Question

For the angle of minimum deviation of a prism to be equal to its refracting angle, the prism must be made of a material whose refractive index $ \mu $ is:
A) $ \sqrt 2 < \mu < 2 $
B) $ \mu < 1 $
C) $ 1 < \mu < \sqrt 2 $
D) $ \mu > \sqrt 2 $

Answer 1

Hint : In this solution, we will use the relation of the refractive index of the prism with the angle of prism and angle of minimum deviation. By taking the assumption that the angle of minimum deviation of a prism is equal to its refracting angle, we will obtain a range of possible refractive index for the prism.

Formula used: In this solution, we will use the following formula:
 $\Rightarrow \mu = \dfrac{{sin\left( {\dfrac{{A + {\delta _m}}}{2}} \right)}}{{sin\left( {\dfrac{A}{2}} \right)}} $ where $ \mu $ is the refractive index, $ A $ is the angle of the prism, and $ {\delta _m} $ is the angle of minimum deviation.

Complete step by step answer
The minimum deviation $ {\delta _m} $ in a prism occurs when the entering angle and the exiting angle are the same. In this situation, the ray of light when inside the prism is parallel to the base of the prism. The angle of refraction inside the prism on either side of the prism is also the same in this situation.
Given that we know that angle of minimum deviation and the angle of the prism, we can find the refractive index of the prism $ \mu $ as:
 $\Rightarrow \mu = \dfrac{{sin\left( {\dfrac{{A + {\delta _m}}}{2}} \right)}}{{sin\left( {\dfrac{A}{2}} \right)}} $
Since we’ve been told for the angle of minimum deviation of a prism ( $ {\delta _{min}} $ ) to be equal to its refracting angle ( $ A $ ), we can write $ {\delta _m} = A $ . On substituting it in the above equation, we can write
 $\Rightarrow \mu = \dfrac{{sin\left( {\dfrac{{A + A}}{2}} \right)}}{{sin\left( {\dfrac{A}{2}} \right)}} $
 $ \Rightarrow \mu = \dfrac{{sin\left( A \right)}}{{sin\left( {\dfrac{A}{2}} \right)}} $
Now, we know that $ \sin (A) = 2\sin (A/2)\cos (A/2) $ so we can write,
 $\Rightarrow \mu = \dfrac{{2\sin (A/2)\cos (A/2)}}{{sin\left( {\dfrac{A}{2}} \right)}} $
 $ \Rightarrow \mu = 2\cos (A/2) $
The angle of the prism $ A $ can only lie between $ 0^\circ $ and $ 90^\circ $ , so the minimum value of the refractive index will be for $ 90^\circ $ and will be,
 $\Rightarrow {\mu _{min}} = 2\cos (45^\circ ) $
 $ \Rightarrow {\mu _{\min }} = \dfrac{2}{{\sqrt 2 }} = \sqrt 2 $
And the maximum value will be for $ 0^\circ $ which will be
 $\Rightarrow {\mu _{min}} = 2\cos (0^\circ ) $
 $ \Rightarrow {\mu _{\min }} = 2 $
So, $ \sqrt 2 < \mu < 2 $ which corresponds to option A.

Note
While calculating the limits of the refractive index, the limits depend on the angle of the prism which can only lie between $ 0^\circ $ and $ 90^\circ $ and a prism cannot have any angle outside this range. However, the range of refractive index that we obtained is specific for the case where the angle of minimum deviation of a prism is equal to its refracting angle otherwise a prism can have a wide range of refractive index depending on the material it is made up of.