Question

A ray of light is incident at an angle of incidence i, on one face of a prism of angle A (assumed to be small) and emerges normally from the opposite face. If the refractive index of the prism is \[\mu \]¸ the angle of incidence $i$ is nearly equal to
A) \[\dfrac{A}{\mu }\]
B) \[\dfrac{A}{{2\mu }}\]
C) \[\mu A\]
D) \[\dfrac{{\mu A}}{2}\]

Answer 1

Hint: As we know snell’s law is all about the relationship between the angle of incidence and angle refraction so Here, we will be using the formula of Snell’s law and will do some approximation wherever needed.

Formula used:
We will use Snell’s law here which is
\[\mu_1\sin i = \mu_2\sin r\]
Where $i$ is the angle of incidence and r is the angle of refraction. And \[\mu_1\& \mu_2\] are the refractive indexes of the first and second medium through which the light passes.

Complete step by step solution:
Given, the Angle of incidence of the light is $i$ on the prism. The angle of the prism is A which is to be assumed small. The ray of light emerges normally from the opposite face of the prism. And the refractive index of the prism is given which is \[\mu \].

In the figure, consider triangle PQO
As we know, the sum of all angles of a triangle is equal to \[{180^ \circ }\]. So,
\[\angle {\text{PQO + }}\angle {\text{QPO + }}\angle {\text{POQ = 18}}{0^ \circ }\]
Putting \[\angle {\text{PQO = A,}}\angle {\text{QPO = }}\alpha ,\angle {\text{POQ = 9}}{{\text{0}}^ \circ }\]
So \[ \Rightarrow A + \alpha + {90^ \circ } = {180^ \circ }\]
Therefore after calculation we will get
  \[ \Rightarrow \alpha = {90^ \circ } - A\]
\[ \Rightarrow A = {90^ \circ } - \alpha \] (1)
Since, CD is perpendicular on the face of the prism, So
\[\alpha + r = {90^ \circ }\]
From equation (1)
$
   \Rightarrow r = {90^ \circ } - \alpha \\
   \Rightarrow r = A \\
$
Therefore angle prism is equal to angle of refraction
Now, From Snell’s law,

For small angle,
i = \[\mu A\]. since sin x = x, for small angles.
Thus, we can see the angle of incidence is nearly equal to \[\mu A\].

Hence, option C is correct.

Note:
The approximation of sin function done in this solution is applicable in case of small angles only. The precisely made diagrams will give you a better understanding of the given condition.