A ray of light is incident from air on a glass plate having thickness $\sqrt{3} cm$ and refractive index $\sqrt{2}$ . The angle of incidence of a ray is equal to the critical angle for glass-air interface. The lateral displacement of the ray when it passes through the plate is $. . . \times 10^{- 2} cm$ (given $\sin 15 ° = 0.26$ )

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answer is 52.

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Detailed Solution

$\sin i = \sqrt{2} sinr$
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$\begin{array}{l} \sin r = \frac{1}{2} \Rightarrow r = 30^{0} \\ μ = \frac{1}{\sin c} \\ c = 45^{\circ} \\ Lateral displacement, x = \frac{tsin (i - r)}{\cos r} \\ x = \frac{\sqrt{3} \sin (45 - 30)}{\cos 30^{0}} \\ = \frac{\sqrt{3} \times \sin 15^{\circ}}{(\sqrt{3} / 2)} = 2 \times 0.26 = 0.52 cm \\ = 52 \times 10^{- 2} cm \end{array}$