Question

The angle of minimum deviation measured with a prism is ${30}^{\circ }$ and the angle of prism is ${60}^{\circ }$. The refractive index of the prism material is
A. $\sqrt{2}$
B. 2
C. ${\displaystyle \frac{3}{2}}$
D. ${\displaystyle \frac{4}{3}}$

Answer 1

Hint: The refractive index of a material is a dimensionless quantity that determines the ratio of the speed of light in vacuum to the speed of light in medium. A prism is a transparent medium with two polished and one grounded faces. The angle between the polished faces is called the refracting angle.

Formula used: $\mu ={\displaystyle \frac{\sin {\displaystyle \frac{A+D}{2}}}{\sin {\displaystyle \frac{A}{2}}}}$

Complete step by step answer:
The refractive index of the prism is given by the relation
$\mu ={\displaystyle \frac{\sin {\displaystyle \frac{A+D}{2}}}{\sin {\displaystyle \frac{A}{2}}}}$
Where, μ represents the refractive index
     D is the angle of minimum deviation
     A is the angle of prism
Angle of prism is the angle between two lateral faces of the prism. When a light ray is passed through the prism, it makes an emergent ray and bends at an angle to the direction of incident light rays. This angle is known as the angle of deviation of the prism.

The smallest angle through which a light is bent by the element is known as minimum deviation. It is denoted by ‘D’. In prism, the deviation angle is minimum if the incident rays form equal angles with the prism faces. It is dependent upon the material of the prism, the angle and wavelength of incident rays.
The given angle of minimum deviation is ${30}^{\circ }$ and the given angle of prism is ${60}^{\circ }$
$\mu ={\displaystyle \frac{\sin {\displaystyle \frac{A+D}{2}}}{\sin {\displaystyle \frac{A}{2}}}}$
Lets substitute the given values in the above formula.
$\mu ={\displaystyle \frac{\sin \left({\displaystyle \frac{60+30}{2}}\right)}{\sin {\displaystyle \frac{60}{2}}}}$
$\mu ={\displaystyle \frac{\sin \left({\displaystyle \frac{90}{2}}\right)}{\sin 30}}$
$\mu ={\displaystyle \frac{\sin 45}{\sin 30}}$
We know, $\sin {45}^{\circ }={\displaystyle \frac{1}{\sqrt{2}}}$ and $\sin {30}^{\circ }={\displaystyle \frac{1}{2}}$
Therefore, $\mu ={\displaystyle \frac{1}{\sqrt{2}}}\times {\displaystyle \frac{2}{1}}$
$\mu =\sqrt{2}$
The correct answer for the given question is option (A).

Note: The refractive index formula is given by applying the angle of incidence and angle of emergent ray in the Snell’s rule. Snell’s law is usually used to describe the relationship between the angles of incidence and refraction between two isotropic media.