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Q.

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 as shown in the figure. If the angle of incidence of ray is $θ$ , the value of is $θ$ :

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a

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

b

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

c

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

d

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

answer is C.

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Detailed Solution

$\begin{array}{l} \frac{μ_{2}}{μ_{1}} = \frac{\sin i}{\sin r}; \frac{1.8}{1.6} = \frac{\sin θ}{\sin r} or \\ \sin r = \frac{16}{18} \sin θ; r = c; s e c o n d m e d i i u m a n d t h i r d m e d i u m \frac{1.3}{1.8} = \frac{16}{18} \sin θ \\ θ = \sin^{- 1} (\frac{13}{16}) \end{array}$

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