Question

If the electric field of a plane wave is

$\overrightarrow{E}(z,t)=\hat{x}3cos(\omega t-kz+{30}^{\circ })$

$-\hat{y}4sin(\omega t-kz+{45}^{\circ })\left(mV/m\right),$

the polarization state of the plane wave is

• A. right elliptical
• B. right circular
• C. left elliptical
• D. left circular

Answer 1

The correct option is C left elliptical

Ans : (a)

$\overrightarrow{E}(z,t)=3\hat{x}cos(\omega t-kz+{30}^{\circ })-4\hat{y}sin(\omega t-kz+{45}^{\circ })$

The plane wave is travelling in +z direction

${\hat{E}}_x=3cos(\omega t-kz+{30}^{\circ })$

${\hat{E}}_y=-4sin(\omega t-kz+{45}^{\circ })$

At z = 0, tracing E locus for various times

At t = 0, $E_x=3cos{30}^{\circ }=3\cdot \frac{1}{2}=\frac{3}{2}=1.5$

$E_y=-4sin{45}^{\circ }=-2\sqrt{2}=-2.83$

At t = $\frac{T}{6}$

$E_x=3cos(\frac{2\pi }{T}\times \frac{T}{6}+{30}^{\circ })=0$

$E_y=-4sin({60}^{\circ }+{45}^{\circ })=-3.86$

This is left elliptical polarized plane-wave.