Question

The electric field of an electromagnetic wave propagating in the positive \(z\)-direction is given by : \[ \vec{E}=\hat{a}_x \sin(\omega t-\beta z)+\hat{a}_y \sin\left(\omega t-\beta z+\frac{\pi}{2}\right) \] The wave is :

• A. Linearly polarized in \(z\)-direction
• B. Elliptically polarized
• C. Left-hand circularly polarized
• D. Right-hand circularly polarized

Hint:

Conditions for circular polarization:
• Equal amplitudes
• Phase difference \(=\pm90^\circ\) Sign of phase difference determines: \[ \text{LHCP or RHCP} \]

Answer 1

The Correct Option is C

Concept: Polarization describes the orientation and rotation of the electric field vector. For circular polarization:
• Orthogonal field components must have equal amplitudes.
• Phase difference must be: \[ \pm 90^\circ \] Given: \[ E_x=\sin(\omega t-\beta z) \] \[ E_y=\sin\left(\omega t-\beta z+\frac{\pi}{2}\right) \] Thus:
• Amplitudes are equal.
• Phase difference is: \[ +90^\circ \] Hence the wave is circularly polarized.

Step 1: Check amplitude equality. The amplitudes of: \[ E_x \] and \[ E_y \] are both unity. Thus: \[ |E_x|=|E_y| \] This satisfies the condition for circular polarization.

Step 2: Check phase difference. Phase of: \[ E_x=(\omega t-\beta z) \] Phase of: \[ E_y=\left(\omega t-\beta z+\frac{\pi}{2}\right) \] Therefore: \[ \Delta\phi=+\frac{\pi}{2} \] which equals: \[ +90^\circ \]

Step 3: Determine polarization type. Equal amplitudes + phase difference of \(90^\circ\) imply: \[ \text{Circular polarization} \] For a wave propagating in \(+z\)-direction:
• \(E_y\) leading \(E_x\) by \(90^\circ\)
• corresponds to left-hand circular polarization Hence: \[ \text{Wave is LHCP} \]

Step 4: Write the final answer. Therefore, the wave is: \[ \boxed{\text{Left-hand circularly polarized}} \] Hence, the correct option is: \[ \boxed{(C)\ \text{Left-hand circularly polarized}} \]