Question

The electric field of a monochromatic plane wave travelling in a lossless isotropic and homogenous medium is given by
\[ \vec{E}(z,t)=E_0\left[\hat{x}\cos(\omega t-kz)+\hat{y}\sin(\omega t-kz)\right] \]
in a right-handed orthogonal co-ordinate system.
Which of the following is the correct polarization of the electromagnetic wave?

• A. Right-handed circularly polarized
• B. Left-handed circularly polarized
• C. Linearly polarized
• D. Linearly polarized with \(-45^{\circ}\) angle to \(\hat{x}\)

Hint:

Since \(E_x=E_0\cos\phi\) and \(E_y=E_0\sin\phi\) have equal amplitude and a 90 degree phase difference, the wave is circularly polarized; use the right-hand rule along the propagation direction to fix the handedness.

Answer 1

The Correct Option is A

Step 1: Identify the direction of propagation and the field components.
The phase term \(\omega t-kz\) shows the wave travels in the \(+\hat{z}\) direction. Writing \(\phi=\omega t-kz\), the two transverse components are
\[ E_x=E_0\cos\phi,\qquad E_y=E_0\sin\phi \]
Both components have the same amplitude \(E_0\) and are \(90^{\circ}\) out of phase with each other, which already rules out linear polarization (options C and D), since a linearly polarized wave needs its two components either in phase or exactly out of phase by \(180^{\circ}\), tracing a straight line rather than a circle.

Step 2: Track how the tip of the field vector moves in time at a fixed plane.
Fix \(z=0\) and watch the field as \(t\) increases. At \(\omega t=0\), \((E_x,E_y)=(E_0,0)\), pointing along \(+\hat{x}\). At \(\omega t=90^{\circ}\), \((E_x,E_y)=(0,E_0)\), pointing along \(+\hat{y}\). So as time increases, the tip of \(\vec{E}\) sweeps from \(+\hat{x}\) toward \(+\hat{y}\).

Step 3: Apply the handedness rule for the direction of propagation \(+\hat{z}\).
Using the standard right-hand (corkscrew) convention: point the right thumb along the direction of propagation, \(+\hat{z}\). The fingers of the right hand then curl from \(+\hat{x}\) toward \(+\hat{y}\) as seen by an observer standing further along \(+\hat{z}\) and looking back toward the source, that is, looking in the \(-\hat{z}\) direction. This is exactly the sense of rotation found in Step 2, sweeping from \(+\hat{x}\) to \(+\hat{y}\) as \(t\) grows.

Step 4: Conclude the handedness.
Since the field's rotation matches the right-hand curl direction for propagation along \(+\hat{z}\), the wave is right-handed circularly polarized.

Step 5: Rule out the remaining option.
Left-handed circular polarization would require the field to sweep the opposite way, from \(+\hat{x}\) toward \(-\hat{y}\) as \(t\) increases, which would correspond to a minus sign in front of the \(\sin\phi\) term instead of the plus sign actually given.

Final Answer:
\[ \boxed{\text{Right-handed circularly polarized}} \]