Question

Identify the polarisation of the wave given, \(E_x=2\cos\omega t\), \(E_y=2\sin\omega t\) and the phase difference is \(+90^\circ\)

• A. Left hand circularly polarized
• B. Right hand circularly polarized
• C. Left hand elliptically polarized
• D. Right hand elliptically polarized

Hint:

Equal amplitudes + phase difference of \(90^\circ\) \[ \Rightarrow \] Circular polarization.

Answer 1

The Correct Option is A

Concept: For circular polarization:
• Magnitudes of \(E_x\) and \(E_y\) must be equal.
• Phase difference must be \(\pm90^\circ\).

Step 1: Check amplitude condition.
Given \[ E_x=2\cos\omega t \] \[ E_y=2\sin\omega t \] Both components have equal amplitudes. \[ |E_x|=|E_y|=2 \] Hence circular polarization is possible.

Step 2: Check phase difference.
\[ E_y=2\sin\omega t = 2\cos(\omega t-90^\circ) \] Thus \(E_y\) lags \(E_x\) by \(90^\circ\). This corresponds to left-hand circular polarization.

Step 3: Final Answer.
\[ \boxed{\text{Left Hand Circular Polarization}} \] Hence, \[ \boxed{\text{Correct Option (A)}} \]