Circularly Polarized Light through a Quarter-Wave Plate

Question

I'm trying to understand what happens to the intensity and polarization of light that passes through a quarter-wave plate. I believe I know what happens when unpolarized light and linearly polarized light pass through. However, I'm not sure what happens when circularly polarized light passes through.

According to a lecture I'm looking at from my university, the light should emerge linearly polarized. Unfortunately, there is no explanation as to why (I would expect it to be still circularly polarized) or how the intensity is changed (if at all).

I suspect it has something to do with the shift that occurs when the light passes through the plate, but I'm not sure. Note that this is all in the context of monochromatic waves.

Answer 1

A waveplate is a birefringent crystal. Birefringence is a particular kind of anisotropy where the refractive index depends on the plane of polarization of the light: there are two, orthogonal linear polarization planes which have a relatively lower ("fast axis") and higher ("slow axis") refractive index. An waveplate of angle $\phi$ is of such a thickness that the phase delay differs for these two polarization states by $\phi$ as light passes through the crystal.

Therefore, the way to analyze the action of such a crystal is to translate the above into equations. We represent a general polarization state by a $2\times2$ vector:

$$\left(\begin{array}{c}\alpha\,\cos(\omega\,t+\delta)\\\beta\,\cos(\omega\,t+\epsilon)\end{array}\right)$$

where $\alpha,\,\beta,\,\delta,\,\epsilon$ encode the magnitude and phase of the component of the electric field vector along the fast and slow axes.

In a circular polarization state, $\beta = \alpha = 1$ and $\epsilon = \delta\pm \pi/2$, which means that the fast polarization state leads/ lags the slow by $\pi/2$. This in turn means that if the fast polarization amplitude oscillates as $\cos\omega\,t$, the slow oscillates by $\sin\omega\,t$. It's not hard to show that the head of the vector:

$$\left(\begin{array}{c}\cos(\omega\,t)\\\cos(\omega\,t\pm\frac{\pi}{2})\end{array}\right)=\left(\begin{array}{c}\cos(\omega\,t)\\\mp\sin(\omega\,t)\end{array}\right)$$

undergoes uniform circular motion and that this is therefore circular polarization.

If such a state is input to a quarter wave plate, then the plate imparts a further $\pi/2$ phase delay between the states. So now our state becomes

$$\left(\begin{array}{c}\cos(\omega\,t)\\\cos(\omega\,t\pm\frac{\pi}{2}\pm\frac{\pi}{2})\end{array}\right)=\left(\begin{array}{c}\cos(\omega\,t)\\\pm\cos(\omega\,t)\end{array}\right)$$

and the fast and slow components are now either in or exactly out of phase. It's not hard to see that the head of this vector undergoes simple harmonic rectillinear motion, and that therefore we're now dealing with linear polarization.

Answer 2

Circularly polarised light can be thought of as being produced by two simple harmonic motions of the same frequency and amplitude oscillating at right angles to one another with a phase difference of $90^\circ$.

The introduction of a quarter wave plate changes the phase of one of the simple harmonic motions by $90^\circ$ so that the phase difference between the two simple harmonic motions becomes either $0^\circ$ or $180^\circ$ and this is linear polarisation.

There is a nice simulation of the addition of two shms to be found at this website.

The simulation of the addition of shms with a phase difference of $90^\circ$ yields:

And for phase difference of $0^\circ$ and $180^\circ$: