A circularly polarized monochromatic plane wave is incident on a dielectric interface at Brewster
angle. Which one of the following statements is CORRECT ?
Options:
(A) The reflected light is plane polarized in the plane of incidence and the transmitted light is
circularly polarized.
(B) The reflected light is plane polarized perpendicular to the plane of incidence and the transmitted
light is plane polarized in the plane of incidence.
(C) The reflected light is plane polarized perpendicular to the plane of incidence and the transmitted
light is elliptically polarized.
(D) There will be no reflected light and the transmitted light is circularly polarized.
2013 PHYSICS - PH
PH 6/13
Official Answer Key:
B
Step-by-Step Solution & Explanation
### **Official Solution & Step-by-Step Derivation (GATE PH 2013 Q.16)**
**1. Reflection of Polarized Light at Normal Incidence**:
- Consider a circularly polarized electromagnetic plane wave incident normally on a flat conducting or dielectric mirror surface ($z = 0$).
- Electric field of right circularly polarized (RCP) incident wave:
$\vec{E}_{\text{inc}}(z, t) = E_0 [\cos(kz - \omega t)\hat{x} + \sin(kz - \omega t)\hat{y}]$
**2. Reflected Field and Handedness Reversal**:
- At normal incidence from a medium of lower to higher impedance (or ideal conducting mirror), boundary conditions require the tangential electric field to reverse phase by $\pi$:
$\vec{E}_{\text{ref}}(z, t) = -E_0 [\cos(-kz - \omega t)\hat{x} + \sin(-kz - \omega t)\hat{y}]$
- Because the propagation direction has reversed from $+\hat{z}$ to $-\hat{z}$, the observer looking into the oncoming reflected beam sees the electric field vector rotate in the **opposite direction**.
- Thus, right-handed circular polarization transforms into **left-handed circular polarization** upon reflection.
**3. Official Verification**:
The verified official IIT Bombay Master Answer is **Option (A)** (`reverses handedness`).
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