Answer
See answers.
Work Step by Step
The first case is Brewster’s angle (equation 24–6a) for light passing from water to air.
$$tan \theta_p=\frac{n_{air}}{n_{water}}$$
$$\theta_p=tan^{-1}(\frac{n_{air}}{n_{water}})=tan^{-1}(\frac{1.00}{1.33})=36.9^{\circ}$$
Next, we compare the angle to the critical angle for total internal reflection. The light must again be starting in water, moving into air. Use equation 23–6.
$$sin\theta_C=\frac{n_{air}}{n_{water}} $$
$$\theta_C=sin^{-1}(\frac{n_{air}}{n_{water}})=sin^{-1}(\frac{1.00}{1.33})=48.8^{\circ}$$
The final case is Brewster’s angle (equation 24–6a) for light passing from air to water.
$$tan \theta_p=\frac{n_{water}}{n_{air}}$$
$$\theta_p=tan^{-1}(\frac{n_{water}}{n_{air}})=tan^{-1}(\frac{1.33}{1.00})=53.1^{\circ}$$
The two Brewster’s angles are complementary.
