficient for fiber

# Given data
n=1.46#                                      # Refractive index
p=0.286#                                     # Average photoelastic coefficient
B=7.25*10**-11#                               # Isothermal compressibility
k=1.38*10**-23#                               # Boltzmann's constant
T=1350#                                      # Fictive temperature
l1=1*10**-6#                                  # Wavelength 1
l2=1.3*10**-6#                                # Wavelength 2
L=10**3#                                      # Length

# Rayleigh scattering coefficient for length 1
y1=8*(pi)**3*(n)**8*(p)**2*B*k*T/(3*(l1)**4)#
# Rayleigh scattering coefficient for length 2
y2=8*(pi)**3*(n)**8*(p)**2*B*k*T/(3*(l2)**4)#
y1=y1#
y2=y2#
#Attenuation 1
T1=exp(-(y1*L))#
#Attenuation 2
T2=exp(-(y2*L))#

print"First Rayleigh scattering coefficient = %0.6f m**-1 "%(y1)#
print"Second Rayleigh scattering coefficient = %0.6f m**-1 "%(y2)#

print"Attenuation (@ Length 1) = %0.2f dB/km"%(T1)#
print"Attenuation (@ Length 2) = %0.2f dB/km"%(T2)#

First Rayleigh scattering coefficient = 0.000189 m**-1
Second Rayleigh scattering coefficient = 0.000066 m**-1
Attenuation (@ Length 1) = 0.83 dB/km
Attenuation (@ Length 2) = 0.94 dB/km


## Example 2.8 Page no 52¶

In :
# Calculation of threshold power of stimulated Brillouin scattering and Raman Scattering

# Given data
A=0.5#                       # Attenuation
d=5#                         # Core diameter
lamda=1.3#                  # Operating wavelength
v=0.7#                       # Bandwith of laser diode

# Threshold power of stimulated Brillouin scattering
Pb=4.4*10**-3*d**2*lamda**2*A*v#
Pb=Pb*10**3#

#Threshold power stimulated Raman Scattering
Pr=5.9*10**-2*d**2*lamda*A#

print"Threshold power of stimulated Brillouin scattering = %0.2f mW"%(Pb)#
print"Threshold power stimulated Raman Scattering = %0.2f W"%(Pr)#

Threshold power of stimulated Brillouin scattering = 65.06 mW
Threshold power stimulated Raman Scattering = 0.96 W


## Example 2.1 Page no. 479¶

In :
# Compuatation of mode parameter

#Given data
n1=1.503# # refractive index of core
n2=1.50# # refractive index of cladding
lamda=1*10**-6# # light wavelength

# Mode parameter computation
V=(2*pi*a*sqrt(n1**2-n2**2))/(lamda)#

#Displaying the result in command window
print"Mode parameter is = %0.3f "%(V)#

# The answer vary due to round off error

Mode parameter is = 2.385


## Example 2.2 Page no. 479¶

In :
# Calculation of numerical aperature

#Given data
v=2.111#                  # Mode parameter
a=4.01*10**-6#             # Core radius in m
lamda=1.3*10**-6#         # Wavelength of laser light m

#Numerical aperture computation
NA=(v*lamda)/(2*pi*a)#

#Displaying the result in command window
print"Numerical aperature = %0.2f"%(NA)#

Numerical aperature = 0.11


## Example 2.3 Page no 480¶

In :
from math import log
# Calculation of potential difference

# Given data
na=10**24#                     # Accepter impurity level
nd=10**22#                     # Donor impurity level
ni=2.4*10**19#                 # Intrinsic electron
T=290#                        # Room temperature
e=1.602*10**-19#               # Electric charge
K=1.38*10**-23#                # Boltzmann constant

#Potential difference
V=(K*T)/e*(log(na*nd/(ni)**2))#

print"Potential difference = %0.2f V "%(V)#
# The potential difference varies with the variation of Na, Nd and ni

Potential difference = 0.42 V


## Example 2.4 Page no 480¶

In :
# Calculation of (a) Numerical aperature and (b) critical angle

# Given data
n1=1.5#                   # Refractive index of core
n2=1.47#                  # Refractive index of cladding

# (a) Numerical aperature
NA= sqrt(n1**2-n2**2)#

# (b) Critical angle
theatha=degrees(asin(n2/n1))#

print"Numerical aperature = %0.2f "%(NA)#
print"Critical angle = %0.1f degrees "%(theatha)#

Numerical aperature = 0.30
Critical angle = 78.5 degrees


## Example 2.5 Page no 480¶

In :
import math
# Computation of (a) normalized frequency and (b) no. of guided modes

#Given data
lamda=0.85*10**-6#      # wavelength of fiber
a=40*10**-6#             # core diameter of fiber
delta=0.015#            # relative refractive index
n1=1.48#                # refractive index of core

# (a) Normalized frequency
v=(2*pi*a*n1*(2*delta)**(1/2))/lamda#
#(b) Number of guided modes
m=v**2/2#
m=math.ceil(m)#
#Displaying results in the command window
print"Normalized frequency is = %0.1f  "%(v)#
print"Number of guided modes = %0.0f  "%(m)#

Normalized frequency is = 75.8
Number of guided modes = 2873


## Example 2.6 Page no 480¶

In :
# Computation of normalized frequency and no of guided modes

#Given data
lamda=1.30*10**-6#        # Wavelength of fiber
a=25*10**-6#               # Core diameter of fiber
delta=0.01#               # Relative refractive index
n1=1.50#                  # Refractive index of core

# (a) Normalized frequency
v=((2*pi*a*n1)/(lamda))*((2*delta)**(1/2))#
#(b) Number of guided modes
m=v**2/2#
#m=ceil(m)#

#Displaying results in the command window
print"Normalized frequency = %0.2f  "%(v)#
print"Number of guided modes = %0.0f  "%(m)#

#Answer varies due to round off error

Normalized frequency = 25.63
Number of guided modes = 329


## Example 2.7 Page no 481¶

In :
# Calculation of normalized frequency and no of guided modes

#Given data
lamda=1.55*10**-6#      # Wavelength of fiber
a=30*10**-6#             # Core diameter of fiber
delta=0.015#            # Relative refractive index
n1=1.48#                # Refractive index of core

# (a) Normalized frequency
v=(2*pi*a*n1*(2*delta)**(1/2))/lamda#
#(b) Number of guided modes
m=v**2/2#

#Displaying results in the command window
print"Normalized frequency = %0.2f  "%(v)#
print"Number of guided modes = %0.0f  "%(m)#
# The answers vary due to round off error

Normalized frequency = 31.17
Number of guided modes = 486


## Example 2.8 Page no 481¶

In :
# Calculation of normalized frequency and no of guided modes

#Given data
lamda=1.55*10**-6#         # Wavelength of fiber
a=4*10**-6#                 # Core diameter of fiber
delta=0.01#                # Relative refractive index
n1=1.48#                   # Refractive index of core
# (a) Normalized frequency
v=(2*pi*a*n1*(2*delta)**(1/2))/lamda#
#(b) Number of guided modes
m=v**2/2#

#Displaying results in the command window
print"Normalized frequency = %0.3f  "%(v)#
print"Number of guided modes = %0.0f  "%(m)#
# The answers vary due to round off error

Normalized frequency = 3.394
Number of guided modes = 6


## Example 2.9 Page no 481¶

In :
# Calculation of Core radius

#Given data
lamda=0.85*10**-6#           # Wavelength of fiber
delta=0.015#                 # Relative refractive index
n1=1.48#                     # Refractive index of core
v=2.403#                     # Normalized frequency for single mode fiber
a=v*lamda/(2*pi*n1*sqrt(2*delta))#
a=a*10**6#

print"Radius of core = %0.1f micrometer "%(a)#

Radius of core = 1.3 micrometer


## Example 2.10 Page no 482¶

In :
# Calculation of Cut off wavelength

# Given data
V=2.403#              # Normalized frequency
delta=0.25#           # Refractive index of core
n1=1.46#              # Relative refractive index

# Cut off wavelenth
lamda=(2*pi*a*n1*(sqrt(2*delta)))/V#

print"Cut off wavelenth = %0.0f nm "%(lamda*10**8)#

# The answers vary due to round off error

Cut off wavelenth = 1215 nm


## Example 2.11 Page no 482¶

In :
#Calculation of (a) reflection and (b) loss of light signal at joint areas.

# Given data
n1=1.5#                  # Refractive index of core
n=1#                     # Refractive index of air

# (a) Reflection at the fiber air interface
R=((n1-n)/(n1+n))**2#

# (b) Light loss due to fiber air interface
l= -10*log10(1-R)#

print"Reflection at the fiber air interface = %0.2f  "%(R)#
print"Light loss due to fiber air interface = %0.2f dB "%(l)#

Reflection at the fiber air interface = 0.04
Light loss due to fiber air interface = 0.18 dB


## Example 2.12 Page no 482¶

In :
# Computation of (a) numerical aperature and (b) maximum angle of entrance

#Given data
n1=1.48#             # Refractive index of core
n2=1.46#             # Refractive index of cladding

# (a) Numerical Aperture
NA=sqrt(n1**2-n2**2)#

#(b) Maximum angle of entrance
theata=degrees(asin(NA))

#Displaying result in the command window
print"Numerical Aperture  = %0.3f  "%(NA)#
print"Maximum angle of entrance = %0.0f degress "%(theata)#

# Final answer in the book is wrong. Please refer example 2.11 of
# Fiber Optic Communication by Gerd Keiser book.


Numerical Aperture  = 0.242
Maximum angle of entrance = 14 degress


## Example 2.13 Page no 483¶

In :
# Calculation of (a) core radius and (b) maximum value of angle of acceptance of the fiber

#Given data
lamda=1320*10**-9#          # Wavelength of fiber
delta=0.077#                # Relative refractive index
n1=1.48#                    # Refractive index of core
n2=1.478#                   # Refractive index of cladding
v=2.403#                    # Normalized frequency

a=v*lamda/(2*pi*delta)#
a=a*10**6#

#Numerical Aperture
NA=sqrt(n1**2-n2**2)#

# (b) Angle of acceptance
theata = degrees(asin(NA))

print"Radius of core = %0.1f micrometer "%(a)#
print"Numerical aperture = %0.5f  "%(NA)#
print"Angle of acceptance = %0.0f degrees "%(theata)#

# The answers vary due to round off error

Radius of core = 6.6 micrometer
Numerical aperture = 0.07692
Angle of acceptance = 4 degrees


## Example 2.14 Page no 483¶

In :
# Calculation of critical wavelength

#Given data
a=3*10**-6#            # Core diameter of fiber
delta=0.15#           # Relative refractive index
v=2.405#              # Normalized frequency

# Critical wavelength
lamda=(2*pi*a*delta)/v#
lamda*=10**9#

print"Critical wavelength = %0.0f nm "%(lamda)#

# The answers vary due to round off error

Critical wavelength = 1176 nm