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Question: The refractive angle of the prism is \(\theta \) and the refractive index of the material of the pri...

The refractive angle of the prism is θ\theta and the refractive index of the material of the prism is cotθ2\cot \dfrac{\theta }{2}. The angle of minimum deviation is:
A. 1802θ180^\circ - 2\theta
B. 90θ90^\circ - \theta
C. 180+2θ180^\circ + 2\theta
D. 1803θ180^\circ - 3\theta

Explanation

Solution

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 = μ=cotθ2\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,
μ=sin(m+θ2)sin(θ2)\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(θ2)=sin(m+θ2)sin(θ2)\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(θ2)=cos(θ2)sin(θ2)\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-
sin(m+θ2)sin(θ2)=cos(θ2)sin(θ2)\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(m+θ2)=cos(θ2)\sin \left( {\dfrac{{\partial m + \theta }}{2}} \right) = \cos \left( {\dfrac{\theta }{2}} \right)…………………………………………………………………… (V)
But we know,
cosθ2=sin(π2θ2)\cos \dfrac{\theta }{2} = \sin \left( {\dfrac{\pi }{2} - \dfrac{\theta }{2}} \right) ……………………………………………………………………… (VI)
Hence from equation (V) and (VI) we can write:
sin(m+θ2)=sin(π2θ2)\sin \left( {\dfrac{{\partial m + \theta }}{2}} \right) = \sin \left( {\dfrac{\pi }{2} - \dfrac{\theta }{2}} \right)
Hence,
m+θ2=π2θ2\dfrac{{\partial m + \theta }}{2} = \dfrac{\pi }{2} - \dfrac{\theta }{2}
m+θ=πθ\partial m + \theta = \pi - \theta
m=π2θ\partial m = \pi - 2\theta
We know that value of π\pi has value 180180^\circ
m=1802θ\partial m = 180^\circ - 2\theta
Hence, the angle of minimum deviation = m=1802θ\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.