Question

A ray of light is incident at an angle of ${60^ \circ }$ on one face of a prism of angle ${30^ \circ }$ . The ray emerging out of the prism makes an angle of ${30^ \circ }$ with the incident ray. The emergent ray is
(A) Normal to the face through which it emerges
(B) Inclined at ${30^ \circ }$ , to the face through which it emerges
(C) Inclined at ${60^ \circ }$ , to the face through which it emerges
(D) None of the above

Answer 1

To solve this question, we need to use the formula for the angle of deviation of a light as it passes through a prism. Substituting the values of the angles of incidence, angle of prism, and angle of deviation from the given question, we will get the value of the angle of emergence. From that, we can determine the final angle made by the emergent ray with the face of the prism.

Formula used The formula used to solve this question is given by
$\delta = i + e - A$ , here $\delta$ is the total angle of deviation of a light incident on a prism of angle $A$ , at an angle of incidence of $i$ , and $e$ is the angle of emergence.

Complete step-by-step solution
According to the question, the ray of light is incident at an angle of ${60^ \circ }$ . This means that the angle of incidence is equal to ${60^ \circ }$ . So we have
$i = {60^ \circ }$ .............................(1)
The angle of the prism is given to be equal to ${30^ \circ }$ . So we have
$A = {30^ \circ }$.............................(2)
Also, the ray emerges out of the prism by making an angle of ${30^ \circ }$ with the incident ray. We know that the angle made by the emergent ray with the incident ray is termed as the total angle of deviation of the light. So in this case, we have the angle of deviation of ${30^ \circ }$ . So we can write
$\delta = {30^ \circ }$ .............................(3)
Now, we know that the angle of deviation is related to the angle of incidence, the angle of prism, and the angle of emergence by the relation
$\delta = i + e - A$
Substituting (1), (2) and (3) in the above equation, we get
${30^ \circ } = {60^ \circ } + e - {30^ \circ }$
On solving the above equation we get
$e = {0^ \circ }$
So the angle of emergence comes out to be equal to ${0^ \circ }$ . We know that the angle of emergence is the angle made by the emergent ray with the normal to the face of the prism. So the angle made by the emergent ray with the face of the prism can be given by
$E = {90^ \circ } - e$
$\Rightarrow E = {90^ \circ } - {0^ \circ } = {90^ \circ }$
So the emergent ray is normal to the face of the prism through which it emerges.
Hence, the correct answer is option (A).

Note
Do not misunderstand the angle of the incident ray to be made with the face of the prism. The angle of incidence is always defined with respect to the normal to the surface at which the ray of light is incident.