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A prism of refractive index n and angle A is placed in minimum deviation position. If the angle of minimum deviation is equal to the angle A, then the value of A is

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$\sin^{- 1} (\frac{\sqrt{2}}{2})$

$2 \sin^{- 1} [\sqrt{\frac{4 - n^{2}}{4}}]$

$\sin^{- 1} (\frac{n}{2})$

$2 \sin^{- 1} [\sqrt{\frac{1 - n^{2}}{2}}]$

answer is D.

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Detailed Solution

$n = \frac{\sin (\frac{A + A}{2})}{\sin \frac{A}{2}}$

$\Rightarrow \sin A = n \sin \frac{A}{2}$

$\Rightarrow 2 \sin \frac{A}{2} \cos \frac{A}{2} = n \sin \frac{A}{2}$

$\Rightarrow \cos \frac{A}{2} = \frac{n}{2}$

$\Rightarrow \sin \frac{A}{2} = \sqrt{1 - \frac{n^{2}}{4}}$

$\Rightarrow A = 2 \sin^{- 1} [\sqrt{\frac{4 - n^{2}}{4}}]$

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