Question

The refractive angle of the prism is $\theta $ and the refractive index of the material of the prism is $\cot \dfrac{\theta }{2}$. The angle of minimum deviation is:
A. $180^\circ - 2\theta $
B. $90^\circ - \theta $
C. $180^\circ + 2\theta $
D. $180^\circ - 3\theta $

Answer 1

Hint: The angle of minimum deviation is the smallest angle through which light is bent by an optical element or system and the angle between a refracted ray and the perpendicular drawn at the point of incidence of the ray to the interface at which refraction occurs is the refractive angle.

Complete step by step answer:
Given:
Refractive angle of prism = $\mu = \cot \dfrac{\theta }{2}$…………………………………………………… (I)
The angle between a refracted ray and the perpendicular drawn at the point of incidence of the ray to the interface at which refraction occurs is the refractive angle.
The angle of minimum deviation is the smallest angle through which light is bent by an optical element or system.
Angle of deviation is a minimum if the incident and exciting rays form equal angles with the prism faces.
Angle of deviation is denoted as:
We know,
\[\mu = \dfrac{{\sin \left( {\dfrac{{\partial m + \theta }}{2}} \right)}}{{\sin \left( {\dfrac{\theta }{2}} \right)}}\]………………………………………………………………………....... (II)
Now from equation (I) and (II) we can write,
\[\cot \left( {\dfrac{\theta }{2}} \right) = \dfrac{{\sin \left( {\dfrac{{\partial m + \theta }}{2}} \right)}}{{\sin \left( {\dfrac{\theta }{2}} \right)}}\]…………………………………………………………………… (III)
But we know that,
\[\cot \left( {\dfrac{\theta }{2}} \right) = \dfrac{{\cos \left( {\dfrac{\theta }{2}} \right)}}{{\sin \left( {\dfrac{\theta }{2}} \right)}}\]………………………………………………………………………... (IV)
Now from equation (III) and (IV) we can write-
\[\dfrac{{\sin \left( {\dfrac{{\partial m + \theta }}{2}} \right)}}{{\sin \left( {\dfrac{\theta }{2}} \right)}} = \dfrac{{\cos \left( {\dfrac{\theta }{2}} \right)}}{{\sin \left( {\dfrac{\theta }{2}} \right)}}\]
The dividend of both terms are same so we can cancel both and can write-
\[\sin \left( {\dfrac{{\partial m + \theta }}{2}} \right) = \cos \left( {\dfrac{\theta }{2}} \right)\]…………………………………………………………………… (V)
But we know,
$\cos \dfrac{\theta }{2} = \sin \left( {\dfrac{\pi }{2} - \dfrac{\theta }{2}} \right)$ ……………………………………………………………………… (VI)
Hence from equation (V) and (VI) we can write:
$\sin \left( {\dfrac{{\partial m + \theta }}{2}} \right) = \sin \left( {\dfrac{\pi }{2} - \dfrac{\theta }{2}} \right)$
Hence,
$\dfrac{{\partial m + \theta }}{2} = \dfrac{\pi }{2} - \dfrac{\theta }{2}$
$\partial m + \theta = \pi - \theta $
$\partial m = \pi - 2\theta $
We know that value of $\pi $ has value $180^\circ $
$\partial m = 180^\circ - 2\theta $
Hence, the angle of minimum deviation = $\partial m = 180^\circ - 2\theta $

Therefore option (A) is the correct answer.

Note: The angle between a refracted ray and the perpendicular drawn at the point of incidence of the ray to the interface at which refraction occurs is the refractive angle. The point is significant comparative with crystal spectroscopes since it tends to be effortlessly decided.