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

Options

A. `60^(@)`
B. `45^(@)`
C. `90^(@)`
D. `180^(@)`

Answer

c.

Explanation

`mu_(g)sintheta_(c)=mu_(1)sin90^(@)`
or `mu_(g)sintheta_(c)=1`
When water is poured,
`mu_(w)sinr=mu_(g)sintheta_(c)`
or `mu_(w)sinr=1`
Again, `mu_(a)sintheta=mu_(w)sinr`
or `mu_(a)sintheta=1`
or `sintheta=1 or theta=90^(@)`