A light ray falls on a glass surface of refra | vedclass

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(B) Method $(i)$:Using Snell's law,$n_1 \sin i = n_2 \sin r$.Given $n_1 = 1$ (air),$n_2 = \sqrt{3}$ (glass),and $i = 60^{\circ}$.$1 \cdot \sin 60^{\ci

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(B) Method $(i)$:Using Snell's law,$n_1 \sin i = n_2 \sin r$.Given $n_1 = 1$ (air),$n_2 = \sqrt{3}$ (glass),and $i = 60^{\circ}$.$1 \cdot \sin 60^{\circ} = \sqrt{3} \cdot \sin r$$\frac{\sqrt{3}}{2} = \sqrt{3} \sin r$$\sin r = \frac{1}{2} \implies r = 30^{\circ}$.The angle between the reflected ray and the normal is $i = 60^{\circ}$.The angle between the refracted ray and the normal is $r = 30^{\circ}$.The angle between the reflected ray and the refracted ray is $180^{\circ} - (i + r) = 180^{\circ} - (60^{\circ} + 30^{\circ}) = 180^{\circ} - 90^{\circ} = 90^{\circ}$.Method $(ii)$:Since the angle of incidence $i = 60^{\circ}$ satisfies the condition for Brewster's angle,where $	an i_p = \mu = \sqrt{3}$,it implies $i_p = 60^{\circ}$.At Brewster's angle,the reflected and refracted rays are perpendicular to each other,so the angle between them is $90^{\circ}$.

$A$ light ray falls on a glass surface of refractive index $\sqrt{3}$,at an angle of $60^{\circ}$. The angle between the refracted and reflected rays would be ....... $^o$.

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