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(C) The formula for lateral displacement $x$ through a glass slab is given by:$x = \frac{t \sin(i - r)}{\cos r}$Given: thickness $t = 15 \, cm$,angle

Vedclass · Accepted Answer

$1.732$

Vedclass · Answer

(C) The formula for lateral displacement $x$ through a glass slab is given by:$x = \frac{t \sin(i - r)}{\cos r}$Given: thickness $t = 15 \, cm$,angle of incidence $i = 60^{\circ}$,and lateral displacement $x = 5\sqrt{3} \, cm$.Substituting the values:$5\sqrt{3} = \frac{15 \sin(60^{\circ} - r)}{\cos r}$$\frac{5\sqrt{3}}{15} = \frac{\sin 60^{\circ} \cos r - \cos 60^{\circ} \sin r}{\cos r}$$\frac{\sqrt{3}}{3} = \sin 60^{\circ} - \cos 60^{\circ} 	an r$$\frac{1}{\sqrt{3}} = \frac{\sqrt{3}}{2} - \frac{1}{2} 	an r$$\frac{1}{2} 	an r = \frac{\sqrt{3}}{2} - \frac{1}{\sqrt{3}} = \frac{3 - 2}{2\sqrt{3}} = \frac{1}{2\sqrt{3}}$$	an r = \frac{1}{\sqrt{3}} \implies r = 30^{\circ}$.Now,the refractive index $\mu$ is given by Snell's law:$\mu = \frac{\sin i}{\sin r} = \frac{\sin 60^{\circ}}{\sin 30^{\circ}} = \frac{\sqrt{3}/2}{1/2} = \sqrt{3} \approx 1.732$.

$A$ ray of light is incident on one face of a transparent slab of thickness $15 \, cm$. The angle of incidence is $60^{\circ}$. If the lateral displacement of the ray on emerging from the parallel plane is $5\sqrt{3} \, cm$,the refractive index of the material of the slab is:

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