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(A) Given,refractive index of the glass plate,$\mu = \sqrt{3}$.According to Brewster's law,the polarising angle $	heta_p$ is related to the refractiv

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$30$

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(A) Given,refractive index of the glass plate,$\mu = \sqrt{3}$.According to Brewster's law,the polarising angle $	heta_p$ is related to the refractive index by the formula $\mu = 	an 	heta_p$.Substituting the value,$\sqrt{3} = 	an 	heta_p$,which gives $	heta_p = 60^{\circ}$.Since the ray is incident at the polarising angle,the angle of incidence $i = 	heta_p = 60^{\circ}$.Using Snell's law,$\mu = \frac{\sin i}{\sin r}$,we have $\sqrt{3} = \frac{\sin 60^{\circ}}{\sin r}$.Substituting $\sin 60^{\circ} = \frac{\sqrt{3}}{2}$,we get $\sqrt{3} = \frac{\sqrt{3}/2}{\sin r}$.This simplifies to $\sin r = \frac{1}{2}$,which implies $r = 30^{\circ}$.Alternatively,at the polarising angle,the reflected and refracted rays are perpendicular,so $i + r = 90^{\circ}$. Thus,$r = 90^{\circ} - 60^{\circ} = 30^{\circ}$.

$A$ ray of light is incident on the surface of a glass plate of refractive index $\sqrt{3}$ at the polarising angle. The angle of refraction of the ray is (in $^{\circ}$)

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