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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?
Options:
Solution:
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Verified Answer
The correct answers are:
The ray gets totally internally reflected at face $C D$
,
The ray comes out through face $A D$
,
The angle between the incident ray and the emergent ray is $90^{\circ}$
The ray gets totally internally reflected at face $C D$
,
The ray comes out through face $A D$
,
The angle between the incident ray and the emergent ray is $90^{\circ}$
$$
\begin{aligned}
& \sqrt{3}=\frac{\sin 60^{\circ}}{\sin r} \\
& \therefore r=30^{\circ}
\end{aligned}
$$
$$
\theta_C=\sin ^{-1}\left(\frac{1}{\sqrt{3}}\right)
$$
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).
\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).
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