Brewster angle

We can see the point $r_p = 0$

$$-n_1 \cos\theta'' + n_2 \cos\theta = -n_1 \left( \cos\theta'' - \frac{\sin\theta}{\sin\theta'' } \cos\theta \right) =0$$ (114)

$$(\cos\theta'' \sin\theta'' - \cos\theta\sin\theta) = \frac{e^{i\theta'' } + e^{-i\theta'' }}{2} \frac{e^{i\theta'' } -... ...i} - \frac{e^{i\theta} + e^{-i\theta}}{2} \frac{e^{i\theta} - e^{-i\theta}}{2i}$$

$$= \frac{e^{2i\theta'' }-e^{-2i\theta'' }}{4i} - \frac{e^{2i\theta}-e^{-2i\theta}}{4i}$$

$$= \frac{2i\sin(2\theta'' )- 2i\sin(2\theta)}{4i}=0$$

Then

$$\theta'' = \frac{\pi}{2}- \theta$$ (115)

The angle between the reflected light and the transmitted light is orthogonal.

$$n_2 \sin\theta'' = n_2 \sin (\pi/2 - \theta) = n_2\cos\theta = n_1 \sin\theta$$ (116)

$$\theta_{\rm Brewster} = \tan^{-1} \left(\frac{n_2}{n_1}\right)$$ (117)

From air to water reflection, the Brewster angle is 53.10 degree, and from water to air the angle is 36.90 degree.