# Chapter 15 - Fibre Optics¶

## Example 1 - pg 524¶

In [1]:
#pg 524
#Calculate the acceptance angle
#Given :
import math
n0 = 1;#refractive index of outer medium
n1 = 1.5025; # refractive index of core
n2 = 1.4975; # refractive index of cladding
#calculations
NA = math.sqrt(n1**2 - n2**2); # Numerical aperture with cladding
alpha_c = math.asin(NA/n0)*57.3; # acceptance angle in degrees
NA1 = math.sqrt(n1**2 - n0**2);# Numerical aperture without cladding
#results
print "With cladding , NA  and Acceptance angle =",round(NA,4),"and",round(alpha_c,3),"degrees"
print "Without cladding , NA = ",round(NA1,4)

With cladding , NA  and Acceptance angle = 0.1225 and 7.035 degrees
Without cladding , NA =  1.1214


## Example 2 - pg 525¶

In [2]:
#pg 525
#calculate the Ls and reflections per m
#Given :
import math
n1 = 1.5025;# refractive index of core
delta = 0.0033; #
a = 50.; # core radius in mu_m
#calculations
Ls = a*math.sqrt(2/delta);# skip distance in mu_m
# 1 mu_m = 1.0*10**-6 m
R = 1/(Ls*10**-6);# reflections per m
#results
print "Ls (mu_m) = ",round(Ls,1)
print "Reflections per m = ",round(R,0)

Ls (mu_m) =  1230.9
Reflections per m =  812.0


## Example 3 - pg 526¶

In [3]:
#pg 526
#calculate the Limiting diameter
#Given :
import math
lambd = 1.25; # wavelength in mu_m
n1 = 1.462; # refractive index of core
n2 = 1.457; # refractive index of cladding
#calculations
# Single mode propogation : (2*pi*a*sqrt(n1**2 - n2**2))/lambd < 2.405
a  = (2.405*lambd)/(2*math.pi*math.sqrt(n1**2 - n2**2)); # radius in mu_m
d = a*2; # diameter in mu_m
#results
print "Limiting diameter (mu_m) = ",round(d,2)

Limiting diameter (mu_m) =  7.92


## Example 4 - pg 527¶

In [4]:
#pg 527
#calculate the Attenuation
#Given :
import math
n1 = 1.525; # refractive index of core
n2 = 1.500; # refractive index of cladding
d = 30.; # core diameter in mu_m
#calculations
ab = 0.00001/100; # percentage absorbed
a = d/2.; # core radius in mu_m
delta = (n1-n2)/n1;
Ls = a*math.sqrt(2/delta);# skip distance in mu_m
#1 mu_m = 1.0*10^-6 m
R = 1000/(Ls*10**-6); # reflections per km (1000 m)
red_p = 1 - ab; # reduced power for each reflection
#Power P1km = P0*red_p^(6*10^6)
# A = 10*log10[P0/P1km] , P0 in the numerator and denominator will cancel each other
A = 10*math.log10(1/(red_p)**(R));
#results
print "Attenuation (dB/km) = ",round(A,1)

Attenuation (dB/km) =  2.6


## Example 5 - pg 533¶

In [5]:
#pg 533
#calculate the maximum delay and bandwidth
#Given :
n1 = 1.5025; # refractive index of core
n2 = 1.4975; # refractive index of cladding
L = 1; # length in m
F = 2*10**6; # frequency in Hz
c = 3*10**8;# light speed in m/s
#calculations
delta_t = (n1*L/c)*((n1/n2)-1);# maximum delay in s;
f = 1/(2*delta_t); # bandwidth for 1 m propogation
L1 = 1/(2*F*delta_t); # distance for 2MHz bandwidth
#results
print "Maximum delay (ps) = ",round(delta_t*10**12,1)
print "Bandwidth of 2MHz can propogate a distance of (km) =  ",round(L1*10**-3,0);

Maximum delay (ps) =  16.7
Bandwidth of 2MHz can propogate a distance of (km) =   15.0