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 $\sin 15^{\circ}=0.25$)
Answer (integer)
2
Solution
<p>$$\begin{aligned}
& \mu=\sqrt{2} \\
& \sin \theta_C=\frac{1}{\sqrt{2}} \\
& Q_C=45^{\circ} \\
& i=Q_C=45^{\circ}
\end{aligned}$$</p>
<p>$$\begin{aligned}
& \text { ( } \phi \text { ) lateral displacement }=\frac{t \sin (i-r)}{\cos r} \\
& \sin 45^{\circ}=\sqrt{2} \sin r \\
& \Rightarrow \quad r=30^{\circ} \\
& \therefore \quad d=\frac{4 \sqrt{3} \sin \left(45^{\circ}-30^{\circ}\right)}{\cos 30^{\circ}} \\
& =\frac{4 \sqrt{3} \times \frac{1}{4}}{\frac{\sqrt{3}}{2}}=2
\end{aligned}$$</p>
About this question
Subject: Physics · Chapter: Optics · Topic: Refraction and Snell's Law
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