Question 84 of 295

A light ray is incident on a glass slab of thickness $4 \sqrt{3} \mathrm{~cm}$ and refractive index $\sqrt{2}$ . The angle of incidence is equal to the critical angle for the glass slab with air. The lateral displacement of ray after passing through glass slab is _____ $\mathrm{cm}$. $\left(\right.$ Given $\left.\sin 15^{\circ}=0.25\right)$

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Answer: 2

$\begin{aligned}
& i=\theta_c \\
& \Rightarrow i=\sin ^{-1}\left(\frac{1}{\mu}\right) \\
& \Rightarrow i=45^{\circ}
\end{aligned}$
and according to snell's law
$\begin{aligned}
& 1 \sin 45^{\circ}=\sqrt{2} \sin \mathrm{r} \\
& \Rightarrow \mathrm{r}=30^{\circ}
\end{aligned}$
Lateral displacement $\Delta=\frac{\mathrm{t} \sin (\mathrm{i}-\mathrm{r})}{\cos \mathrm{r}}$
$\begin{aligned}
& \Rightarrow \Delta=\frac{4 \sqrt{3} \times \sin 15^{\circ}}{\cos 30^{\circ}} \\
& \Rightarrow \Delta=2 \mathrm{~cm}
\end{aligned}$