Question 280: Ray Optics

Question

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 ______ $$\mathrm{cm}$$.
(Given $$\sin 15^{\circ}=0.25$$)

Accepted Answer

$$\begin{aligned}
& \mu=\sqrt{2} \
& \sin 	heta_C=\frac{1}{\sqrt{2}} \
& Q_C=45^{\circ} \
& i=Q_C=45^{\circ}
\end{aligned}$$
$$\begin{aligned}
& 	ext { ( } \phi 	ext { ) lateral displacement }=\frac{t \sin (i-r)}{\cos r} \
& \sin 45^{\circ}=\sqrt{2} \sin r \
& \Rightarrow \quad r=30^{\circ} \
& 	herefore \quad d=\frac{4 \sqrt{3} \sin \left(45^{\circ}-30^{\circ}ight)}{\cos 30^{\circ}} \
& =\frac{4 \sqrt{3} 	imes \frac{1}{4}}{\frac{\sqrt{3}}{2}}=2
\end{aligned}$$

Ray Optics

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