I have $532\,\mathrm{nm}$-laser passing through a quarter-wave plate polarizer.
How could I check whether the light out from the quarter-wave plate is circularly polarized?
[Physics] How to check if a laser’s light is circularly polarized
opticspolarization
Related Solutions
With circularly polarised waves, the effect can only be observed interferometrically, because the Faraday rotation simply becomes the imposition of different phase delays on the two circularly polarised components. To understand this, witness that the Faraday rotation on a general polarisation state expressed with linear polarisation state basis is:
$$\left(\begin{array}{c}x\\y\end{array}\right)\mapsto \left(\begin{array}{cc}\cos\theta&-\sin\theta\\\sin\theta&\cos\theta\end{array}\right)\left(\begin{array}{cc}x\\y\end{array}\right)$$
and then we transform to circular polarisation basis states $f_+,\,f_-$ by:
$$\left(\begin{array}{c}f_+\\f_-\end{array}\right)= \frac{1}{\sqrt{2}}\left(\begin{array}{cc}1&i\\1&-i\end{array}\right)\left(\begin{array}{cc}x\\y\end{array}\right)$$
so that our Faraday rotation through angle $\theta$ above becomes:
$$\left(\begin{array}{c}f_+\\f_-\end{array}\right)\mapsto \left(\begin{array}{cc}1&i\\1&-i\end{array}\right)\left(\begin{array}{cc}\cos\theta&-\sin\theta\\\sin\theta&\cos\theta\end{array}\right)\left(\begin{array}{cc}1&i\\1&-i\end{array}\right)^{-1}\left(\begin{array}{cc}f_+\\f_-\end{array}\right) = \left(\begin{array}{cc}e^{i\,\theta}&0\\0&e^{-i\,\theta}\end{array}\right)\left(\begin{array}{cc}f_+\\f_-\end{array}\right)$$
Indeed, this is the essence of the spin-1 particle: its eigenstates pick up a phase of $e^{i\,\theta}$ on rotation through angle $\theta$ (unlike an electron, which would pick up a phase of $e^{i\,\frac{\theta}{2}}$).
Edit: Your understanding expressed in your question seems to be right. In the above I give the Jones matrix, i.e. the linear transformation wrought on the vector of superposition weights, for the two cases where (1) linear polarisation states and (2) circular polarisation states are used as the state space basis: in the former case, the effect is a rotation of the plane of polarisation through $\theta$ radians, in the latter the right handed circular polarised component's phase is increased by $\theta$ radians, the left hand component's phase is decreased by $\theta$ radians. In all cases, this $\theta$ is given by $\theta = V\, B_\parallel \,L$ where $L$ is the crystal's length, $B_\parallel$ the magnetic induction component along the optical axis and $L$ the crystal's length.
Edit 2: Well, the results for circularly polarised loght follow from: (1) experimental measurements for linearly polarised light and (2) experimental confirmation of the system's linearity; once we have (2) then circular / linear description is simply a change of basis. Moreover, you can do the experiments for circularly polarised light, you're simply looking for changes in phase. So you can do the experiment interferometrically. Or, you can do an experiment as below with waveplates and the rotator.
The rotator works exactly the same with linearly polarised input light whether or not the waveplates are in place, confirming the linearity of the system. So it is acting on the circularly polarised light states exactly as my basis trasnformation equations above foretell. At a deeper level, the waveplate's action as a transformer to and from circularly polarised light can be checked by the following experiment in 1936 by Richard Beth:
The intensity is changed when circularly polarized light passes through a polarizer, though. The light is changed from circularly polarized to linearly polarized. If you draw the evolution of the waves with respect to an axis, any part of the circularly polarized wave that is not aligned with the linear polarizer will have its intensity reduced.
Best Answer
The quarter wave plate doesn't destroy the polarization - it just changes its character. Circularly polarized light has an electrical vector that rotates without changing its magnitude; if you pass the light through a linear polarizing filter, and rotate the filter, the intensity of the observed light should not change.
Contrast this with the original light from the laser pointer: if you do the same experiment, you will see the intensity after the linear polarizer go from "almost zero", to "almost fully transmitted".
In the intermediate stage of elliptical polarization, you would see the intensity change without ever fully being extinguished. This happens when the initial light is not polarized at exactly 45° to the fast/slow axis of the quarter wave plate.