Chapter Eight : Fiber Optical Waveguides

Example 8.1, Page Number 364

In [1]:
from math import sqrt
from math import sin
from math import exp
from math import pi

n1=1.5 #Refraction index of glass/air interface
n2=1 #Refraction index of air

c1=1*2*3.14*n2
s1=(sin(pi/3))**2  #By equation 8.8
s2=(n1/n2)**2
s3=(s1*s2)-1
s4=sqrt(s3)
s5=s4*c1

#assuming wavelength = field distance

F=exp(-1*s5) #Where F is the field decay factor
F=round(F,4) #Rounding out the value to equivalent figure

#at theta1=42 degrees

c1=1*2*3.14*n2
theta1=0.733038
#ta1=theta1*(180/math.pi)
#heta1=round(theta1,1)

s1=(sin(theta1))**2 #By equation 8.8
s2=(n1/n2)**2
s3=(s1*s2)-1
s4=sqrt(s3)
c4=s4*c1

#assuming wavelength = field distance

F1=exp(-1*c4)#where F1 is the field decay factor at 42 degrees
F1=round(F1,3)

print "The Field Decay factor at 60 degrees is "+str(F)
print "The Field Decay factor at 42 degrees is "+str(F1)
The Field Decay factor at 60 degrees is 0.0055
The Field Decay factor at 42 degrees is 0.583

Example 8.2, Page Number 369

In [2]:
from math import sqrt

w=100*(10**-6) #Thickness of the WG
n1=1.48
n2=1.46
l=1*(10**-6) #Wavelength of light
PI=3.14

V=((PI*w)/l)*sqrt((n1**2)-(n2**2))
V1=(2*V)/PI

#from equation 8.15
N=2*(1+abs(V1)) #Where N is the total number of possible modes
N=round(N,0)
N=N-1
print "The Total number of possible Modes is "+str(N)
The Total number of possible Modes is 98.0

Example 8.3, Page Number 370

In [3]:
from math import sqrt

n1=1.48 #Given refraction index
n2=1.46
l=1*(10**-6) #Wavelength

#from equation 8.16

d=l*(1/(2*sqrt((n1**2)-(n2**2))))

print "The Waveguide core thickness must be less than %.2e meter"%(d)
The Waveguide core thickness must be less than 2.06e-06 meter

Example 8.4, Page Number 375

In [4]:
from math import sqrt
n1=1.48 #Given refraction index
n2=1.46
l=900*(10**-9) #Wavelength in meter
r=100*(10**-6) #Core radius in meter
PI=3.14

V=(2*PI*r*sqrt((n1**2)-(n2**2)))/l #Where V is a given Parameter
V=round(V,1)
N=(V**2)/2 #Where N is the Number of Modes
N=round(N,1)
print "The Given V Parameter is "+str(V)
print "The Number of Modes able to propagate in a fiber is "+str(N)
The Given V Parameter is 169.2
The Number of Modes able to propagate in a fiber is 14314.3

Example 8.5, Page Number 377

In [5]:
from math import asin
from math import degrees
from math import sqrt

n1=1.48
n2=1.46 #Given refraction Index
no=1

theta=asin(n2/n1)
theta1=degrees(theta)

#Using equation 8.22
delta=((n1**2)-(n2**2))/(2*(n1**2))
delta=round(delta,4)

#using expression of 8.22a
delta1=(n1-n2)/n1
delta1=round(delta1,4)

NA=sqrt((n1**2)-(n2**2))
NA=round(NA,3)

a=asin(NA) #where a is another fiber parameter

a1=degrees(a)
a1=round(a1,2)

print "The Delta Parameter is "+str(delta)+" & "+str(delta1)
print "The Numerical aperture is "+str(NA)
print "The fiber acceptance angle is "+str(a1)+" degrees"
The Delta Parameter is 0.0134 & 0.0135
The Numerical aperture is 0.242
The fiber acceptance angle is 14.0 degrees

Example 8.6, Page Number 380

In [6]:
n1=1.48 #Given refractive index
n2=1.46
L=1*(10**3) #length of fiber
c=3*(10**8) #speed of light

#Using equation 8.24
t=((l*n1)/(c*n2))*(n1-n2) #where t is the time difference due tp intermodal dispersion

print "The Time difference due to Intermodal dispersion is %.2e seconds"%(t)
The Time difference due to Intermodal dispersion is 6.08e-17 seconds

Example 8.7, Page Number 383

In [7]:
n1=1.48 #Given refraction index
n2=1.46
l=10**3 #length of the fiber
c=3*(10**8) #Speed of light

delta=((n1**2)-(n2**2))/(2*(n1**2)) #Where delta is one of the fiber parameter
delta=round(delta,4)
t=(l*n1*(delta**2))/(c*8) #where t is the minimum pulse broadening

print "The Delta Paramter is "+str(delta)
print "The Minimum Pulse Broadening is %.2e seconds"%(t)
The Delta Paramter is 0.0134
The Minimum Pulse Broadening is 1.11e-10 seconds

Example 8.8, Page Number 387

In [8]:
n1=1.48 #Given refraction indices
n2=1.46 
NA=0.242
l=1.5*(10**-6) #Wavelength of radiation

#From equation 8.27
a=(2.405*l)/(2*3.14*NA) #where a is the maximum core radius
print "The Maximum Core Radius is less than or is %.2e meter"%(a)

#With NA=0.1
NA=0.1
a=(2.405*l)/(2*3.14*NA) #where a is the maximum core radius
print "The Maximum Core radius is less than or is %.2e meter"%(a)

#At V=2 The Diameter would 9.5 micro meter
The Maximum Core Radius is less than or is 2.37e-06 meter
The Maximum Core radius is less than or is 5.74e-06 meter

Example 8.9, Page Number 387

In [9]:
V=2

#From equation 8.29a
w=0.65+(1.619*(V**(-1.5)))+(2.879*(V**-6))
w=round(w,3)
print "The Mode Field Irradiance Diameter is "+str(w)

#Answer is a bit Miscalculated here
The Mode Field Irradiance Diameter is 1.267

Example 8.10, Page Number 390

In [10]:
%matplotlib inline
import math
from matplotlib.pyplot import plot,suptitle,xlabel,ylabel

X = [0,0.25,0.5,0.75,1,1.25,1.5,1.55,1.75,2,2.5]
V = [1,0.08,0.07,0.04,0.02,0.0,-0.007,-0.015,-0.03,-0.04,-0.06]
plot(X,V);
xlabel("Wavelength in micrometer")
ylabel("D (Dimensionless Quantity)")
suptitle("Fig 8.26")

l=10.0**3 #length of the fiber
w1=850.0*(10**-9) #Given Wavelength
lw=50.0*(10**-9) #Linewidth in meter
w2=1550.0*(10**-9) #Given Wavelength 2
lw2=3.0*(10**-9) #Linewidth 2 in meter
c=3.0*(10**8) #Speed of light
       
print""
print "From Figure 8.26 We can calculate the dimensionless quantity "
       
d1=2.14*(10**-2) #For w1 after observation
d2=-1.02*(10**-2) #For w2 after observation

#From equation 8.34
       
t1=(l/c)*d1*(lw/w1) #Where t1 is the material dispersion effects for w1
t2=(l/c)*d2*(lw2/w2) #Where t2 is the material dispersion effects for w2

print " "
print " (a) The Material Dispersion Effect Parameter for the LED is %.2e seconds"%(t1)
print " (b) The Material Dispersion Effect Paramter for the Laser is %.2e seconds"%(-1*t2)
From Figure 8.26 We can calculate the dimensionless quantity 
 
 (a) The Material Dispersion Effect Parameter for the LED is 4.20e-09 seconds
 (b) The Material Dispersion Effect Paramter for the Laser is 6.58e-11 seconds

Example 8.11, Page Number 396

In [11]:
from math import exp
from math import log10

l=1*(10**-6) #Given Wavelength
l1=10**3 #length of the fiber
n=1.45 #Refractive Index
p=0.286
B=7*(10**-11) 
T=1400 #Temperature in kelvin
PI=3.14
k=1.38*(10**-23) #Boltzman Constant

#From equation 8.38

ar=((8*(PI**3))/(3*(l**4)))*((n**8)*B*T*k*(p**2))    #Where ar is the attenuation due to raleigh scattering in per meter
ar=round(ar,6)
print "The Attenuation due to Raleigh Scattering in a silica fiber is "+str(ar)+" /m"

i1=(ar*-1*l1)
j=exp(i1)
Loss=-10*log10(j) #Where Loss is the given loss generated from attenuation
Loss=round(Loss,2)
print "The Total loss due to Attenuation is "+str(Loss)+" db/km"

#at a wavelength of 1.55 micro meter we have
l=1.55*(10**-6) #new Wavelength

ar=((8*(PI**3))/(3*(l**4)))*((n**8)*B*T*k*(p**2)) #Where ar is the attenuation due to raleigh scattering in per meter
ar=round(ar,6)
print "The Attenuation due to Raleigh Scattering in a silica fiber is "+str(ar)+" /m"

Loss=-10*log10(exp(ar*-1*l1))    #Where Loss is the given loss generated from attenuation
Loss=round(Loss,2)
print "The Total loss due to Attenuation is "+str(Loss)+" db/km"
The Attenuation due to Raleigh Scattering in a silica fiber is 0.000178 /m
The Total loss due to Attenuation is 0.77 db/km
The Attenuation due to Raleigh Scattering in a silica fiber is 3.1e-05 /m
The Total loss due to Attenuation is 0.13 db/km

Example 8.12, Page Number 398

In [12]:
from math import log10
n=1.48 #Refraction index of fiber
n0=1  #refraction index between the fibers in air

#From equation 8.39
Rf=(((n-1)/(n+1))**2) #Where Rf is the fraction of light
Rf=round(Rf,4)

Tf=((1-Rf)**2)  #Where Tf is the total transmission for each face due to fresnals reflection
Tf=round(Tf,4)

L=-10*log10(Tf) #where L is the Transmission Loss
L=round(L,2)

print "The Fraction of light reflected back at each end is "+str(Rf)
print "The Total Transmission for each face due to Fresnals Reflection is "+str(Tf)
print "The Total Transmission loss is "+str(L)+" db"
The Fraction of light reflected back at each end is 0.0375
The Total Transmission for each face due to Fresnals Reflection is 0.9264
The Total Transmission loss is 0.33 db

Example 8.13, Page Number 399

In [13]:
from math import acos
from math import log10
from math import sqrt

l=0.1 #where l=D/2a and occurs due to lateral misalignment where D is the lateral displacement and a is the fiber core radius
PI=3.14

#From equation 8.40
T=(2/PI)*(acos(l)-l*(sqrt(1-(l**2))))
L=-10*log10(T) #Where L is the Transmission loss
L=round(L,2)

#taking tha calculation of Fresnels loss from above example also we have

LT=L+0.332
print "The Total Transmission loss is "+str(L)+" db"
print "The Total Transmission loss including Fresnels loss is "+str(LT)+" db"
The Total Transmission loss is 0.59 db
The Total Transmission loss including Fresnels loss is 0.922 db

Example 8.14, Page Number 404

In [14]:
from math import log10

d1=200*(10**-6) #in meter
d2=250*(10**-6) #in meter
# d2 & d1 where d1 is the core diameter and d2 is the core+cladding diameter
#The Mixing rod has a diameter of 3d2
d3=3*d2
#Given Power Levels P1=(B*3.14*(3d2**2))/4 & P2=(B*3.14*(d1**2)/)/4 where B is a constant

L=-10*log10((d1**2)/(d3**2)) #Where L is the Insertion Loss
Le=-10*log10((7*(d1**2))/(d3**2)) #Where Le is the Excess loss
L=round(L,1)
Le=round(Le,2)
print "The Insertion loss is "+str(L)+" db"
print "The Excess Loss is "+str(Le)+" db"
The Insertion loss is 11.5 db
The Excess Loss is 3.03 db