A ray of light traveling in glass `(mu=3//2)` is incident on a horizontal glass-air surface at the critical angle `theta_(C)`. If a thin layer of water `(mu=4//3)` is now poured on the glass-air surfac, the angle at which the ray emerges into air at the water-air surface is

Explanation

Critical angle for glass-air surface
`mu_gsinC=1xxsin90^@`
`sinC=(1)/(mu_g)=(1)/(3//2)=2/3`
Refraction at AB
`mu_gsinC=mu_omegasinr`
`3/2*2/3=4/3sinr`
`sinr=3/4`
Refraction at water-air surface
`mu_omegasinr=1xxsintheta`
`4/3xx3/4=sintheta=1`
`theta=90^@`
The ray will emerge parallel to water-air surface.