Physics - Geometrical Optics Question with Solution | TestHub

$$\begin{array}{rl}& \sqrt{3}=\frac{\sin 60^{\circ }}{\sin r}\\ & \therefore r=30^{\circ }\end{array}$$
$${\theta }_C={\sin }^{-1}\left(\frac{1}{\sqrt{3}}\right)$$ or $\sin {\theta }_C=\frac{1}{\sqrt{3}}=0.577$ At point $Q$, angle of incidence inside the prism is $i=45^{\circ }$. Since $\sin i=\frac{1}{\sqrt{2}}$ is greater than $\sin {\theta }_C=\frac{1}{\sqrt{2}}$, ray gets totally internally reflected at face $CD$. Path of ray of light after point $Q$ is shown in figure. From the figure, we can see that angle between incident ray $OP$ and emergent ray $RS$ is $90^{\circ }$. Therefore, correct options are (a), (b) and (c).