A ray $O P$ of monochromatic light is incident on the face $A B$ of prism $A B C D$ near vertex $B$ at an incident angle of $60^{\circ}$ (see figure). If the refractive index of the material of the prism is $\sqrt{3}$, which of the following is (are) correct?

• Correct answer
  A

  The ray gets totally internally reflected at face $C D$
• Correct answer
  B

  The ray comes out through face $A D$
• Correct answer
  C

  The angle between the incident ray and the emergent ray is $90^{\circ}$
• D

  The angle between the incident ray and the emergent ray is $120^{\circ}$

Solution

$$ \begin{aligned} & \sqrt{3}=\frac{\sin 60^{\circ}}{\sin r} \\ & \therefore r=30^{\circ} \end{aligned} $$
$$ \theta_C=\sin ^{-1}\left(\frac{1}{\sqrt{3}}\right) $$ 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 $C D$. Path of ray of light after point $Q$ is shown in figure. From the figure, we can see that angle between incident ray $O P$ and emergent ray $R S$ is $90^{\circ}$. Therefore, correct options are (a), (b) and (c).

Hence, the correct answer is options: 'A', 'B', 'C'.