Question

A ray $\text{OP}$ of a monochromatic light is incident on the face $\text{AB}$ of prism $\text{ABCD}$ near vertex $\text{B}$ at an incident angle of ${60}^{\circ }$ as shown in figure. If the refractive index of the material of the prism is $\sqrt{3}$, which of the following is $($are$)$ correct?
[Assume, Prism is surrounded by air medium]

• A. The ray gets totally internally reflected at face $\text{CD}$.
• B. The ray comes out through face $\text{AD}$.
• 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 ${120}^{\circ }$.

Answer 1

Answer: (A), (B), (C)

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

Critical angle of glass-air interface,

$C={\sin }^{-1}\left(\frac{1}{\mu }\right)$

From the data given in the question,

$C={\sin }^{-1}\left(\frac{1}{\sqrt{3}}\right)$

The ray diagram for the given problem is as shown below,


Applying Snell's law for the refraction at face $\text{AB}$,

${\mu }_{\text{air}}\sin i={\mu }_{\text{prism}}\sin r$

Substituting the given values we get,

$1\sin {60}^{\circ }=\sqrt{3}\sin r$

$\Rightarrow 1\times \frac{\sqrt{3}}{2}=\sqrt{3}\times \sin r$

$\Rightarrow \sin r=\frac{\sqrt{3}/2}{\sqrt{3}}=\frac{1}{2}$

$\therefore r={30}^{\circ }$

From figure,
$\therefore \angle BPQ={90}^{\circ }+r={90}^{\circ }+{30}^{\circ }={120}^{\circ }$

In a quadrilateral $\text{BPQC}$,

$\therefore \angle PQC={360}^{\circ }-({60}^{\circ }+{120}^{\circ }+{135}^{\circ })={45}^{\circ }$

Hence, refracted ray goes along $\text{PQ}$ and meets the face $\text{CD}$ at angle of incidence $i^{\prime }={45}^{\circ }$.

We know ,
Condition for $\text{TIR}$ : $i^{\prime }>C$

$\therefore {45}^{\circ }>{\sin }^{-1}\frac{1}{\sqrt{3}}$

So, $\frac{1}{\sqrt{2}}>\frac{1}{\sqrt{3}}$

Therefore, the ray $\text{PQ}$ meets the face $\text{CD}$ at an angle of incidence greater than critical angle.

Hence, total internal reflection takes place at face $\text{CD}$ of prism.The ray goes along $\text{QR}$ and meets the face $\text{DA}$ at an angle of incidence $i^{\prime \prime }={30}^{\circ }$ (from ray diagram).

Applying Snell's law at face $\text{AD}$

${\mu }_{\text{prism}}\sin i^{\prime \prime }={\mu }_{\text{air}}\sin r^{\prime }$

$\Rightarrow \sqrt{3}\sin {30}^{\circ }=\sin r^{\prime }$

$\Rightarrow r^{\prime }={60}^{\circ }$

The emergent ray $\text{RT}$ will go making an angle of emergence ${60}^{\circ }$as shown in figure.

The angle of deviation between incident ray $\text{OP}$ and emergent ray $\text{RT}$ will be $-{30}^{\circ }+{90}^{\circ }+{30}^{\circ }={90}^{\circ }.$

Thus, options (a) , (b) and (c) are the correct answers.