Question

A ray of light is incident on one of the faces of the prism with an angle $ 40{}^\circ $ with the surface. The angle of the prism is $ 60{}^\circ $ . The emergent ray is deviated through an angle $ 38{}^\circ $ . The angle of emergence is:
(A) $ 38{}^\circ $
(B) $ 48{}^\circ $
(C) $ 52{}^\circ $
(D) $ 58{}^\circ $

Answer 1

Hint: We know that the relationship between angle of incidence, angle of prism, angle of deviation and angle of emergence is given as:
 $ \text{i}+\text{e}=\text{A+}\delta $
The values of i, A and $ \delta $ are known. So using the above equation, the value of e can be found.

Complete step by step solution
It is given that:
Angle of incidence $ =50{}^\circ $
Angle of prism A $ =60{}^\circ $
Angle of deviation $ \delta $ = $ 38{}^\circ $
We have to find the angle of emergence
We know that,
 $ \begin{align}
  & \text{i+e}=\text{A+}\delta \\
 & \text{50+e=60+38} \\
 & \text{ e=98-50} \\
 & \text{ e}=48{}^\circ \\
\end{align} $
Hence, angle of emergence is $ 48{}^\circ $
Correct option is (B).

Note
The equation $ \text{A}+\text{ }\delta =\text{i}+\text{e} $ tells us that when a ray of light passes through a prism, the sum of angle of prism and the angle of deviation is equal to sum of angle of incidence and angle of emergence.
The deviation produced in the path of ray of light depends upon:
-Angle of prism.
-Material of prism
-Angle of incidence
If the angle of incidence is increased gradually, the angle of deviation first decreases and attains a minimum value and then starts increasing again.