If the angle of incidence is twice the angle of refraction in a medium of refractive index ‘μ’, then the angle of incidence is

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${2sin}^{- 1} (\frac{μ}{2})$

${2cos}^{- 1} (\frac{μ}{2})$

${sin}^{- 1} (\frac{μ}{2})$

${cos}^{- 1} (\frac{μ}{2})$

answer is C.

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Detailed Solution

$μ = \frac{\sin i}{\sin r}$
i = 2r
r = i/2
$\begin{array}{l} μ = \frac{\sin i}{\sin \frac{i}{2}} = \frac{2 \sin \frac{i}{2} \cos \frac{i}{2}}{\sin \frac{i}{2}} \\ μ = 2 \cos \frac{i}{2} \\ \cos \frac{i}{2} = \frac{μ}{2} \\ \frac{i}{2} = \cos^{- 1} (\frac{μ}{2}) \\ i = 2 \cos^{- 1} (\frac{μ}{2}) \end{array}$

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