Question

The angle of minimum deviation produced by a $60^\circ $prism is$40^\circ $. Calculate the refractive index of the material of the prism.
(A) $1.33$
(B) $1.53$
(C) $1.63$
(D) $1.44$

Answer 1

Hint
The question is based on the angle of minimum deviation produced by a prism. A $60^\circ $prism refers to a prism having an angle of prism equal to $60^\circ $. We will make use of the formula that relates the angle of minimum deviation, refractive index and the angle of prism to find the refractive index of the material of the prism.
$\Rightarrow \mu = \dfrac{{\sin \left( {\dfrac{{A + \delta }}{2}} \right)}}{{\sin \left( {\dfrac{A}{2}} \right)}}$
Where $A$is the angle of the prism,
$\delta $is the angle of minimum deviation and
$\mu $is the refractive index

Complete step by step answer
The formula that relates the angle of minimum deviation, refractive index and the angle of the prism is as follows.
$\Rightarrow \mu = \dfrac{{\sin \left( {\dfrac{{A + \delta }}{2}} \right)}}{{\sin \left( {\dfrac{A}{2}} \right)}}$
The angle of a prism is the angle between the lateral faces of the prism. When passing through the prism, the light ray gets deviated from its original path. This deviation in the path of the ray is called the angle of deviation.
In the question we are provided that the angle of prism equal to $60^\circ $and the angle of minimum deviation equal to $40^\circ $. So on substituting these values in the formula mentioned above we get the refractive index as,
$\Rightarrow \mu = \dfrac{{\sin \left( {\dfrac{{60^\circ + 40^\circ }}{2}} \right)}}{{\sin \left( {\dfrac{{60^\circ }}{2}} \right)}}$
On doing the calculation of the angles inside the brackets,
$\Rightarrow \mu = \dfrac{{\sin \left( {50^\circ } \right)}}{{\sin \left( {30^\circ } \right)}}$
Now, the values of $\sin \left( {50^\circ } \right)$ is equal to, $0.7660$ and the value of $\sin \left( {30^\circ } \right)$ is $0.5$
Thus, substituting the values we have,
$\Rightarrow \mu = \dfrac{{0.7660}}{{0.5}}$
This on calculating gives the refractive index as, $\mu = 1.53$
$\therefore $ The refractive index of the material of the $60^\circ $prism is $1.53$.
Thus, option (B) is correct.

Note
The ray of light while travelling through the prism gets deviated because; the speed of light in different materials is different. The speed of light is maximum in vacuum and as the refractive index of a medium increases, the speed of light in that medium decreases.