Chapter 7.3, Problem 118AYU

To determine

Determine the Brewster angle for a light beam travelling through water.

Answer to Problem 118AYU

  $\boxed{{\theta }_B=49°}$

Explanation of Solution

Given information:

If the angle of incidence and angle of refraction are complementary angles, the angle of incidence is referred to as the Brewster angle ${\theta }_B$ . The Brewster angle is related to the index of refraction of the two media, $n_1$ and $n_2$ by the equation $n_1\sin {\theta }_B=n_2\cos {\theta }_B$ , where, $n_1$ is the index of refraction of the incident medium and $n_2$ is the index of refraction of the refractive medium . Determine the Brewster angle for a light beam travelling through water at $\left(20°C\right)$ that makes an angle of incidence with a smooth, flat slab of crown glass.

Calculation:

Consider the expression,

  $n_1\sin {\theta }_B=n_2\cos {\theta }_B$

Where, $n_1$ is the index of refraction of the incident medium and $n_2$ is the index of refraction of the refractive medium.

  ${\theta }_B$ is the Brewster angle

A light beam travelling through water at $\left(20°C\right)$ that makes an angle of incidence with a smooth , flat slab of crown glass.

The index of refraction of the water is $n_1=1.33$

The index of refraction of the flat slab of crown glass is $n_2=1.52$

Now,

  $\begin{array}{l}\frac{\sin {\theta }_B}{\cos {\theta }_B}=\frac{n_2}{n_1}=\\ \tan {\theta }_B=\frac{1.52}{1.33}\\ \tan {\theta }_B=1.143\\ {\theta }_B={\tan }^{-1}\left(1.143\right)\\ {\theta }_B=49°\end{array}$

Hence, the Brewster angle is $\boxed{{\theta }_B=49°}$