Why does it seem that linearly polarized light does not always stay linearly polarized after reflecting off a mirror or beamsplitters set at different angles? What is this phenomenon?

I have a source of linear polarized light (an Lcd) which I'm reflecting twice - once off a beamsplitter set at 45 degrees to the lcd, and again off of a front-surface mirror set at about 24 degrees from the beamsplittersketch describing this setup

Using a linear polarizing filter set perpendicular to the original polarized light source, I would have expected the light to be totally blocked, but it isn't.( I can get some light blocking by rotating the polarizing filter about 30 to 45degrees.) I checked and the light maintains linear polarization coming off the 45 degree beamsplitter as expected, but the polarization seems to get lost after reflecting off the shallower 24 deg angle of the 2nd mirror. If I rotate the 2nd mirror to be parallel to the beamsplitter polarizing is maintained though.

Is there some ideal angle range of a mirror to preserve linear polarization? Are there maybe other factors at play here? Edit to clarify: lcd native linear polarization is 45 degrees. Polarization should flip 90 degrees each time it is reflected, which it does cleanly after first beamsplitter, but not 2nd mirror. Polarizing filter is at 135 degrees and it doesn't block light as expected after second mirror. If rotate polarizing filter, it will begin to block at 30-45 degrees, though not to expected degree.

  • $\begingroup$ Do you know the direction of polarization for light coming out of the source? For example is it polarized either out of the plane in the above image or is it polarized in the plane of the above image? Or is it polarized at an angle in-between in-plane and out of plane? This will tell us whether the light is s or p polarized or something in-between. Can you measure the power of the beam after each optic? Do you see a loss of power at all? When you say light passes through your polarizing how much light? What fraction of the initial intensity is able to pass through the polarizer? $\endgroup$ – jgerber Apr 9 '19 at 19:07
  • $\begingroup$ It sounds like you are seeing some sort of birefringence of the mirror coatings but that explanation is not quite consistent with the fact that you say the polarization remains unchanged if the 23 deg mirror is removed. An important test is to rotate the final polarizer around 360 degrees and monitor the minimum and peak values of the the transmitted light. This will tell you the "polarization purity" of the beam, or how linearly polarized it is. Hopefully the ratio is very high for light coming of the polarized source. It sounds like you are saying it decreases after the 23 deg mirror. $\endgroup$ – jgerber Apr 9 '19 at 19:09
  • $\begingroup$ If this polarization purity parameter decreases it likely doesn't mean your light is becoming "unpolarized" but that it is becoming circularly polarized. It is well known that dielectric mirror exhibit linear birefringence which can turn linearly polarized light which is a superposition of s and p into the circularly polarized light which is a superposition of s and p. But again, I'm not sure about the angle dependence of this effect and it seems inconsistent with your statement that polarization is preserved after the first 45 deg mirror. $\endgroup$ – jgerber Apr 9 '19 at 19:11
  • $\begingroup$ The light from the source is at 45 degrees linear polarization angle (the standard direction for a TN lcd panel). I expected the light should flip 90 degrees each time it is reflected. It does flip cleanly after the 1st beamsplitter (I can't measure power of transmission but looks it good). After the second mirror, I have the polarization filter iset at 135 degrees, ( which should extinguish the light from lcd, being perpendicular). It doesn't do block at this angle though... I found if I rotate the polarizing filter 30-45 degrees it begins to block, though it is not as strong of an effect. $\endgroup$ – Nickelaus Apr 9 '19 at 19:49
  • $\begingroup$ What wavelength of light are you using? What kind of mirrors are you using? Metal mirrors or dielectric or something else? $\endgroup$ – jgerber Apr 9 '19 at 20:01

Placing as an answer what should be a comment (not enough reputation yet). In which direction is the polarization of incident light?

The "safe" directions are only two, "p" and "s" polarized with respect to the planes of the mirrors; that means, horizontal and vertical.

Edit: now this is becoming an answer.

If your mirror is metallic and the incidence angle is different from zero, the phase shifts for the s- and p-polarized components of the field are different.

Possible solutions:

  • rotate the polarization to be s or p before you enter the beamsplitter-mirror setup (if you have a waveplate that does it)
  • rotate the source by $45^{\circ}$.
  • $\begingroup$ Incident light angle is 45 degrees linear. $\endgroup$ – Nickelaus Apr 9 '19 at 20:10
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    $\begingroup$ Then the explanation of what happens is that the second mirror has different phase shifts for vertical ("s") and horizontal ("p") polarizations (is it perhaps metallic? I think I recall that they do that). What you can do is rotate the polarization to be s or p before you enter the beamsplitter-mirror setup - if you have a waveplate that do it, or rotate the source. There are also dielectric mirrors which have a low difference in phase between s- and p-reflections (I am not sure on whether all dielectric mirrors approximate this very well), but they cost more than metallic mirrors usually. $\endgroup$ – JTS Apr 9 '19 at 20:17
  • $\begingroup$ I think you might be right. From reading some more, sounds like all plate beam splitters have an angle of incidence (AOI) generally of 45 deg. I'm reading that they make dichroic beam splitters which can handle a range of angles. $\endgroup$ – Nickelaus Apr 9 '19 at 21:55
  • $\begingroup$ I learnt that this happens by having the same problem :-) I agree that the beamsplitter may be designed to handle all polarizations. $\endgroup$ – JTS Apr 9 '19 at 21:57
  • $\begingroup$ See also the answer of @S. McGrew on keeping the beam in one plane, even if maybe in this case you are already satisfying that condition. $\endgroup$ – JTS Apr 9 '19 at 22:24

A crucial factor is to keep the beam in one plane. Any time you let it wander out of the plane it can rotate the polarization. You will see this if you simply divert the beam upward, then to the right: the polarization will rotate 90 degrees.


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