As shown in the figure, a ray AB of unpolarized light enters from water of refractive index $n_w=\frac{4}{3}$ into a medium of refractive index $n_p=\frac{4}{\sqrt{3}}$ after passing through a glass plate of refractive index n_g = 1.5 and a layer of water. At a particular incident angle i the reflected ray CD is polarized in the direction as shown in the figure. The value of i (in degrees) is:

Correct Answer: 60

Step-by-Step Explanation:

1. Understanding the Physical Setup from the Figure:
From the given diagram, light travels through parallel media layers:
- Bottom layer: Water of refractive index $n_w=\frac{4}{3}$ where ray AB is incident at an angle i to the normal.
- Middle layer: Glass plate of refractive index $n_g=1.5$.
- Next layer: Water layer of refractive index $n_w=\frac{4}{3}$.
- Top medium: Medium of refractive index $n_p=\frac{4}{\sqrt{3}}$.

2. Polarization by Reflection (Brewster's Law):
The ray reflects from the boundary between the water layer ($n_w=\frac{4}{3}$) and the top medium ($n_p=\frac{4}{\sqrt{3}}$). The reflected light CD is polarized perpendicular to the plane of incidence (indicated by dots along CD). This means reflection takes place at the Brewster angle (polarizing angle $i_p$) at this boundary.

According to Brewster's law, at the polarizing angle $i_p$ for refraction from medium 1 (water) to medium 2 (top medium):

$\tan \left(i_p\right)=\frac{n_p}{n_w}$

Substituting the given refractive indices:

$\tan \left(i_p\right)=\frac{4/\sqrt{3}}{4/3}=\frac{3}{\sqrt{3}}=\sqrt{3}$

Therefore, the angle of incidence at the top water-medium boundary is:

$i_p={\tan }^{-1}\left(\sqrt{3}\right)=60°$

3. Applying Snell's Law across Parallel Interfaces:
Since all boundary interfaces are parallel to each other, by Snell's law, the product of the refractive index of a medium and the sine of the angle made by the ray with the normal remains constant across all parallel layers:

$n_w\cdot \sin \left(i\right)=n_w\cdot \sin \left(i_p\right)$

Dividing both sides by $n_w$:

$\sin \left(i\right)=\sin \left(i_p\right)$

$i=i_p=60°$

Hence, the value of the incident angle i in degrees is 60.