# Chapter 9: Fibre Optics¶

## Example 9.1, Page 463¶

In :
from math import *

#Variable declaration
mu_1 = 1.55;  # Refractive index of the core
mu_2 = 1.50;  # Refractive indices of cladding

#Calculations
NA = mu_1*sqrt(2*(mu_1-mu_2)/mu_1);
print "The NA of the optical fibre = %5.3f"%NA
theta_a = degrees(asin(NA));    # The acceptance angle of optical fibre, degrees

#Result
print "The acceptance angle of the optical fibre is = %.1f degrees"%theta_a

The NA of the optical fibre = 0.394
The acceptance angle of the optical fibre is = 23.2 degrees


## Example 9.2, Page 463¶

In :
from math import *

#Variable declaration
mu_1 = 1.50;  # Refractive index of the core
mu_2 = 1.45;  # Refractive index cladding

#Calculations&Results
NA = mu_1*sqrt(2*(mu_1-mu_2)/mu_1);     # Numerical aperture of optical fibre
print "The NA of the optical fibre = %5.3f"%NA
theta_a = degrees(asin(NA));        # The acceptance angle of optical fibre, degrees
print "The acceptance angle of the optical fibre = %5.2f degrees"%theta_a
theta_c = degrees(asin(mu_2/mu_1)); # The critical angle of the optical fibre, degrees
print "The acceptance angle of the optical fibre = %4.1f degrees"%theta_c

The NA of the optical fibre = 0.387
The acceptance angle of the optical fibre = 22.79 degrees
The acceptance angle of the optical fibre = 75.2 degrees


## Example 9.3, Page 464¶

In :
from math import *

#Variable declaration
mu0 = 1;  # Refactive index of fibre in air
mu2 = 1.59;  # Refactive index of the cladding
NA = 0.2;  # Numerial aperture of optical fibre

#Calculations
mu1 = sqrt(NA**2+mu2**2);  # Refractive index of core
mu0 = 1.33;     # Refactive index of fibre in water
NA = sqrt(mu1**2-mu2**2)/mu0;    # Numerial aperture of optical fibre in water
theta_a = degrees(asin(NA));    # Acceptance angle for the fibre in water

#Result
print "The acceptance angle for the optical fibre in water = %3.1f degrees"%theta_a


The acceptance angle for the optical fibre in water = 8.6 degrees


## Example 9.4, Page 464¶

In :
from math import *

#Variable declaration
mu0 = 1;        # Refractive index of air
mu1 = 1.50;  # Refractive index of glass core
delta = 0.005;  # Fractional change in refractive index

#Calculations&Results
mu2 = mu1*(1-delta);      # Refractive index of cladding
print "The refractive index of cladding =%6.4f"%mu2
theta_c = degrees(asin(mu2/mu1));   # Critical angle, degrees
print "The critical angle = %5.2f degrees"%theta_c
theta_a = degrees(asin(sqrt(mu1**2-mu2**2)/mu0));     # Acceptance angle, degrees
print "The value of acceptance angle is = %4.2f degrees"%theta_a
NA = mu1*sqrt(2*delta);      # Numerical aperture of optical fibre
print "The NA of the optical fibre = %4.2f"%NA   #incorrect answer in the textbook

The refractive index of cladding =1.4925
The critical angle = 84.27 degrees
The value of acceptance angle is = 8.62 degrees
The NA of the optical fibre = 0.15


## Example 9.5, Page 465¶

In :
from math import *

#Variable declaration
NA = 0.22;      # Numerical aperture of the optical fibre
delta = 0.012;  # Fractional difference between the refractive index of core and cladding

#Calculations
mu1 = NA/sqrt(2*delta);   # The refractive index of core of optical fibre
print "The refractive index of core = %4.2f"%mu1
mu2 = mu1*(1-delta);  # The refractive index of cladding of optical fibre

#Result
print "The refractive index of cladding = %4.2f"%mu2

The refractive index of core = 1.42
The refractive index of cladding = 1.40


## Example 9.6, Page 466¶

In :
from math import *

#Variable declaration
mu1 = 1.466;   # Refractive index of core
mu2 = 1.460;   # Refractive index of cladding
v = 2.4;  # Cut-off parameter of the optical fibre
lamda = 0.8e-006;  # Operating wavelength, m

#Calculations&Results
NA = sqrt(mu1**2-mu2**2);
print "The NA of optical fibre = %4.2f"%NA
# Asthe cut-off parameter v of the optical fibre, v = 2*%pi*a*sqrt(mu1^2-mu2^2)/lambda, solving for a
a = lamda*v/(2*pi*sqrt(mu1**2-mu2**2));
print "The core radius of the optical fibre = %4.2f micrometer"%(a/1e-006)

The NA of optical fibre = 0.13
The core radius of the optical fibre = 2.31 micrometer


## Example 9.7, Page 466¶

In :
from math import *

#Variable declaration
mu1 = 1.54;    # The refractive index of core
mu2 = 1.50;    # The refractive index of cladding
lamda = 1.3e-006;  # Operating wavelength of optical fibre, m
a = 25e-006;  # Radius of fibre core, m

#Calculations
v = 2*pi*a*sqrt(mu1**2-mu2**2)/lamda;   # V-number of optical fibre
print "The cut-off parameter of the optical fibre = %5.2f"%v
n = v**2/2;      # The number of modes supported by the fibre

#Result
print "The number of modes supported by the fibre = %3d"%(ceil(n))

The cut-off parameter of the optical fibre = 42.14
The number of modes supported by the fibre = 888


## Example 9.8, Page 466¶

In :
from math import *

#Variable declaration
mu1 = 1.54;  # Refractive index of core
v = 2.405;  # Cut-off parameter of optical fibre
lamda = 1.3e-006;  # Operating wavelength of optical fibre, m
a = 1e-006;  # Radius of the core,

#Calculations
NA = v*lamda/(2*pi*a);    # Numerical aperture of optical fibre
delta = 1./2*(NA/mu1)**2;   #  Fractional change in refractive index of core and cladding
print "The fractional difference of refractive indices of core and cladding = %7.5e"%delta
mu2 = mu1*(1-delta);     # Maximum value of refractive index of cladding
print "The maximum refractive index of cladding = %5.3f"%mu2

The fractional difference of refractive indices of core and cladding = 5.22018e-02
The maximum refractive index of cladding = 1.460


## Example 9.9, Page 467¶

In :
from math import *

#Variable declaration
mu1 = 1.45;  # Index of refraction of core
NA = 0.16;  # Numerical aperture of step index fibre
a = 3e-006;  # Radius of the core, m
lamda = 0.9e-006;  # Operating wavelength of optical fibre, m

#Calculations
v = 2*pi*a*NA/lamda;      # The normalized frequency or v-number of optical fibre

#Result
print "The normalized frequency of the optical fibre = %5.2f"%v

The normalized frequency of the optical fibre =  3.35


## Example 9.10, Page 467¶

In :
from math import *

#Variable declaration
mu1 = 1.52;  # Refractive index of core
a = 14.5e-006;  # Radius of the fibre core, m
delta = 0.0007;  # Fractional index difference
lamda = 1.3e-006;  # Operating wavelength of optical fibre, m

#Calculations&Results
mu2 = mu1*(1-delta);      # Refractive index of cladding
v = 2*pi*a*sqrt(mu1**2-mu2**2)/lamda;     # Cut-off parameter v of the optical fibre
print "The cut-off parameter of the optical fibre = %5.3f"%v
#The is number of modes supported by the fibre given by,
n = v**2/2;
print "The number of modes supported by the fibre = %d"%(ceil(n))
#Incorrect answer in the textbook for mu2. Hence the difference in answers

The cut-off parameter of the optical fibre = 3.985
The number of modes supported by the fibre = 8


## Example 9.11, Page 468¶

In :
from math import *

#Variable declaration
alpha = 3.5;  # Attenuation of the optical fibre, dB/km
Pi = 0.5;  # Input power of optical fibre, mW
L = 4;      # Distance through the optical wave transmits through the fibre, km

#Calculations
# As alpha = 10/L*log10(Pi/Po), solving for Po
Po = Pi/exp(alpha*L*2.3026/10); # Output power of optical fibre, mW

#Result
print "The output power of optical fibre = %4.1f micro-watt"%(Po/1e-003)

The output power of optical fibre = 19.9 micro-watt


## Example 9.12, Page 468¶

In :
from math import *

#Variable declaration
Pi =1;  # Input power of optical fibre, mW
Po = 0.85;  # Outptu power of optical fibre, mW
L = 0.5;  #The distance through the optical wave transmits through the fibre, km

#Calculations
alpha = (10/L)*log10(Pi/Po);      # The attenuation of power through the optical fibre

#Result
print "The attenuation of power through the optical fibre = %5.3f dB/km"%alpha

The attenuation of power through the optical fibre = 1.412 dB/km


## Example 9.13, Page 469¶

In :
from math import *

#Variable declaration
C = 0.8;    # Connector loss per km, dB
F = 1.5;    # Fibre loss per km, dB
alpha = C + F;  # Attenuation of power the optical fibre, dB/km
Po = 0.3e-006;  # Output power of optical fibre, W
L = 15;  # The distance through the optical wave transmits through the fibre, km

#Calculations
#As the attenuation, alpha = 10/L*log(Pi/Po), solving for Pi
Pi = Po*exp(2.3026*alpha*L/10);     # Input power of optical fibre, mW

#Result
print "The minimum input power to optical fibre = %5.3f mW"%(Pi/1e-003)

The minimum input power to optical fibre = 0.846 mW
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