Monochromatic Light Ray Incident on Quadrilateral Prism

To determine the correct options, we analyze the path of the monochromatic light ray step-by-step as it traverses the quadrilateral prism $ABCD$.

1. Refraction at Face $AB$

From the given geometry of the prism:

• $\angle A = 90^\circ$
• $\angle B = 60^\circ$
• $\angle C = 135^\circ$
• $\angle D = 75^\circ$
• Refractive index of the prism, $\mu = \sqrt{3}$.

Let face $AD$ lie along the positive $x$-axis and face $AB$ along the positive $y$-axis. The normal to face $AB$ is parallel to the $x$-axis.

The incident ray $OP$ hits face $AB$ at an angle of incidence $i_1 = 60^\circ$. By Snell's Law at face $AB$:
$1 \cdot \sin(i_1) = \mu \sin(r_1)$
$\sin(60^\circ) = \sqrt{3} \sin(r_1)$
$\frac{\sqrt{3}}{2} = \sqrt{3} \sin(r_1) \implies \sin(r_1) = \frac{1}{2} \implies r_1 = 30^\circ$

The unit vector in the direction of the incident ray is:
$\hat{v}_{\text{inc}} = \cos(60^\circ)\hat{i} - \sin(60^\circ)\hat{j} = \frac{1}{2}\hat{i} - \frac{\sqrt{3}}{2}\hat{j}$

The unit vector of the refracted ray inside the prism is:
$\hat{v}_{\text{ref1}} = \cos(30^\circ)\hat{i} - \sin(30^\circ)\hat{j} = \frac{\sqrt{3}}{2}\hat{i} - \frac{1}{2}\hat{j}$

2. Propagation towards Face $CD$

The direction of face $BC$ relative to face $AB$ (which is along $+\hat{j}$) makes an angle of $60^\circ$ downwards, given by the unit vector:
$\hat{u}_{BC} = \sin(60^\circ)\hat{i} - \cos(60^\circ)\hat{j} = \frac{\sqrt{3}}{2}\hat{i} - \frac{1}{2}\hat{j}$

Since $\hat{v}_{\text{ref1}} = \hat{u}_{BC}$, the refracted ray travels parallel to face $BC$. Because point $P$ is near vertex $B$, the ray does not intersect face $BC$ and directly reaches face $CD$.

3. Total Internal Reflection at Face $CD$

The angle of line $CD$ relative to the $x$-axis is $-75^\circ$. Since the ray travels at an angle of $-30^\circ$ with the $x$-axis, the angle between the ray and face $CD$ is:
$\theta = |-30^\circ - (-75^\circ)| = 45^\circ$

Thus, the angle of incidence at face $CD$ is:
$i_{CD} = 90^\circ - 45^\circ = 45^\circ$

The critical angle $\theta_c$ for the prism-air interface is given by:
$\sin(\theta_c) = \frac{1}{\mu} = \frac{1}{\sqrt{3}} \approx 0.577$

Since $\sin(i_{CD}) = \sin(45^\circ) = \frac{1}{\sqrt{2}} \approx 0.707 > \sin(\theta_c)$, we have $i_{CD} > \theta_c$.
Therefore, the ray undergoes Total Internal Reflection (TIR) at face $CD$ (Option A is correct).

4. Reflection and Emergence at Face $AD$

The angle of reflection at face $CD$ is $r_{CD} = 45^\circ$. The reflected ray undergoes a total direction shift of $2 \times 45^\circ = 90^\circ$ downwards/inwards.

The direction vector of the reflected ray inside the prism is:
$\hat{v}_{\text{refl}} = \cos(-120^\circ)\hat{i} + \sin(-120^\circ)\hat{j} = -\frac{1}{2}\hat{i} - \frac{\sqrt{3}}{2}\hat{j}$

This reflected ray travels towards face $AD$ (which lies on $y = 0$). Hence, the ray comes out through face $AD$ (Option B is correct).

The normal to face $AD$ is along $-\hat{j}$. The angle of incidence at face $AD$ is:
$\cos(i_{AD}) = \hat{v}_{\text{refl}} \cdot (-\hat{j}) = \frac{\sqrt{3}}{2} \implies i_{AD} = 30^\circ$

By Snell's Law at face $AD$:
$\mu \sin(i_{AD}) = 1 \cdot \sin(e)$
$\sqrt{3} \sin(30^\circ) = \sin(e) \implies \sin(e) = \frac{\sqrt{3}}{2} \implies e = 60^\circ$

The direction unit vector of the emergent ray in air is:
$\hat{v}_{\text{emergent}} = -\sin(60^\circ)\hat{i} - \cos(60^\circ)\hat{j} = -\frac{\sqrt{3}}{2}\hat{i} - \frac{1}{2}\hat{j}$

5. Angle Between Incident and Emergent Rays

Taking the scalar product of the incident ray vector $\hat{v}_{\text{inc}}$ and the emergent ray vector $\hat{v}_{\text{emergent}}$:
$\hat{v}_{\text{inc}} \cdot \hat{v}_{\text{emergent}} = \left(\frac{1}{2}\right)\left(-\frac{\sqrt{3}}{2}\right) + \left(-\frac{\sqrt{3}}{2}\right)\left(-\frac{1}{2}\right) = 0$

Since the dot product is zero, the angle between the incident ray and the emergent ray is $90^\circ$ (Option C is correct).

Conclusion

The correct options are A, B, and C.