I know a quarter wave plate inserts a phase difference $ \Delta \phi = \frac{\pi}{2}$ or path difference $\frac{\lambda}{4}$. But how does it convert linearly polarized light into circularly polarized or elliptically polarized light? Also, why does circular polarization become elliptical polarization when the angle between the linear polarization and the quarter wave plate is not $\frac{\pi}{4}$?
[Physics] What does a quarter wave plate actually do
opticspolarization
Best Answer
A quarter wave plate has two perpendicular axes, x and y, each with a different index of refraction. Light polarized along one axis, Ex, comes out with a phase shifted by a quarter wave relative to the phase of light polarized along the other axis, Ey.
Now, any plane polarized light can be decomposed into components, Ex and Ey, polarized along these axes. If the quarter wave plate optical axes are oriented at 45 degrees with respect to the plane polarization of an incoming beam, then the incoming beam has equal components, Ex and Ey, along the optical axes of the quarter wave plate. The phase of one of these components gets shifted by one quarter wave with respect to the phase of the other component.
When you add two equal perpendicular components, Ex and Ey, and there is a phase difference of a quarter wave between the two components, the net vector sum, E, sweeps out a circle. This is circular polarization.
If the quarter wave plate is not oriented at 45 degrees to the polarization of the incoming beam, then the incoming components, Ex and Ey, are not equal. One component is still shifted with respect to the other.
When you add two unequal perpendicular components, Ex and Ey, and there is a phase difference between the two components, the net vector sum, E, sweeps out an ellipse. This is elliptic polarization.