Snell's Law

(A) $$\sin i_1 = \sin e_1 = n_1 \sin \frac{A_1}{2}$$

$$\sin i_2 = \sin e_2 = n_2 \sin \frac{A_2}{2}$$

$$e_1 = i_2 \Rightarrow n_1 \sin \frac{A_1}{2} = n_2 \sin \frac{A_2}{2}$$

(B) $$i_2 = e_2 = e_1$$

$$\sin i_2 = n_2 \sin \frac{A_2}{2}$$

But $$i_1 \neq i_2$$

(C) $$\delta_1 = (\mu_1 - 1) A_1, \delta_2 = (\mu_2 - 1) A_2$$

$$\frac{1}{2} \left( \frac{\delta_1}{\mu_1 - 1} \right) + \frac{1}{2} \left( \frac{\delta_2}{\mu_2 - 1} \right) = \frac{1}{2} \frac{A_1}{2} + \frac{A_2}{2} = \theta$$

(D) $$i_1 = e_1 = i_2$$

$$\sin i_1 = n_1 \sin \frac{A_1}{2}$$

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:

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

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

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