A parallel beam of light is incident from air | vedclass

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(A) Applying Snell's law at surface $PQ$:$1 	imes \sin \alpha = n 	imes \sin \beta$Given $\alpha = 45^{\circ}$ and $n = \sqrt{2}$,we have:$\sin 45^{

Vedclass · Accepted Answer

$15$

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(A) Applying Snell's law at surface $PQ$:$1 	imes \sin \alpha = n 	imes \sin \beta$Given $\alpha = 45^{\circ}$ and $n = \sqrt{2}$,we have:$\sin 45^{\circ} = \sqrt{2} \sin \beta$$\frac{1}{\sqrt{2}} = \sqrt{2} \sin \beta \implies \sin \beta = \frac{1}{2} \implies \beta = 30^{\circ}$.In the triangle formed by the light path,the angle of incidence at face $PR$ is $\gamma = 90^{\circ} - (	heta + \beta)$.For total internal reflection at face $PR$,the angle of incidence $\gamma$ must be equal to the critical angle $C$ for the minimum $\alpha$.The critical angle $C$ is given by $\sin C = \frac{1}{n} = \frac{1}{\sqrt{2}}$,so $C = 45^{\circ}$.Thus,$\gamma = 45^{\circ}$.Substituting $\gamma = 90^{\circ} - (	heta + \beta)$:$45^{\circ} = 90^{\circ} - (	heta + 30^{\circ})$$45^{\circ} = 60^{\circ} - 	heta$$	heta = 60^{\circ} - 45^{\circ} = 15^{\circ}$.Therefore,the correct option is $A$.

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