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Q.

A ray of light passes from glass having a refractive index of 1.6, to air. The angle of incidence for which the angle of refraction is twice the angle of incidence is

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a

$\sin^{- 1} (\frac{3}{5})$

b

$\sin^{- 1} (\frac{2}{5})$

c

$\sin^{- 1} (\frac{4}{5})$

d

$\sin^{- 1} (\frac{5}{8})$

answer is B.

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Detailed Solution

$\begin{array}{l} _{g} μ_{a} = \frac{\sin i}{\sin r} \Rightarrow \frac{1}{_{a} μ_{g}} = \frac{\sin i}{\sin 2 i} \\ \frac{1}{16} = \frac{\sin i}{2 \sin i \cos i} or \frac{1}{2 \cos i} = \frac{10}{16} = \frac{5}{8} \\ or \frac{1}{\cos i} = \frac{5}{4} or \cos i = \frac{4}{5} \\ Now, \sin i = \sqrt{1 - \cos^{2} i} \\ or \sin r = \sqrt{1 - \frac{16}{265}} = \sqrt{\frac{9}{25}} = \frac{3}{5} \\ or i = \sin^{- 1} (\frac{3}{5}) \end{array}$

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