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JEE Mains · Physics · STD 12 - Ray optics and optical instruments
  • JEE Mains 2024
  • Hard
  • Numerical

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 & Solution

Correct Answer

(B) \(2\)

Step-by-step Solution

Detailed explanation

\(i=\theta_c\)

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

\(\Rightarrow i=45^{\circ}\)

\(\text { and according to snell's law }\)

\(1 \sin 45^{\circ}=\sqrt{2} \sin r\)

\(\Rightarrow \mathrm{r}=30^{\circ}\)

\(\text { Lateral displacement } \Delta=\frac{\mathrm{t} \sin (\mathrm{i}-\mathrm{r})}{\cos \mathrm{r}}\)

\(\Rightarrow \Delta=\frac{4 \sqrt{3} \times \sin 15^{\circ}}{\cos 30^{\circ}}\)

\(\Rightarrow \Delta=2 \mathrm{~cm}\)

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