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 \(\left.\sin 15^{\circ}=0.25\right)\)
Pick an option to check your answer.
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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