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JEE Main 20244 AprilEvening Shift
Exam
JEE Main
Year
2024
Date
4 April
Shift
Evening Shift
Subject
Physics
Chapter
Ray Optics

JEE Main 2024 · 4 April · Evening Shift · Physics · Ray Optics

A light ray is incident on a glass slab of thickness 4 3 ~cm and...

A light ray is incident on a glass slab of thickness 43 cm4 \sqrt{3} \mathrm{~cm} and refractive index 2\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 _____ cm\mathrm{cm}. \left(\right.Givensin⁡15∘=0.25)Given \left.\sin 15^{\circ}=0.25\right)A light ray is incident on a glass slab of thickness 4 3 ~cm and refractive index 2 . The angle of incidence is equal

Correct answer

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Step-by-step solution

A light ray is incident on a glass slab of thickness 4 3 ~cm and refractive index 2 . The angle of incidence is equal
i=θc⇒i=sin⁡−1(1μ)⇒i=45∘\begin{aligned} & i=\theta_c \\ & \Rightarrow i=\sin ^{-1}\left(\frac{1}{\mu}\right) \\ & \Rightarrow i=45^{\circ} \end{aligned} and according to snell's law 1sin⁡45∘=2sin⁡r⇒r=30∘\begin{aligned} & 1 \sin 45^{\circ}=\sqrt{2} \sin \mathrm{r} \\ & \Rightarrow \mathrm{r}=30^{\circ} \end{aligned}
Lateral displacement Δ=tsin⁡(i−r)cos⁡r\Delta=\frac{\mathrm{t} \sin (\mathrm{i}-\mathrm{r})}{\cos \mathrm{r}}⇒Δ=43×sin⁡15∘cos⁡30∘⇒Δ=2 cm\begin{aligned} & \Rightarrow \Delta=\frac{4 \sqrt{3} \times \sin 15^{\circ}}{\cos 30^{\circ}} \\ & \Rightarrow \Delta=2 \mathrm{~cm} \end{aligned}$$
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