A ray of light is incident at an angle of `60^(@)` on one face of a prism of angle `30^(@)`. The ray emerging out of the prism makes an angle of `30^(@)` with the incident ray. The emergent ray is

Options

A. `0^(@)`
B. `90^(@)`
C. `30^(@)`
D. `45^(@)`

Answer

B

Explanation

Emergent ray makes an angle `30^(@)` with the incident ray, hence angle of deviation `(delta)` is `30^(@)`.
If `i` is angle of incidence, e is angle of emergence and A is angle of prism then we can write angle of deviation as follows:
`delta= i+e-A`
Substituting values we get the following :
`30=60+e-30^(@)`
`rArr e=0`
Angle of emergence is zero hence light, ray emerges perpendicular to the surface.
Hence option (b ) is correct.