Answer to Question #182276 in Optics for reen

Question

Internal critical angle in core at core-cladding interface of a step index fiber is 78.6°. The refractive index of cladding is 1.45. find acceptance angle

Answer 1

GIVEN:-

critical angle in core at core-cladding interface of a step index fiber is 78.6° = $\theta _C$ ......(1)

 refractive index of cladding = 1.45

To Find :-

acceptance angle

solution :-

acceptance angle = $\boxed{\theta_A=sin^{-1}(N_A)}$ .......(A)

N_A= Numerical apertuture

And we know that

$\boxed{\N_A= (\mu_1^2-\mu_2^2)}$

$\mu_1 =$ cladding rifrective index

$\mu_2=$ core refractive index

and

$\boxed{sin\theta_c= {\mu_2 \over\mu_1}}$

From (1)

78.6° = $\theta _C$

$\implies$

$sin76^o = {1.45\over\mu_2}$

$0.97029573= {1.45\over\mu_2}$

$\mu_2={1.45\over 0.97029573 }$

$\mu_2 = 1.49438975682$ ...........(2)

$\implies$

and

$\boxed{\N_A= (\mu_1^2-(1.45)^2)}$

from (2)

$\boxed{\N_A= ((1.49438975682)^2-(1.45)^2)}$

$\boxed{N_A=0.129}$ ...........(3)

Now

acceptance angle =

$\boxed{\theta_A=sin^{-1}(N_A)}$

Answer:- acceptance angle

$\boxed{\theta_A=7.41181^o}$