# Chapter 5:Fiber Optics¶

## Example 5.1, Page number 5.28¶

In [125]:
import math
from __future__ import division

#variable declaration
n1=1.50          #Core refractive index

#Calculations
C_a=math.asin(n2/n1)        #Critical angle
N_a=(n1**2-n2**2)**(1/2)
A_a=math.asin(N_a)

#Results
print "The Critical angle =",round(C_a*180/math.pi,1),"degrees"
print "The numerical aperture =",round(N_a,2)
print "The acceptance angle =",round(A_a*180/math.pi,1),"degrees"

The Critical angle = 78.5 degrees
The numerical aperture = 0.3
The acceptance angle = 17.4 degrees


## Example 5.2, Page number 5.28¶

In [126]:
import math
from __future__ import division

#variable declaration
d=50                #diameter
N_a=0.2             #Numerical aperture
lamda=1             #wavelength

#Calculations
N=4.9*(((d*10**-6*N_a)/(lamda*10**-6))**2)

#Result
print "N =",N
print "Fiber can support",N,"guided modes"
print "In graded index fiber, No.of modes propogated inside the fiber =",N/2,"only"

N = 490.0
Fiber can support 490.0 guided modes
In graded index fiber, No.of modes propogated inside the fiber = 245.0 only


## Example 5.3, Page number 5.29¶

In [1]:
import math
from __future__ import division

#variable declaration
d=50                #diameter
n1=1.450
n2=1.447
lamda=1             #wavelength

#Calculations
N_a=(n1**2-n2**2)   #Numerical aperture
N=4.9*(((d*10**-6*N_a)/(lamda*10**-6))**2)

#Results
print "Numerical aperture =",N_a
print "No. of modes that can be propogated =",round(N)

Numerical aperture = 0.008691
No. of modes that can be propogated = 1.0


## Example 5.4, Page number 5.29¶

In [34]:
import math
from __future__ import division

#variable declaration
delta=0.05
n1=1.46

#Calculation
N_a=n1*(2*delta)**(1/2)     #Numerical aperture

#Result
print "Numerical aperture =",round(N_a,2)

Numerical aperture = 0.46


## Example 5.5, Page number 5.29¶

In [40]:
import math
from __future__ import division

#variable declaration
a=50
n1=1.53
n2=1.50
lamda=1             #wavelength

#Calculations
N_a=(n1**2-n2**2)   #Numerical aperture
V=((2*math.pi*a)/lamda)*N_a**(1/2)

#Result
print "V number =",round(V,2)
print "maximum no.of modes propogating through fiber =",round(N)

V number = 94.72
maximum no.of modes propogating through fiber = 4486.0


## Example 5.6, Page number 5.29¶

In [64]:
import math
from __future__ import division

#variable declaration
a=100
N_a=0.3               #Numerical aperture
lamda=850             #wavelength

#Calculations
V_n=(2*(math.pi)**2*a**2*10**-12*N_a**2)/lamda**2*10**-18
#Result
print "Number of modes =",round(V_n/10**-36),"modes"
print "No.of modes is doubled to account for the two possible polarisations"
print "Total No.of modes =",round(V_n/10**-36)*2

Number of modes = 24589.0 modes
No.of modes is doubled to account for the two possible polarisations
Total No.of modes = 49178.0


## Example 5.7, Page number 5.29¶

In [88]:
import math

#variable declaration
a=5;
n1=1.48;
delta=0.01;
V=25;

#Calculation
lamda=(math.pi*(a*10**-6)*n1*math.sqrt(2*delta))/V   # Cutoff Wavelength

#Result
print "Cutoff Wavellength =",round(lamda*10**7,3),"micro m."

Cutoff Wavellength = 1.315 micro m.


## Example 5.8, Page number 5.30¶

In [87]:
import math

#variable declaration
V=2.405
lamda=1.3
N_a=0.05

#Calculations
a_max=(V*lamda)/(2*math.pi*N_a)

#Result

Maximum core radius= 9.95 micro m


## Example 5.9, Page number 5.30¶

In [2]:
import math
from __future__ import division

#variable declaration
N_a=0.3
gamma=45

#Calculations
theta_a=math.asin(N_a)
theta_as=math.asin((N_a)/math.cos(gamma))

#Results
print "Acceptance angle, theta_a =",round(theta_a*180/math.pi,2),"degrees"
print "For skew rays,theta_as ",round(theta_as*180/math.pi,2),"degrees"
print"#Answer given in the textbook is wrong"

Acceptance angle, theta_a = 17.46 degrees
For skew rays,theta_as  34.83 degrees
#Answer given in the textbook is wrong


## Example 5.10, Page number 5.30¶

In [115]:
import math
from __future__ import division

#variable declaration
n1=1.53
delta=0.0196

#Calculations
N_a=n1*(2*delta)**(1/2)
A_a=math.asin(N_a)
#Result
print "Numerical aperture =",round(N_a,3)
print "Acceptance angle =",round(A_a*180/math.pi,2),"degrees"

Numerical aperture = 0.303
Acceptance angle = 17.63 degrees


## Example 5.11, Page number 5.30¶

In [4]:
import math
from __future__ import division

#variable declaration
n1=1.480
n2=1.465
V=2.405
lamda=850*10**-9

#Calculations
delta=(n1**2-n2**2)/(2*n1**2)
a=(V*lamda*10**-9)/(2*math.pi*n1*math.sqrt(2*delta))

#Results
print "delta =",round(delta,2)

delta = 0.01
Core radius,a = 1.55 micro m


## Example 5.12, Page number 5.31¶

In [147]:
import math
from __future__ import division

#variable declaration
n1=1.5
n2=1.49
a=25

#Calculations
C_a=math.asin(n2/n1)           #Critical angle
L=2*a*math.tan(C_a)
N_r=10**6/L

#Result
print "Critical angle=",round(C_a*180/math.pi,2),"degrees"
print "Fiber length covered in one reflection=",round(L,2),"micro m"
print "Total no.of reflections per metre=",round(N_r)
print "Since L=1m, Total dist. travelled by light over one metre of fiber =",round(1/math.sin(C_a),4),"m"

 Critical angle= 83.38 degrees
Fiber length covered in one reflection= 430.84 micro m
Total no.of reflections per metre= 2321.0
Since L=1m, Total dist. travelled by light over one metre of fiber = 1.0067 m


## Example 5.13, Page number 5.31¶

In [155]:
import math
from __future__ import division

#variable declaration
alpha=1.85
lamda=1.3*10**-6
a=25*10**-6
N_a=0.21

#Calculations
V_n=((2*math.pi**2)*a**2*N_a**2)/lamda**2
N_m=(alpha/(alpha+2))*V_n

print "No.of modes =",round(N_m,2),"=155(approx)"
print "Taking the two possible polarizations, Total No.of nodes =",round(N_m*2)

No.of modes = 154.69 =155(approx)
Taking the two possible polarizations, Total No.of nodes = 309.0


## Example 5.14, Page number 5.32¶

In [2]:
import math
from __future__ import division

#variable declaration
P_i=100
P_o=2
L=10

#Calculations
S=(10/L)*math.log(P_i/P_o)
O=S*L

#Result
print "a)Signal attention per unit length =",round(S,1),"dB km**-1"
print "b)Overall signal attenuation =",round(O),"dB"
print "#Answer given in the textbook is wrong"

a)Signal attention per unit length = 3.9 dB km**-1
b)Overall signal attenuation = 39.0 dB
#Answer given in the textbook is wrong


## Example 5.15, Page number 5.32¶

In [1]:
import math
from __future__ import division

#variable declaration
L=10
n1=1.55
delta=0.026
C=3*10**5

#Calculations
delta_T=(L*n1*delta)/C
B_W=10/(2*delta_T)

#Result
print "Total dispersion =",round(delta_T/10**-9,1),"ns"
print "Bandwidth length product =",round(B_W/10**5,2),"Hz-km"
print "#Answer given in the text book is wrong"

Total dispersion = 1343.3 ns
Bandwidth length product = 37.22 Hz-km
#Answer given in the text book is wrong