The refracting angle of a prism is 'A' and its refractive index is ((A)/(2)). The minimum deviation produced by the prism is

Correct answer: B. (180)^ ^ - 2A The refractive index of a prism is given by the formula: μ = ( ((A + _m)/(2)))/( ((A)/(2))) Given that μ = ((A)/(2)) , we can write: ((A)/(2)) = ( ((A)/(2)))/( ((A)/(2))) Equating the two expressions for μ : ( ((A + _m)/(2)))/( ((A)/(2))) = ( ((A)/(2)))/( ((A)/(2))) ((A + _m)/(2)) = ((A)/(2)) Using the trigonometric identity (θ) = (90^ - θ) , we get: ((A + _m)/(2)) = (90^ - (A)/(2)) Equating the angles: (A + _m)/(2) = 90^ - (A)/(2) A + _m = 180^ - A _m = 180^ - 2A Answer: (180)^ - 2A

MHT CET 202624 AprilEvening Shift

Exam: MHT CET
Year: 2026
Date: 24 April
Shift: Evening Shift
Subject: Physics
Chapter: Ray Optics

MHT CET 2026 · 24 April · Evening Shift · Physics · Ray Optics

The refracting angle of a prism is 'A' and its refractive index is...

The refracting angle of a prism is

'A' and its refractive index is

\cot

\left(\dfrac{A}{2}\right).

The minimum deviation produced by the prism is

Options

A $(90)^{}^{\circ} - \dfrac{3A}{2}$
B $(180)^{}^{\circ} - 2A$
C $(180)^{}^{\circ} - A$
D $(90)^{}^{\circ} - 2A$

Correct answer

B. $(180)^{}^{\circ} - 2A$

Step-by-step solution

The refractive index of a prism is given by the formula:

\mu = \dfrac{\sin\left(\dfrac{A + \delta_m}{2}\right)}{\sin\left(\dfrac{A}{2}\right)}

Given that

\mu = \cot\left(\dfrac{A}{2}\right)

, we can write:

\cot\left(\dfrac{A}{2}\right) = \dfrac{\cos\left(\dfrac{A}{2}\right)}{\sin\left(\dfrac{A}{2}\right)}

Equating the two expressions for

\mu

\dfrac{\sin\left(\dfrac{A + \delta_m}{2}\right)}{\sin\left(\dfrac{A}{2}\right)} = \dfrac{\cos\left(\dfrac{A}{2}\right)}{\sin\left(\dfrac{A}{2}\right)}

\sin\left(\dfrac{A + \delta_m}{2}\right) = \cos\left(\dfrac{A}{2}\right)

Using the trigonometric identity

\cos(\theta) = \sin(90^{\circ} - \theta)

, we get:

\sin\left(\dfrac{A + \delta_m}{2}\right) = \sin\left(90^{\circ} - \dfrac{A}{2}\right)

Equating the angles:

\dfrac{A + \delta_m}{2} = 90^{\circ} - \dfrac{A}{2}

A + \delta_m = 180^{\circ} - A

\delta_m = 180^{\circ} - 2A

Answer:

(180)^\circ - 2A

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