Question

A ray of light is incident normally on the surface of an equilateral prism made up of material with refractive index 1.5. The angle of deviation is
(A) $ 30^\circ $
(B) $ 45^\circ $
(C) $ 60^\circ $
(D) $ 75^\circ $

Answer 1

Hint: If the angle of incidence is greater than the critical angle of the medium then the light ray is totally reflected back. First, we will use Snell’s law to find the angle of refraction and observe the behavior of the ray at the second face.
Angle of deviation $ \delta = i + e - A $

Complete step by step solution:
We can observe the angle of incidence $ i = 0^\circ $ since it is given that the ray of light is normally incident on the surface of the prism. Also, the angle of the prism $ A = 60^\circ $ .
We know that from Snell’s law $ \sin {r_1} = \dfrac{{\sin i}}{n} $ .
 $ \Rightarrow \sin {r_1} = \dfrac{{\sin 0^\circ }}{n} = 0 $
 $ \Rightarrow {r_1} = 0^\circ $
Now, we also know that for a prism $ A = {r_1} + {r_2} $ .
 $ \Rightarrow {r_2} = A - {r_1} = 60^\circ - 0^\circ $
 $ \Rightarrow {r_2} = 60^\circ $
We also know that if $ c $ is the critical angle of the material of the prism, then $ \sin c = \dfrac{1}{n} $ .
 $ \Rightarrow \sin c = \dfrac{1}{{1.5}} = \dfrac{1}{{\left( {\dfrac{3}{2}} \right)}} = \dfrac{2}{3} $
 $ \Rightarrow c = {\sin ^{ - 1}}\left( {\dfrac{2}{3}} \right) $
 $ \Rightarrow c = 41.81^\circ $
So, the critical angle of the material of the prism is $ 41.81^\circ $ .
Here, the angle of incidence at the second face of the prism $ {r_2} = 60^\circ $ is greater than the critical angle of the prism. So, total internal reflection happens at the second face of the prism. That is, the light ray is totally internally reflected back into the prism from the second face. The angle of reflection will be $ 60^\circ $ as $ {r_2} = 60^\circ $ . This ray will be incident normally on the base face of the prism. Hence, it will emerge normally perpendicular to the prism. That is, it will emerge with the angle of emergence $ e = 0^\circ $ .
Now we know that $ \delta = i + e - A $ .
 $ \Rightarrow \delta = 0 - 0 - 60. $
So, the angle of deviation will be $ 60^\circ $. Hence, the correct option is (C).

Note:
The angle of deviation steadily decreases as the angle of incidence is increased. But after attaining a certain minimum value, it starts increasing with increase in the angle of incidence. This minimum value is called the angle of minimum deviation and is denoted by $ D $.