JEE MAIN - Physics (2023 - 25th January Morning Shift - No. 20)
A ray of light is incident from air on a glass plate having thickness $$\sqrt3$$ cm and refractive index $$\sqrt2$$. 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 $$\sin 15^\circ = 0.26$$)
Answer
52
Explanation
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$\Rightarrow$ at point (1)
$\mu \sin r=\sin i=\frac{1}{\sqrt{2}}$
$\sin r=\frac{1}{2} \quad \Rightarrow \quad r=30^{\circ}$
Lateral displacement
$=\frac{t}{\cos r} \sin \left(15^{\circ}\right)=\frac{\sqrt{3}}{\left(\frac{\sqrt{3}}{2}\right)} \times 0.26$
$=2 \times 0.26$
$=0.52 \mathrm{~cm}$ $=52 \times 10^{-2} \mathrm{~cm}$
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