A ray of light is incident on a right-angled | vedclass

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(A, C) The critical angle $	heta_{C} = \sin^{-1}\left(\frac{1}{\sqrt{2}}ight) = 45^{\circ}$.Applying Snell's law at the incident surface $S$ $(n_{1

Vedclass · Accepted Answer

The ray emerging at $R$ is parallel to the ray incident at $S$.

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(A, C) The critical angle $	heta_{C} = \sin^{-1}\left(\frac{1}{\sqrt{2}}ight) = 45^{\circ}$.Applying Snell's law at the incident surface $S$ $(n_{1} \sin 	heta_{1} = n_{2} \sin 	heta_{2})$:$1 \cdot \sin 45^{\circ} = \sqrt{2} \cdot \sin 	heta \implies \sin 	heta = \frac{1}{2} \implies 	heta = 30^{\circ}$.At point $P$, the angle of incidence is $90^{\circ} - 15^{\circ} = 75^{\circ}$. Since $75^{\circ} > 45^{\circ}$, total internal reflection $(TIR)$ occurs at $P$.At point $Q$, the angle of incidence is $15^{\circ}$. Since $15^{\circ} The ray emerging at $R$ is parallel to the incident ray at $S$ because the geometry of the prism and the path of the ray result in a net deviation of $0^{\circ}$ (or $360^{\circ}$), meaning they are parallel. Thus, statement $A$ and $C$ are correct.

$A$ ray of light is incident on a right-angled isosceles prism parallel to its base as shown in the figure. The refractive index of the material of the prism is $\sqrt{2}$. Which of the following statement$(s)$ is/are true?

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