A beam of unpolarized light in a medium with dielectric constant $\epsilon_1$ is reflected from a plane interface formed with another medium of dielectric constant $\epsilon_2 = 3\epsilon_1$. The two media have identical magnetic permeability. If the angle of incidence is $60^\circ$, then the reflected light
is plane polarized perpendicular to the plane of incidence
This question examines the polarization of light after reflection from an interface between two different dielectric media. The key factors are the dielectric constants of the media, the magnetic permeability, and the angle of incidence.
The refractive index ($n$) is related to the dielectric constant ($\epsilon$) and magnetic permeability ($\mu$) by the formula $n = \sqrt{\epsilon \mu}$.
Given that the magnetic permeability is identical for both media ($\mu_1 = \mu_2$), the ratio of the refractive indices depends solely on the ratio of the square roots of their dielectric constants.
Let $n_1$ and $n_2$ be the refractive indices of the first and second media, respectively.
We have $\epsilon_2 = 3\epsilon_1$. Therefore:
$ \frac{n_2}{n_1} = \sqrt{\frac{\epsilon_2 \mu_2}{\epsilon_1 \mu_1}} = \sqrt{\frac{3\epsilon_1 \mu_1}{\epsilon_1 \mu_1}} = \sqrt{3} $
Brewster's angle ($\theta_B$) is the angle of incidence where the reflected light is completely polarized in one direction.
For light traveling from medium 1 to medium 2, Brewster's angle is defined by:
$ \tan(\theta_B) = \frac{n_2}{n_1} $
Using the calculated ratio:
$ \tan(\theta_B) = \sqrt{3} $
Solving for $\theta_B$:
$ \theta_B = \arctan(\sqrt{3}) = 60^\circ $
The given angle of incidence ($\theta_i$) is $60^\circ$. This is exactly equal to the Brewster's angle ($\theta_B$) calculated for this interface ($\theta_i = \theta_B = 60^\circ$).
At Brewster's angle:
Therefore, when the angle of incidence is $60^\circ$, which equals the Brewster's angle, the reflected light is plane polarized perpendicular to the plane of incidence.
A monochromatic plane wave is incident normally from a dielectric medium $A$ onto another dielectric medium $B$. The indices of refraction satisfy $n_A < n_B$. One-fourth of the incident energy is reflected back into medium $A$. Let $E$ be the resultant electric field due to the superposition of the incident wave and the reflected wave. Then, the ratio of the two time-averages $\langle \vec{E}^2 \rangle_{\text{min}}/\langle \vec{E}^2 \rangle_{\text{max}}$ is