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Thanks to Kaveh's thesis, this has been increadibly useful when you cannot rotate your laser and you need to rotate the polarization of your laser for overhead/side reference lighting to achieve brewsters.

That's a great way, but can be a bit 'lossy' if the mirrors are inneficient.

If you can get hold of a halfwave plate designed for the wavelength of light you're using that's a much better method. Plus you get the added bonus of being able to set the polarisation at any arbitary angle by simply rotating the waveplate

Probably not as cheap as two mirrors though

Jem

Jem wrote:That's a great way, but can be a bit 'lossy' if the mirrors are inneficient.

Not just lossy, it's also very difficult to align. The second mirror amplifies the error on the first mirror since the angle of refelction is twice the angle of the mirrors. So, if the first mirror is off by delta(theta), then the beam from the second will be off by 4*delta(theta) and the error in the polarisation will be off by cos(4*delta(theta)). This will give a reduction in efficiency of cos^2(4*delta(theta)). This is not very important in display holography, since the object depolarises anyway, but it becomes important for HOEs.

Correct, much cheaper.
Correct, not for HOE's but fine for display.

Very easy to use, works great.

As I was looking at this method I saw a way to use the same principle to get two vertically polarized beams from a polarizing cube beamsplitter if you don't have an output waveplate and with only one additional mirror. If you position the cube so that the the 90 deg output is traveling out vertically, then the straight through beam will have vertical polarization. A mirror placed just above the beamsplitter to turn the vertical beam to travel horizontally (still 90 deg to the input beam), will leave that beam with vertical polarization as well. I made crude sketch to show the idea:

The beam coming from the top of the cube is s polarised, whereas the beam coming from the mirror above the cube is p polarised. I believe the beam coming from the mirror above the cube will be horizontally polarised, since the s or p state is preserved.

If a coordinate system is set up so that the plane of your drawing is the x-y plane, with the y axis being vertical, then the positive z axis comes out of the plane of the drawing (out of "the paper"). Your drawing shows the beam coming in from the negative z direction towards the positive z axis through the origin, being split by the cube into two beams: one continuing on into the positive z direction and one going along the positive y axis. The beam going along the positive y axis ("up") has the y-z plane as the plane of incidence while the polarisation vector is along the x axis, and so the beam as shown is s polarised. When the beam hits the mirror atop the cube, the beam changes direction to go along the positive x direction. Now the plane of incidence is the x-y plane and the polarisation vector is along the y axis, which means it's p polarised. I think the s polarisation state of the input beam will be preserved and the reflected beam from the mirror above the cube will have a polarisation vector along the z axis, ie horizontally.

Kate is right. It is the exact equivalent of the image John was referring to. Attached is better diagram. Wish I had SketchUp in the old days.

For your info, the original diagram (pre personal computers) was programmed in Fortran, submitted to microfilm (35mm BW film), then printed on photo paper in a darkroom!!

kaveh1000 wrote:Kate is right.

Why?

Since the output beam from the top mirror can be both horizontally and vertically polarised, what principle or law states which one it must be. There are three possibilities:
Both permitted choices occur, in which case, the bam is once more unpolarised
S polarisation is maintained, in which case it's horizontally polarised
P polarisation is created, in which case it's vertically polarised.
What law or equation determines which case occurs?

http://www.directindustry.com/prod/opto ... 96302.html

That's why!

The portion of the beam reflected by the prism is s-polarized with respect to the prism face, but p with respect to the mirror.