Physics · Geometrical Optics

JEE Main 2024 — 4 April, Shift 2 — Question 55

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

Answer: 2

Numerical answer — enter this value.

Step-by-step solution

$\mathbf{i}=\theta_{\mathrm{C}}$

$\Rightarrow \mathrm{i}=\sin ^{-1}\left(\frac{1}{\mu}\right)$

$\Rightarrow \mathrm{i}=45^{\circ}$

and according to snell's law

$1 \sin 45^{\circ}=\sqrt{2} \sin \mathrm{r}$

$\Rightarrow \mathrm{r}=30^{\circ}$

Lateral displacement

$\Delta=\frac{\mathrm{t} \sin (i-r)}{\cos r}$

$\Rightarrow \Delta=\frac{4 \sqrt{3} \times \sin 15^{\circ}}{\cos 30^{\circ}}$ $\Rightarrow \Delta=2 \mathrm{~cm}$

Answer key and solution verified before publishing.

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Exam: JEE Main 2024
Paper: 4 April, Shift 2
Subject: Physics
Chapter: Geometrical Optics
Topic: Apparent Depth, Glass Slab, Prism and Dispersion