JEE MAIN - Physics (2024 - 4th April Evening Shift - No. 27)
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 $$\sin 15^{\circ}=0.25$$)
Explanation
$$\begin{aligned} & \mu=\sqrt{2} \\ & \sin \theta_C=\frac{1}{\sqrt{2}} \\ & Q_C=45^{\circ} \\ & i=Q_C=45^{\circ} \end{aligned}$$
$$\begin{aligned} & \text { ( } \phi \text { ) lateral displacement }=\frac{t \sin (i-r)}{\cos r} \\ & \sin 45^{\circ}=\sqrt{2} \sin r \\ & \Rightarrow \quad r=30^{\circ} \\ & \therefore \quad d=\frac{4 \sqrt{3} \sin \left(45^{\circ}-30^{\circ}\right)}{\cos 30^{\circ}} \\ & =\frac{4 \sqrt{3} \times \frac{1}{4}}{\frac{\sqrt{3}}{2}}=2 \end{aligned}$$
Comments (0)
