Question

The electric field of a plane electromagnetic wave is \[ E = \mathbf{a_x}\,C_{1x}\cos(\omega t - \beta z) + \mathbf{a_y}\,C_{1y}\cos(\omega t - \beta z + \theta)\ \text{V/m}. \] Which of the following combination(s) will give rise to a left handed elliptically polarized (LHEP) wave?

• A. $C_{1x}=1,\ C_{1y}=1,\ \theta=\pi/4$
• B. $C_{1x}=2,\ C_{1y}=1,\ \theta=\pi/2$
• C. $C_{1x}=1,\ C_{1y}=2,\ \theta=3\pi/2$
• D. $C_{1x}=2,\ C_{1y}=1,\ \theta=3\pi/4$

Hint:

For propagation along \(+z\): \(0 < \theta < \pi \;\Rightarrow\; \text{LHEP},\) and \(-\pi < \theta < 0 \;\Rightarrow\; \text{RHEP}.\) Circular polarization requires \(C_{1x} = C_{1y}\) and \(|\theta| = \pi/2.\)

Answer 1

The Correct Option is A

For a wave propagating in the +z direction, polarization is set by the relative phase of \(E_y\) with respect to \(E_x\).
Represent the field components as \[ E_x = C_{1x}\cos\psi, \quad E_y = C_{1y}\cos(\psi+\theta), \quad \psi = \omega t - \beta z. \]

Step 1: Condition for handedness
- If \(E_y\) leads \(E_x\) (\(0 < \theta < \pi \; \bmod 2\pi\)), the polarization is left–handed (LHEP).
- If \(E_y\) lags \(E_x\) (\(\pi < \theta < 2\pi\)), it is right–handed (RHEP).

Step 2: Condition for ellipticity
- If amplitudes differ (\(C_{1x}\neq C_{1y}\)) or phase difference \(\theta \neq \pi/2\), the polarization is elliptical.
- If amplitudes are equal and \(\theta = \pi/2\), the polarization is circular.

Step 3: Check the options
(A) \(\theta=\pi/4\) (\(0 < \theta < \pi\)) ⇒ left–handed. Amplitudes equal but \(\theta \neq \pi/2\) ⇒ elliptical. ✔

(B) \(\theta=\pi/2\) (\(0 < \theta < \pi\)), \(C_{1x}\neq C_{1y}\) ⇒ left–handed elliptical (not circular). ✔

(C) \(\theta=3\pi/2 \equiv -\pi/2 \ (\bmod 2\pi)\) ⇒ \(E_y\) lags \(E_x\) ⇒ right–handed. ✘

(D) \(\theta=3\pi/4\) (\(0 < \theta < \pi\)) and unequal amplitudes ⇒ left–handed elliptical. ✔

Final Answer:
\[ \boxed{\text{LHEP cases: (A), (B), and (D)}} \]