A light ray is incident on a prism of angle A | vedclass

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(A) Since the emergent ray grazes the surface,the angle of emergence $e = 90^{\circ}$.Applying Snell's law at the second surface:$\mu \sin r_{2} = 1 \

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$\sin^{-1} \left( \frac{\sqrt{3} - 1}{2} \right)$

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(A) Since the emergent ray grazes the surface,the angle of emergence $e = 90^{\circ}$.Applying Snell's law at the second surface:$\mu \sin r_{2} = 1 \cdot \sin e$$\sqrt{2} \sin r_{2} = \sin 90^{\circ} = 1$$\sin r_{2} = \frac{1}{\sqrt{2}} \implies r_{2} = 45^{\circ}$.In a prism,the relation between the angles is $r_{1} + r_{2} = A$.Given $A = 60^{\circ}$,we have $r_{1} + 45^{\circ} = 60^{\circ} \implies r_{1} = 15^{\circ}$.Applying Snell's law at the first surface:$\sin i = \mu \sin r_{1}$$\sin i = \sqrt{2} \sin 15^{\circ}$.Using the value $\sin 15^{\circ} = \sin(45^{\circ} - 30^{\circ}) = \sin 45^{\circ} \cos 30^{\circ} - \cos 45^{\circ} \sin 30^{\circ} = \frac{1}{\sqrt{2}} \cdot \frac{\sqrt{3}}{2} - \frac{1}{\sqrt{2}} \cdot \frac{1}{2} = \frac{\sqrt{3} - 1}{2\sqrt{2}}$.Therefore,$\sin i = \sqrt{2} \left( \frac{\sqrt{3} - 1}{2\sqrt{2}} \right) = \frac{\sqrt{3} - 1}{2}$.Thus,$i = \sin^{-1} \left( \frac{\sqrt{3} - 1}{2} \right)$.

$A$ light ray is incident on a prism of angle $A = 60^{\circ}$ and refractive index $\mu = \sqrt{2}$. The angle of incidence at which the emergent ray grazes the surface is given by:

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