Question

(a) at what angle of incidence will the light reflected from water (n = 1.33) be completely polarized? (b) does this angle depend on the wavelength of the light? (a) number units (b)

Explanation:

Step1: Use Brewster's law

Brewster's law is given by $\tan\theta_{B}=n_{2}/n_{1}$. Here, assume air as the first - medium with $n_{1} = 1$ and water as the second - medium with $n_{2}=1.33$.

Step2: Calculate the Brewster's angle

We have $\theta_{B}=\arctan(n_{2}/n_{1})$. Substituting $n_{1} = 1$ and $n_{2}=1.33$, we get $\theta_{B}=\arctan(1.33)$.
Using a calculator, $\theta_{B}\approx53.0^{\circ}$

For part (b), the refractive index $n$ of a medium (in this case, water) is a function of the wavelength of light (dispersion). Since $\theta_{B}=\arctan(n_{2}/n_{1})$ and $n_{2}$ (refractive index of water) depends on wavelength, the Brewster's angle $\theta_{B}$ depends on the wavelength of light.

Answer:

(a) $53.0^{\circ}$
(b) Yes, the angle depends on the wavelength of the light.