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?
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If the laser is not to strong, grab a pair of 3D glasses and see if the light passes through both glasses more or less equally, make sure they're the kinda modern ones from like the last 5 years or so. These are circularly polarized. To quote Wikipedia, yeah I know, it explains below, it also tells you how to make your own:
To present a stereoscopic motion picture, two images are projected superimposed onto the same screen through circular polarizing filters of opposite handedness. The viewer wears eyeglasses which contain a pair of analyzing filters (circular polarizers mounted in reverse) of opposite handedness. Light that is left-circularly polarized is blocked by the right-handed analyzer, while right-circularly polarized light is blocked by the left-handed analyzer. The result is similar to that of stereoscopic viewing using linearly polarized glasses, except the viewer can tilt his or her head and still maintain left/right separation (although stereoscopic image fusion will be lost due to the mismatch between the eye plane and the original camera plane).
*Quick edit: If it passes equally through both it's of course NOT circular (since one of the two should block the light right/left-handed).
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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.
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Finally found the method to check whether light output from circular polarizer is perfectly circularly polarized or not.
Consider that a polarized light is passed through quarter wave plate which is fixed at certain angle and its output is send to linear polarizer.
now measure the power output from linear polarizer by varying its angle,but keep quarter wave plate fixed and plot a graph (linear polarizer angle of rotation vs power) if the plot is perfect sine wave,the light is perfectly polarized(linear polarizer was used only check circular polarization).if not change the quarter wave plate angle to next and repeat the above method ie,varying linear polarizer angle and measuring power.
I did the experiment by keeping quarter wave plate @ 130,132,135 and varied the linear polarizer and measured the power and from the plot,the perfect circular polarisation can be achieved by keeping quarter wave plate at 135 degree.
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$\begingroup$ That's... some pretty suspect data. If the polarization were truly circular, then that graph would be completely flat as you rotate the linear polarizer; moreover, regardless of the polarization state, the variation here should be completely sinusoidal where your blue curve shows a much funkier variation. There seem to be additional factors in your experiment that you likely need to account for (say, noisy variations in your source's intensity?). $\endgroup$ – Emilio Pisanty Jun 5 '18 at 16:54
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$\begingroup$ More importantly, though, "perfectly circular polarization" isn't an achievable goal ─ instead, you need to focus on an accurate production and characterization of a given polarization, where much of the work is directed to characterizing and presenting your uncertainty in the polarization state of the light (usually via polarization tomography, presented using its Stokes parameters), so that you can say just how close to circular your light is, and deciding what is an acceptable deviation from complete circularity. $\endgroup$ – Emilio Pisanty Jun 5 '18 at 16:57
