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?

Options

A. `180^(@)`
B. `0^(@)`
C. `90^(@)`
D. `45^(@)`

Explanation

(c) `sinC=1/(._(a)mu_(g))=2/3`
Here `(sin C)/(sin r_(1))=(._(a)mu_(w))/(._(a)mu_(g))=(4/3)/(3/2)=8/9`
`sin r_(1)=9/8 sin C=9/8 xx2/3 = 3/4`
Let `r_(2)=` angle of refraction for water -air interface
`(sin r_(1))/(sin r_(2))=1/(._(a)mu_(w))=3/4`
`sin r_(2)=4/3 sin r_(1)=4/3xx3/4=1=sin 90^(@)`
`r_(2)=90^(@)`