Chapter 2: Optical Fibers

Example 2.1, Page number 49

In [1]:
'''Determine maximum thickness of film'''

import math

#Variable declaration
n1 = 3.5         #core refractive index
n2 = 3.0         #cladding refractive index
v = 6            #no. of modes
lamda = 1.5      #propagating wavelength(um)

#Calculations
theta_c = math.degrees(math.asin(n2/n1))
h = (2*math.pi*v*lamda)/(2*math.pi*n1*math.cos(math.radians(theta_c)))

#Result
print "The thickness of the film should be less than",round(h),"um"

The thickness of the film should be less than 5.0 um

Example 2.2, Page number 50

In [2]:
'''find the angle of acceptance and crtitical angle'''

import math

#Variable declaration
n1 = 1.53         #core refractive index
n2 = 1.48         #cladding refractive index
n0 = 1            #refractive index for air

#calculations
theta_a = math.degrees(math.asin(((n1**2-n2**2)**0.5)/n0))   

theta_c = math.degrees(math.asin(n2/n1))

#Result
print "Angle of acceptance =",round(theta_a,2),"degrees"
print "Critical angle =",round(theta_c,2),"degrees"

Angle of acceptance = 22.83 degrees
Critical angle = 75.31 degrees

Example 2.3, Page number 50

In [3]:
'''Determine the numerical aperture'''

import math

#Variable declaration
v = 26.6         #frequency(Hz)
lamda = 1.3      #propagating wavelength(um)
a = 25           #core radius(um)

#Calculation
NA = (v*lamda)/(2*math.pi*a)

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

Numerical aperture = 0.22

Example 2.4, Page number 51

In [5]:
'''Determine the numerical aperture'''

import math

#Variable declaration
n1 = 1.4675         #core refractive index
n2 = 1.4622         #cladding refractive index

#Calculation
NA = math.sqrt(n1**2-n2**2)

#Result
print "Numerical aperture =",round(NA,3)

Numerical aperture = 0.125

Example 2.5, Page number 51

In [7]:
'''Determine cut-off wavelength for step index fiber'''

import math

#Variable declaration
n1 = 1.5         #core refractive index
n2 = 1.47        #cladding refractive index
a = 4            #core radius(um)

#Calculation
lamda_c = (2*math.pi*a*((n1**2-n2**2)**0.5))/2.405

#Result
print "The cut-off wavelength is",round(lamda_c,2),"um"

The cut-off wavelength is 3.12 um

Example 2.6, Page number 51

In [10]:
'''Determine maximum diameter of the core for single mode fiber'''

import math

#Variable declaration
n1 = 1.55         #core refractive index
n2 = 1.48         #cladding refractive index
lamda = 1.55      #wavelength(um)

#Calculations
a = (2.405*lamda)/(2*math.pi*(n1**2-n2**2)**0.5)
d = 2*a #diameter

#Result
print "Maximum diameter of the core is",round(d,2),"um"

Maximum diameter of the core is 2.58 um

Example 2.7, Page number 52

In [14]:
'''Determine the number of modes propagating'''

import math

#Variable declaration
n1 = 1.48         #core refractive index
n2 = 0.01         #cladding refractive index
a = 25            #core radius(um)
lamda = 0.84      #Wavelength(um)

#Calculation
m = 2*(2*math.pi/lamda)**2*(a**2/2)*(n1**2-n2**2)
v = math.sqrt(2*m)

#Result
print "Number of modes =",v, "(Calculation mistake in textbook while calculating 'm'. Hence, the answer differs)"

Number of modes = 391.074660134 (Calculation mistake in textbook while calculating 'm'. Hence, the answer differs)

Example 2.8, Page number 52

In [1]:
'''Determine the number of modes for multimode fiber and calculate the same when lamda is changed to 1.3um'''

import math

#Variable declaration
n1 = 1.475       #core refractive index
n2 = 1.472       #cladding refractive index
a = 20           #core radius(um)
lamda = 0.85      #Wavelength(um)

#Calculation
v = (2*math.pi*a*math.sqrt((n1**2-n2**2)))/lamda
M1 = v**2/2

lamda2 = 1.3    #um
v2 = (2*math.pi*a*math.sqrt((n1**2-n2**2)))/lamda2
M2 = v2**2/2

#Results
print "Number of modes when lamda is changed =",round(M1) #v is calculated wrongly in the book and answer for case a M not given
print "Number of modes when lamda is changed =",round(M2)

Number of modes when lamda is changed = 97.0
Number of modes when lamda is changed = 41.0

Example 2.9, Page number 53

In [21]:
'''Determine the numerical aperture'''

import math

#Variable declaration
n1 = 1.5          #core refractive index
n2 = 1.48         #cladding refractive index

#Calculation
NA = math.sqrt(n1**2-n2**2)

#Result
print "Numerical aperture =",round(NA,5)

Numerical aperture = 0.24413

Example 2.10, Page number 53

In [37]:
'''Determine core radius, NA and maximum acceptance angle'''

import math

#Variable declaration
n1 = 1.450       #core refractive index
n2 = 1.447       #cladding refractive index
lamda = 1.3      #Wavelength(um)

#Calculation
v = 2.405
a = (v*lamda)/(2*math.pi*math.sqrt((n1**2-n2**2)))

NA = math.sqrt(n1**2-n2**2)

theta_max = math.degrees(math.asin(NA))

#Results
print "Core radius =",round(a,3),"um"
print "Numerical aperture =",round(NA,4)
print "Maximum acceptance angle =",round(theta_max,3),"degrees"

Core radius = 5.338 um
Numerical aperture = 0.0932
Maximum acceptance angle = 5.349 degrees

Example 2.11, Page number 53

In [38]:
'''Determine critical angle at core cladding interface, NA and acceptance angle'''

import math

#Variable declaration
n1 = 1.50       #core refractive index
n2 = 1.47       #cladding refractive index

#calculations
theta_c = math.degrees(math.asin(n2/n1))

NA = math.sqrt(n1**2-n2**2)

theta_a = math.degrees(math.asin(NA))

#Result
print "Critical angle at core cladding interface =",round(theta_c,1),"degrees"
print "Numerical aperture =",round(NA,2)
print "Maximum acceptance angle =",round(theta_a,1),"degrees"

Critical angle at core cladding interface = 78.5 degrees
Numerical aperture = 0.3
Maximum acceptance angle = 17.4 degrees

Example 2.12, Page number 55

In [39]:
'''Determine the acceptance angle for skew rays'''

#Variable declaration
NA = 0.4    #numerical aperture
#Since skew rays change direction by 100 degrees at each reflection,
r = 50      #degrees

#Calculations
theta_as = math.degrees(math.asin(NA/math.cos(math.radians(r))))

#print
print "Acceptance angle =",round(theta_as,1),"degrees"

Acceptance angle = 38.5 degrees

Example 2.13, Page number 55

In [40]:
'''Determine normalized frequency and number of guided modes'''

import math

#Variable declaration
n1 = 1.48       #core refractive index
lamda = 0.85    #wavelength(um)
a = 80/2        #core radius(um)
delta = 1.5/100 #relative index difference

#Calculations
v = (2*math.pi*a*n1*(2*delta)**0.5)/lamda

M = v**2/2

#Results
print "Normalized frequency =",round(v,1),"Hz"
print "Number of guided modes =",round(M)

Normalized frequency = 75.8 Hz
Number of guided modes = 2872.0

Example 2.14, Page number 56

In [42]:
'''Determine cut off value for normalized frequency and maximum core radius'''

import math

#Variable declaration
n1 = 1.5          #core refractive index
lamda = 1.3       #wavelength(um)
delta = 1./100.   #relative index difference
alpha = 2

#Calculations
v = 2.4*(1+2/alpha)**0.5

a = (v*lamda)/(2*math.pi*n1*(2*delta)**0.5)

#Results
print "Cut off value for normalized frequency =",round(v,2)
print "Maximum core radius =",round(a,2),"um"

Cut off value for normalized frequency = 3.39
Maximum core radius = 3.31 um

Example 2.15, Page number 56

In [43]:
'''Determine cut-off wavelength for step index fiber'''

import math

#Variable declaration
n1 = 1.46         #core refractive index
a = 4.5           #core radius(um)
delta = 0.25/100

#Calculation
lamda_c = (2*math.pi*a*n1*(2*delta)**0.5)/2.405

#Result
print "The cut-off wavelength is",round(lamda_c,3),"um"

The cut-off wavelength is 1.214 um

Example 2.16, Page number 57

In [44]:
'''Determine the numerical aperture and acceptance angle'''

import math

#Variable declaration
n1 = 1.45         #core refractive index
n2 = 1.4          #cladding refractive index

#Calculation
NA = math.sqrt(n1**2-n2**2)

theta_m = math.degrees(math.asin(NA))

#Result
print "Numerical aperture =",round(NA,4)
print "Acceptance angle =",round(theta_m,2),"degrees"

Numerical aperture = 0.3775
Acceptance angle = 22.18 degrees

Example 2.17, Page number 57

In [46]:
'''Calculate cladding index, crtical internal reflection angle, external critical acceptance angle and numerical aperture'''

import math

#Variable declaration
n1 = 1.5       #core refractive index
delta = 0.0005

#Calculations
n2 = n1*(1-delta)

theta_c = math.degrees(math.asin(n2/n1))

n0 = 1   #refractive index for air
theta_m = math.degrees(math.asin(((n1**2-n2**2)**0.5)/n0))

NA = n1*math.sqrt(2*delta)

#Results
print "Cladding index =",round(n2,5)
print "Crtical internal reflection angle =",round(theta_c,1),"degrees"
print "External critical acceptance angle =",round(theta_m,2),"Degrees"
print "Numerical aperture =",round(NA,4)

Cladding index = 1.49925
Crtical internal reflection angle = 88.2 degrees
External critical acceptance angle = 2.72 Degrees
Numerical aperture = 0.0474

Example 2.18, Page number 58

In [47]:
'''Determine acceptance angle for fiber in water'''

#Variable declaration
NA = 0.20    #numerical aperture
n2 = 1.59    #cladding refractive index
n0 = 1.33    #refractive index for water 

#Calculations
n1 = math.sqrt(NA**2+n2**2)
NA = math.sqrt(n1**2-n2**2)/n0
theta_m = math.degrees(math.asin(NA))

#Result
print "Acceptance angle for fiber in water =",round(theta_m,1),"degrees"

Acceptance angle for fiber in water = 8.6 degrees

Example 2.19, Page number 58

In [49]:
'''Determine the numerical aperture and acceptance angle'''

import math

#Variable declaration
n1 = 1.55         #core refractive index
n2 = 1.51         #cladding refractive index

#Calculation
delta = (n1-n2)/n1
NA = 2*math.sqrt(delta)

theta_m = math.degrees(math.asin(NA))

#Result
print "Numerical aperture =",round(NA,4)
print "Acceptance angle =",round(theta_m,2),"degrees"
print "\nCalculation mistakes in textbook. Hence, the answers differ."

Numerical aperture = 0.3213
Acceptance angle = 18.74 degrees

Calculation mistakes in textbook. Hence, the answers differ.

Example 2.20, Page number 59

In [50]:
'''Determine normalized frequency'''

#Variable declaration
n = 1.45        #core refractive index
lamda = 0.1     #wavelength(um)
a = 60/2        #core radius(um)
NA = 0.16       #numerical aperture

#Calculations
v = (2*math.pi*a*NA)/lamda

#Results
print "Normalized frequency =",round(v,1),"(Calculation mistake in textbook)"

Normalized frequency = 301.6 (Calculation mistake in textbook)

Example 2.21, Page number 59

In [51]:
'''Calculate NA and multi path dospersion per unit length'''

#Variable declaration
c = 3.*10**8    #speed of light in vacuum(m/s)
v = 2.*10**8    #speed of light in core(m/s)
theta_c = 75    #cricial angle(degrees)

#Calculations
n1 = c/v
n2 = n1*math.sin(math.radians(theta_c))
NA = math.sqrt(n1**2-n2**2)

delta_n = n1-n2
md = (n1/n2)*(delta_n/c)  #multipath dispersion

#Results
print "Numerical aperture =",round(NA,2)
print "Multi path dospersion per unit length =",round((md/1E-9),3),"*10^-9 s/m"

Numerical aperture = 0.39
Multi path dospersion per unit length = 0.176 *10^-9 s/m

Example 2.22, Page number 60

In [52]:
'''Determine maximum thickness of guide slab'''

#Variable declaration
n1 = 3.6          #core refractive index
n2 = 3.56         #cladding refractive index
lamda = 0.85      #wavelength(um)
#For TE10 mode,
m = 1
n = 0
vc = 2.405        #for planar guide

#Calculation
a = (vc*lamda)/(2*math.pi*math.sqrt(n1**2-n2**2))

#Result
print "Maximum thickness of guide slab =",round(a,3),"um"

Maximum thickness of guide slab = 0.608 um

Example 2.23, Page number 61

In [6]:
'''Calculate diameter of core'''

import math

#Variable declaration
n1 = 1.5          #core refractive index
lamda = 1.3*10**-6       #wavelength(um)
delta = 1./100.   #relative index difference
M = 1100

#Calculations
V = math.sqrt(2*M)

a = (V*lamda)/(2*math.pi*n1*(2*delta)**0.5)

d = 2*a

#Result
print "Diameter =",round(d/1E-5,2),"um(Calculation mistake in textbook)"

Diameter = 9.15 um(Calculation mistake in textbook)

Example 2.24, Page number 62

In [59]:
'''Determine critical angle and numerical aperture'''


#Variable declaration
n1 = 1.50         #core refractive index
n2 = 1.46         #cladding refractive index

#Calculation
theta_c = math.degrees(math.asin(n2/n1))

NA = math.sqrt(n1**2-n2**2)

#Result
print "Critical angle =",round(theta_c,2),"degrees"
print "Numerical aperture =",round(NA,2)

Critical angle = 76.74 degrees
Numerical aperture = 0.34

Example 2.25, Page number 62

In [84]:
'''Determine the acceptance angle for skew rays'''

#Variable declaration
NA = 0.344    #numerical aperture
#Since skew rays change direction by 100 degrees at each reflection,
gamma = 100/2      #degrees

#Calculations
#For meridional rays
theta_a = math.degrees(math.asin(NA))
#For speed rays
theta_as = math.degrees(math.asin(NA/math.cos(math.radians(gamma))))

#print
print "Acceptance angle for meridional rays =",round(theta_a,2),"degrees"
print "Acceptance angle for speed rays =",round(theta_as,2),"degrees"

Acceptance angle for meridional rays = 20.12 degrees
Acceptance angle for speed rays = 32.36 degrees

Example 2.26, Page number 62

In [61]:
'''Calculate the no. of guided modes and cut-off value of normalized frequency'''

import math

#Variable declaration
n1 = 1.5           #core refractive index
lamda = 1.55       #wavelength(um)
delta = 1.3/100.   #relative index difference
alpha = 1.90       #index profile
a = 40/2           #core radius(um)

#Calculations
Mg = (alpha/(alpha+2))*((n1*2*math.pi*a)/lamda)**2*delta

Vc = 2.405*math.sqrt(1+2/alpha)

#Results
print "Number of guided modes =",round(Mg)
print "Cut-off value of normalized frequency =",round(Vc,2)

Number of guided modes = 94.0
Cut-off value of normalized frequency = 3.45