Question

An unpolarized light is incident on the plane interface of air-dielectric medium shown in figure. If the incident angle is equal to Brewster angle, identify the expression representing reflected wave.

• A. \((E_x \hat{i} + E_y \hat{j})\sin (kx - kz - \omega t)\)
• B. \((E_x \hat{i} + E_y \hat{j})\sin (kx + ky - \omega t)\)
• C. \((E_x \hat{j} + E_y \hat{k})\sin (ky + kz - \omega t)\)
• D. \((E_x \hat{i} + E_y \hat{j} + E_z \hat{k})\sin (kx + ky - kz - \omega t)\)

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
According to Brewster's Law, when unpolarized light is incident at the Brewster angle (\(i_p\)), the reflected light is completely plane-polarized. The vibrations of the electric field in the reflected wave are perpendicular to the plane of incidence.

Step 2: Key Formula or Approach:
1. Brewster's Law: \(\tan i_p = \mu\). 2. The reflected ray and refracted ray are perpendicular to each other. 3. The reflected wave must satisfy the wave equation format \(E = E_0 \sin(\vec{k} \cdot \vec{r} - \omega t)\).

Step 3: Detailed Explanation:
1. In the geometry of Brewster's angle reflection, if the plane of incidence is the \(xz\)-plane, the reflected light's electric field vector will be restricted to a single direction (usually \(\hat{j}\) if it's perpendicular to the plane). 2. The propagation vector \(\vec{k}\) for reflection changes direction compared to the incident wave. If incident is in the \(+z\) and \(+x\) direction, reflection usually involves a sign change in the \(z\)-component (moving away from the interface). 3. The expression must only have components perpendicular to the direction of propagation. 4. Option (A) represents a wave where the phase \((kx - kz)\) indicates a specific direction of propagation in the \(xz\) plane, consistent with reflection laws where the angle of incidence equals the angle of reflection.

Step 4: Final Answer:
The correct expression representing the reflected wave is Option (A).