Chapter 12:Fibre Optics

Example 12.1 , Page no:360

In [1]:
import math
from __future__ import division

#given
u1=1.563; #refractive index of core
u2=1.498; #refractive index of cladding

#calculate
d=(u1-u2)/u1; #calculation of fractional index change

#result
print"The fractional index change for a given optical fibre is =",round(d,4);

The fractional index change for a given optical fibre is = 0.0416

Example 12.2 , Page no:360

In [2]:
import math
from __future__ import division

#given
u1=1.55; #refractive index of core
u2=1.50; #refractive index of cladding

#calculate
d=(u1-u2)/u1; #calculation of fractional index change
NA=u1*math.sqrt(2*d); #calculation of numerical aperture
theta=math.asin(NA); #calculation of acceptance angle
theta1=theta*180/3.14;

#result
print"The numerical aperture of the fibre is NA=",round(NA,3);
print"The acceptance angle of the optical fibre is =",round(theta1,2),"degree";

The numerical aperture of the fibre is NA= 0.394
The acceptance angle of the optical fibre is = 23.2 degree

Example 12.3 , Page no:360

In [3]:
import math
from __future__ import division

#given
u1=1.563; #refractive index of core
u2=1.498; #refractive index of cladding

#calculate
NA=math.sqrt(u1**2-u2**2); #calculation of numerical aperture
theta=math.asin(NA); #calculation of acceptance angle
theta1=theta*180/3.14;
#result
print"The numerical aperture of the fibre is NA=",round(NA,4);
print"The acceptance angle of the optical fibre is =",round(theta1,2),"degree";

The numerical aperture of the fibre is NA= 0.4461
The acceptance angle of the optical fibre is = 26.5 degree

Example 12.4 , Page no:360

In [4]:
import math
from __future__ import division

#given
NA=0.39; #numerical aperture of the optical fibre
d=0.05; #difference in the refractive index of the material of the core and cladding

#calculate
#since NA=u1*sqrt(2*d)
#we have u1=NA/sqrt(2*d)
u1= NA/math.sqrt(2*d); #calculation of refractive index of  material of the core

#result
print"The refractive index of  material of the core is u1=",round(u1,3);

The refractive index of  material of the core is u1= 1.233

Example 12.5 , Page no:361

In [5]:
import math
from __future__ import division

#given
u1=1.50; #refractive index of core
u2=1.45; #refractive index of cladding

#calculate
d=(u1-u2)/u1; #calculation of fractional index change
NA=u1*math.sqrt(2*d); #calculation of numerical aperture
theta_0=math.asin(NA); #calculation of acceptance angle
theta_01=theta_0*180/3.14;
theta_c=math.asin(u2/u1); #calculation of critical angle
theta_c1=theta_c*180/3.14;

#result
print"The numerical aperture of the fibre is NA=",round(NA,3);
print"The acceptance angle of the optical fibre is =",round(theta_01,2),"degree";
print"The critical angle of the optical fibre is =",round(theta_c1,2),"degree";

The numerical aperture of the fibre is NA= 0.387
The acceptance angle of the optical fibre is = 22.8 degree
The critical angle of the optical fibre is = 75.2 degree

Example 12.6 , Page no:361

In [6]:
import math
from __future__ import division

#given
NA=0.33; #numerical aperture
d=0.02; #difference in the refractive index of the core and cladding of the material

#calculate
#since NA=u1*sqrt(2*d)
#therefore we have
u1=NA/math.sqrt(2*d); #calculation of refractive index of the core
#since d=(u1-u2)/u2
#therefore we have
u2=(1-d)*u1; #calculation of refractive index of the cladding

#result
print"The refractive index of the core is u1=",round(u1,2);
print"The refractive index of the cladding is u2=",round(u2,3);
print "NOTE: The answer in the textbook is wrong" 

The refractive index of the core is u1= 1.65
The refractive index of the cladding is u2= 1.617
NOTE: The answer in the textbook is wrong

Example 12.7 , Page no:361

In [7]:
import math
from __future__ import division

#given
u1=3.5; #refractive index of core
u2=3.45; #refractive index of cladding
u0=1; #refractive index of the air

#calculate
NA=math.sqrt(u1**2-u2**2); #calculation of numerical aperture
NA1=NA/u0;
alpha=math.asin(NA); #calculation of acceptance angle
alpha1=alpha*180/3.14;
#result
print"The numerical aperture of the fibre is NA=",round(NA1,2);
print"The acceptance angle of the optical fibre is =",round(alpha1,2),"degree";

The numerical aperture of the fibre is NA= 0.59
The acceptance angle of the optical fibre is = 36.14 degree

Example 12.8 , Page no:361

In [8]:
import math
from __future__ import division

#given
u1=1.48; #refractive index of core
u2=1.45; #refractive index of cladding

#calculate
NA=math.sqrt(u1**2-u2**2); #calculation of numerical aperture
theta=math.asin(NA); #calculation of acceptance angle
theta1=theta*180/3.14;
#result
print"The numerical aperture of the fibre is NA=",round(NA,3);
print"The acceptance angle of the optical fibre is =",round(theta1,2),"degree";
print "   (roundoff error)"

The numerical aperture of the fibre is NA= 0.296
The acceptance angle of the optical fibre is = 17.26 degree
   (roundoff error)