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Calculate the angle of minimum deviation for a ray which ...

Calculate the angle of minimum deviation for a ray which is refracted through an equiangular prism of refractive index 1.4

A.

60⁰

B.

29⁰

C.

99⁰

D.

90⁰

View answer & explanation

Correct answer is B

From the relation refractive index = Sin $\left(\frac{A+Dm}{2}\right)$

Where A = refractive angle = 60⁰, Sin $\left(\frac{A}{2}\right)$

$\implies 1.4 = \frac{\text{Sin}\left(\frac{60^0 + Dm}{2}\right)}{\text{Sin}\left(\frac{60}{2}\right)} \\ = \frac{\text{Sin}\left(\frac{60 + Dm}{2}\right)}{\text{Sin}30^0} \\ = \frac{\text{Sin}\left(\frac{60 + Dm}{2}\right)}{0.5}\\ \text{Therefore} 1.4 \times 0.5 = \text{Sin}\left(\frac{60 + Dm}{2}\right) \\ \text{Therefore} \left(\frac{60 + D_m}{2}\right)= \text{Sin}^{-1} 0.7000 \\ = 44^0 12^1 \\ 60^0 + D_m = 88^0 24^1 \\ D_m = 88^0 24^1 - 60^0 \\ = 28^0 24 \\ 29^0$

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