Question

A ray of light enters one end of an optical fiber with an angle of incidence θi = 50.0◦. The index of refraction of the fiber is 1.62.

θ

a. Find the angle θ the ray makes with the normal when it reaches the curved surface of the fiber.
b. Show that the ray inside the fiber is undergoing total internal reflection. Why would you want this inside an optical fiber?

Answer 1

a )

here A = 50^o

B = 90^o  

n_a = 1 , n_b = 1.62

sin B = sin50 / 1.62

B = 28.22^o  

T + T_i + T_f = 180

T = 180 - 90 - 28.22 = 61.75^o  

b )

sin A = 1 / 1.62

A = sin^-1 ( 1 / 1.62 )

A = 38.1^o  

50 > 38.1

so the total internal reflection is this angle...