A ray of light is incident from air on a glas | vedclass

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(D) The critical angle $c$ for the glass-air interface is given by $\sin c = \frac{1}{\mu} = \frac{1}{\sqrt{2}}$.Thus,$c = 45^{\circ}$.Given that the

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$52$

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(D) The critical angle $c$ for the glass-air interface is given by $\sin c = \frac{1}{\mu} = \frac{1}{\sqrt{2}}$.Thus,$c = 45^{\circ}$.Given that the angle of incidence $i = c = 45^{\circ}$.Using Snell's law at the first interface: $1 \cdot \sin i = \mu \cdot \sin r$,where $r$ is the angle of refraction.$\sin 45^{\circ} = \sqrt{2} \cdot \sin r \implies \frac{1}{\sqrt{2}} = \sqrt{2} \sin r \implies \sin r = \frac{1}{2}$.Thus,$r = 30^{\circ}$.The lateral displacement $x$ is given by the formula: $x = t \frac{\sin(i - r)}{\cos r}$.Substituting the values: $x = \sqrt{3} \cdot \frac{\sin(45^{\circ} - 30^{\circ})}{\cos 30^{\circ}}$.$x = \sqrt{3} \cdot \frac{\sin 15^{\circ}}{\cos 30^{\circ}} = \sqrt{3} \cdot \frac{0.26}{\sqrt{3}/2} = 0.26 \cdot 2 = 0.52 \, cm$.Therefore,$x = 52 	imes 10^{-2} \, cm$.

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