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Q. Which of the following diagrams represents the variation of electric field vector with time for a circularly polarized light?

AIIMSAIIMS 2006Wave Optics

Solution:

When two plane-polarised waves are superimposed, then under certain conditions, the resultant light vector rotates with a constant magnitude in a plane perpendicular to the direction of propagation. The tip of the vector traces a circle and the light is said to be circularly polarised.
To form circularly polarised light
$E _{ x }= E _{0} \sin \omega t$
$E _{ y }= E _{0} \cos \omega t = E _{0} \sin \left(\omega t +\frac{\pi}{2}\right)$
where $E _{0}$ is amplitude.
Resultant amplitude
$|\vec{E}|^{2}=E_{0}+E_{0}+2 E_{0} \cdot E_{0} \cos \frac{\pi}{2}$
$|\vec{E}|^{2}=E_{0} \sqrt{2}=$ constant