As shown in the figure, a ray $$AB$$ of unpolarized light enters from water of refractive index $$n_w=4/3$$ into a medium of refractive index $$n_p=4/\sqrt{3}$$ 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:

Using Snell's law across parallel layers from initial water medium to the final polarizing interface:

$$n_w \sin i = n_w \sin r_{\text{water layer}}$$

Let $$r_w$$ be the angle of incidence at the final interface between the second water layer ($$n_w$$) and the upper medium ($$n_p$$). By geometry:

$$n_w \sin i = n_w \sin r_w$$

Using Brewster's law for complete polarization at the final $$n_w \to n_p$$ reflection interface:

$$\tan r_w = \frac{n_p}{n_w}$$

$$\tan r_w = \frac{4/\sqrt{3}}{4/3} = \frac{3}{\sqrt{3}} = \sqrt{3} \implies r_w = 60^\circ$$

$$\sin i = \sin r_w \implies i = r_w = 60^\circ$$