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A ray of light refracts from medium 1 into a thin layer of medium 2, crosses the layer and is incident at the critical angle on the interface between the medium 2 and 3 shown in the figure. If the angle of incidence of ray is $θ,$ the value of $θ$ is

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$\sin^{- 1} (\frac{13}{18})$

$\sin^{- 1} (\frac{13}{16})$

$\sin^{- 1} (\frac{8}{9})$

$\sin^{- 1} (\frac{8}{13})$

answer is C.

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Detailed Solution

$\frac{μ_{r}}{μ_{d}} = \frac{μ_{3}}{μ_{2}} = \sin c;$

$\frac{μ_{2}}{μ_{1}} = \frac{\sin θ}{\sin c}$

$\Rightarrow μ_{1} \sin θ = μ_{2} \sin c = μ_{3} \sin 90$

$\Rightarrow 1.6 \sin θ = 1.8 \sin c = 1.3$

$\Rightarrow \sin θ = \frac{13}{16}$

$\Rightarrow θ = \sin^{- 1} (\frac{13}{16})$

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