If \(\theta_{b}\) is Brewster's angle for light reflected from the top of an interface between two substances, and \(\theta_{\mathrm{b}}^{\prime}\) is Brewster's angle for light reflected from below, prove that \(\theta_{\mathrm{b}}+\theta_{\mathrm{b}}^{\prime}=90.0^{\circ}\).

From Brewster's Law, we have the relation: \( \tan(\theta_{b}) = \frac{n_2}{n_1} \) for the light reflected from the top of the interface, and \( \tan(\theta_{b}^{\prime}) = \frac{n_1}{n_2} \) for the light reflected from below. Where \( n_1 \) and \( n_2 \) are the refractive indexes of the two substances.

We want to show that \( \theta_{b}+\theta_{b}^{\prime} = 90.0^{\circ} \). First, let's take the tangent of the sum of the two angles: \( \tan(\theta_{b} + \theta_{b}^{\prime}) \) Using the tangent addition formula, we get: \( \tan(\theta_{b} + \theta_{b}^{\prime}) = \frac{\tan(\theta_{b}) + \tan(\theta_{b}^{\prime})}{1 - \tan(\theta_{b})\tan(\theta_{b}^{\prime})} \) Now, substitute the values we found from step 1: \( \frac{\frac{n_2}{n_1} + \frac{n_1}{n_2}}{1 - \frac{n_2}{n_1}\frac{n_1}{n_2}} \) Simplify the expression: \( \frac{\frac{n_2^2 + n_1^2}{n_1 n_2}}{1 - \frac{n_1^2 n_2^2}{(n_1 n_2)^2}} \) After further simplification, we get: \( \frac{n_2^2 + n_1^2}{n_1 n_2} \cdot \frac{(n_1 n_2)^2}{n_1^2 n_2^2 - n_1^2 n_2^2} \) This simplifies to: \( \frac{n_2^2 + n_1^2}{n_1 n_2 - n_1 n_2} \) Finally, we get: \( \tan(\theta_{b} + \theta_{b}^{\prime}) = \frac{n_2^2 + n_1^2}{0} \) Since the denominator is 0, this means that the tangent of the sum of the two angles is infinite or undefined. This occurs when the angle is equal to 90º. Therefore, we can conclude that \( \theta_{b}+\theta_{b}^{\prime}=90.0^{\circ} \).