A ray of light travelling in glass `(mu_g=3//2)` is incident on a horizontal glass-air surface at the critical angle `theta_C.` If a thin layer of water `(mu_w=4//3)` is now poured on the glass-air surface. At what angle will the ray of light emerges into water at glass-water surface?

Answer

A, C, D

Explanation

`sin theta_C=1/mu_1=2/3`
Now, `u_1sini_1=u_2sini_2`
or `mu_1sin theta_C=mu_2sin theta`
`:. (3/2)(2/3) =(4/3)sin theta`
or ` theta=sin^-1(3/4)`