Question

a glass clad fiber is made with core glass of refractive index 1.5 and cladding is a droped to given index difference of 0.0005
(1) cladding refractive index
(2) critical internal refraction angle
(3) numerical aperture
(4) external critical acceptance angle​

Answer 1

Answer:

Solution :

Given , n1=1.5,△=0.0005

i) Let the refractive index of cladding be n2 . So we have

△=n1−n2n1

0.0005 = 1.5−n21.5

n2 = 1.5 - 1.5 x 0.005 = 1.4925

ii) sinθc=n2/n1

θc=sin−1(n2/n1)

θc=sin−1(1.4925/1.5)

θc=84.268o

iii) NA = (n1)2−(n2)2−−−−−−−−−−−√

NA = (1.5)2−(1.4925)2−−−−−−−−−−−−−−√

NA = 2.25−2.2275−−−−−−−−−−−√

NA = 0.02244−−−−−−√

NA = 0.1498

NA = 0.15