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If the electric field of a plane wave is $\mathbf{E}(z,t)=\hat{x}\,3\cos(\omega t-kz+30^\circ)-\hat{y}\,4\sin(\omega t-kz+45^\circ)$ mV/m, the polarization state of the plane wave is
- left elliptical
- left circular
- right elliptical
- right circular
Correct answer: right elliptical
Solution
The $x$ and $y$ components have unequal amplitudes (3 and 4), so the wave is elliptically polarized. The phase difference is not $\pm 90^\circ$, and the given phase relation corresponds to right-handed rotation, hence right elliptical polarization.
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