A ray of light refracts from medium 1 into a thin layer of medium 2, crosses the layer and is incident at…
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 $\theta$, the value of $\theta$ is
$\sin ^{-1}\left(\frac{8}{9}\right)$
$\sin ^{-1}\left(\frac{13}{18}\right)$
$\sin ^{-1}\left(\frac{13}{16}\right)$
$\sin ^{-1}\left(\frac{8}{13}\right)$
Solution
As the light ray incident at critical angle,
$\begin{aligned}
& \quad \sin C=\frac{\mu_3}{\mu_2}=\frac{1.3}{1.8} \text { and } C=r \\
& \therefore \quad \sin r=\frac{1.3}{1.8} \\
& \Rightarrow \quad \sin \theta=\frac{1.8}{1.6} \times \frac{1.3}{1.8}=\frac{13}{16} \\
& \theta=\sin ^{-1}\left(\frac{13}{16}\right)
\end{aligned}$
