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Q.

For a prism of refractive index 1.732, the angle of minimum deviation is equal to the angle of the prism. The angle of-the prism is

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a

80°

b

70°

c

60°

d

50°

answer is C.

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Detailed Solution

$\begin{array}{l} μ = \frac{\sin \frac{A + δ_{m}}{2}}{\sin \frac{A}{2}} = \frac{\sin \frac{A + A}{2}}{\sin \frac{A}{2}} = \frac{sinA}{\sin \frac{A}{2}} = \frac{2 \sin \frac{A}{2} \cos \frac{A}{2}}{\sin \frac{A}{2}} = 2 \cos \frac{A}{2} \\ So, \sqrt{3} = 2 \cos \frac{A}{2} \Rightarrow \frac{\sqrt{3}}{2} = \cos \frac{A}{2} \Rightarrow A = 60^{o} \end{array}$

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