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Question

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

The correct answer is

is plane polarized perpendicular to the plane of incidence

Analyzing Light Polarization at an Interface

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.

Calculating Refractive Index Ratio

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} $

Finding Brewster's Angle

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 $

Determining Polarization of Reflected Light

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:

  • Unpolarized incident light is resolved into two components: one parallel and one perpendicular to the plane of incidence.
  • The reflected light consists only of the component perpendicular to the plane of incidence.
  • The component parallel to the plane of incidence is entirely transmitted into the second medium.

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.

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Important Questions from Reflection and Refraction

  1. 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

  2. An electromagnetic wave is incident from vacuum normally on a planar surface of a non-magnetic medium. If the amplitude of the electric field of the incident wave is $E_0$ and that of the transmitted wave is $2E_0/3$, then neglecting any loss, the refractive index of the medium is
  3. Suppose the $yz$-plane forms a chargeless boundary between two media of permittivities $\epsilon_{\text{left}}$ and $\epsilon_{\text{right}}$ where $\epsilon_{\text{left}} : \epsilon_{\text{right}} = 1 : 2$. If the uniform electric field on the left is $\vec{E}_{\text{left}} = c(\hat{i} + \hat{j} + \hat{k})$ (where $c$ is a constant), then the electric field on the right $\vec{E}_{\text{right}}$ is
  4. A leaf appears green in daylight. If this leaf were observed in red light, what colour would it appear to have?
  5. A plane electromagnetic wave from within a dielectric medium (with $\epsilon = 4\epsilon_0$ and $\mu = \mu_0$) is incident on its boundary with air, at $z = 0$. The magnetic field in the medium is $\vec{H} = \hat{j} H_0 \cos(\omega t - kx - k\sqrt{3}z)$, where $\omega$ and $k$ are positive constants. The angles of reflection and refraction are, respectively,
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