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Question

Physics Question on Ray optics and optical instruments

What is the refractive index of a prism whose angle A=60∘A=60{}^\circ and angle of minimum deviation dm=30∘{{d}_{m}}=30{}^\circ ?

A

2\sqrt{2}

B

sin⁡−1(3){{\sin }^{-1}}(\sqrt{3})

C

tan⁡−1(2){{\tan }^{-1}}(\sqrt{2})

D

tan⁡−1(3){{\tan }^{-1}}(\sqrt{3})

Answer

2\sqrt{2}

Explanation

Solution

Refractive index of prism μ=sin⁡A+δm2sin⁡A/2\mu =\frac{\sin \frac{A+\delta m}{2}}{\sin A/2} =sin⁡60∘+30∘2sin⁡602=\frac{\sin \frac{60{}^\circ +30{}^\circ }{2}}{\sin \frac{60}{2}} =sin⁡45∘sin⁡30∘=1212=2=\frac{\sin 45{}^\circ }{\sin 30{}^\circ }=\frac{\frac{1}{\sqrt{2}}}{\frac{1}{2}}=2