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STANDARD Physics Prism MCQ Question

Type: MCQ-conceptual-Medium-Class 12

Angle of minimum deviation δ by a prism (refractive index μ and supposing the angle of prism A to be small) can be given by

A

(μ - 1) A

B

(μ + 1) A

C

δ = sin⁡A+δ2sin⁡A2\frac{\sin \frac{A + \delta}{2}}{\sin \frac{A}{2}}

D

δ = μ−1μ+1A\frac{\mu - 1}{\mu + 1} A

Correct Answer

Option A

Detailed Explanation

For small angles, the angle of minimum deviation δ is approximately given by δ = (μ - 1)A, where μ is the refractive index and A is the angle of the prism.

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