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Polarizer Puzzle

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Dr. Shark

Thu Feb 09 2006, 11:01AM

Registered Member #75 Joined: Thu Feb 09 2006, 09:30AM
Location: Montana, USA
Posts: 711

You might have seen this elsewhere, but I find it very intriguing and we can use some content on the new board, so here it goes:

You have an unpolarized light ray going through two crossed polarizer plates, e.g. first a vetical polarizer and then a horizontal one, as (kind of) pictured below.

|||         =
-->---|||---->----=---------> light ray out 
      |||         =
   filter 1    filter 2

The first one will absorb half of the light and leave only the vertically polarized light, so this will get absorbed completely by the second one. Since a polarizer is basically just a selective absorber, it should not matter if we put a third one between the two, at an angle of 45 degrees. After all, we are only "taking away something" in every step, so we should not expect to absorb less by using more absorbers.

Yet, by absorbing half of the light the middle polarizer actually rotates the light that passes through it by 45deg, so the last polarizer will again pass half of the light, resulting in 1/8 of the original light to exit the apparatus.

This is obvioulsy a contradiction, since both lines of reasoning are perfectly acceptable. So what is the solution to this?
Hint: Quantunm theory comes in handy.

Steve Conner

Thu Feb 09 2006, 11:43AM

Registered Member #30 Joined: Fri Feb 03 2006, 10:52AM
Location: Glasgow, Scotland
Posts: 6706

I didn't believe this at first: From my understanding of polarization, adding the third polarizer should make no difference. So I went through to the lab and borrowed three polarizers from one of our kits (ironically I work for a company that makes optics teaching kits for universities)

Sure enough, adding the third polarizer allows light to pass through, and it does look like about 1/8 of the original light. I handed the three polarizers round my colleagues and challenged them to explain why it happens. None of them could give a satisfactory answer.

The paradox is of course this: To explain the transmission of 1/8 of the light, you have to assume that polarisers rotate the light rather than just blocking out the non-matching component. But if they did do that, then two crossed polarizers should not be completely dark, yet you can see in the experiment that they are.

Neither can I, although I vaguely remember some experiment at Bell Labs back in the 40s where the answer was "1/8" B)

Thanks for posting a great puzzle!

Dr. Slack

Thu Feb 09 2006, 02:03PM

Registered Member #72 Joined: Thu Feb 09 2006, 08:29AM
Location: UK St. Albans
Posts: 1659

The mistake is to think that a photon has polarisation.

What QM says is that you measure the polarisation of a photon by passing it through a polarising filter, and it either passes or it doesn't. If it has previously passed through a polar with the same orientation, then it passes through the next one with 100% probablility (less losses due to absorbtion). If the second is at right angles to the first, then the result is 0%.

It's easy to mistake the second definition as being equivalent to saying that the photon in flight between the two polars has the polarisation of the first.

When the photon encounters the middle polar in your experiment, the new measurement is being made at a different angle to the first, so the result is different. In fact it's the cosine of the angle difference for the amplitude of the beam, or cos[sup]2[/sup] for the power, or number of photons.

If you assume the photon has a polarisation after the first polar, then the filter rotates the half that pass, and blocks the rest, which is a very odd thing for a filter to do. If you assume that the photon has no polarisation until the measurement is made, but that when the measurement is made it remembers its previous measurement, then it's just as weird, but at least that model predicts the correct outcomes when you do the experiment.

Incidently, as cos tends to 1 very quickly as the angle tends to 0, using more intermediate polars will increase the transmission (neglecting the effects of absorbtion), 10 polars at 9 degree steps will give about 78% theoretical transmission, 20 polars 88%.

Steve Conner

Fri Feb 10 2006, 01:01AM

Registered Member #30 Joined: Fri Feb 03 2006, 10:52AM
Location: Glasgow, Scotland
Posts: 6706

My head hurtz! !dodge

I did some Google searching on this and found that the intensity of light transmitted through two polarizers, as a function of the angle between them, is described by Malus' law. (I=Imax*cos^2(theta), as NeilThomas hinted in his last post.) So I went looking for a derivation of Malus' law for three polarizers and couldn't find one. I tried to derive it myself using the amplitudes of the vertical and horizontal field components, then square to get the intensity, and soon got into a very big mess of cos's and sin's. The bottom line though, was that I started with the equation for two polarizers and tried to extend it to three without knowing how it was derived in the first place.

In the light of NeilThomas' last post the proof is pretty simple: you just apply Malus' law twice. pretending that the middle polariser also "rotated the light". The answer is 0.5*cos^2(theta)*cos^2(90-theta): the 0.5 being for the case where you started with randomly polarized light and only half of it gets through the first polarizer. For theta=45 degrees this works out at 1/8.

I'm not sure if there is any classical way of proving that it is valid to just use Malus' law twice like that. I guess there is still the paradox that a photon is a field with vertical and horizontal components, and also a particle that either goes through the polarizer or doesn't. It's not possible for the vertical component to go through and the horizontal component to be blocked.

Dr. Slack

Fri Feb 10 2006, 08:09AM

Registered Member #72 Joined: Thu Feb 09 2006, 08:29AM
Location: UK St. Albans
Posts: 1659

I'm not sure if there is any classical way of proving ...

You don't prove QM results classically (though OTOH classical results can be derived in the large scale limit from QM ones). The "measurement" of the middle polariser "wipes clean" the photon's history, including all previous measurements. It's tempting to say it does that by setting its polarisation, but that's a temptation that must be avoided. If you allow a photon to have a definite polarisation, then Alain Aspect's demonstration of Bell's Inequality should not work, but it does. So while what is actually going on is anybody's guess, it isn't the obvious. :((

Hazmatt_(The Underdog)

Sun Feb 12 2006, 08:16AM

Registered Member #135 Joined: Sat Feb 11 2006, 12:06AM
Location: Anywhere is fine
Posts: 1735

hey! Grab 4 polarizers and set them up at 0 30 45 and 60 and see if you get 1/16 the light transmitted.

Might be fun.

Nik

Sun Feb 12 2006, 04:40PM

Registered Member #53 Joined: Thu Feb 09 2006, 04:31AM
Location: Ontario, Canada
Posts: 638

I will go searching for my answer as i dotn have it right now, but I did this in high school physics and we fogured it out on paper using fairly simple math. It was one of my favorite examples of paradoxical physics.

*edit* they way it was explaned in my phys lass was like this.

Light starts with an x and y component (1,1). The first polarizer takes out either the x or y component of the light like this.
Iy = sin[0]*(1) = 0
Ix = cos[0]*(1) = 1

The next polarizer is at 45 degrees to the first havign this effect.
Iy = sin[45]*(1) = 0
Ix = cos[45]*(1) = 0.5

The third polarizer is at 90 degrees to the first but 45 to the second, so it is just repeatign the previous step.
Iy = sin[45]*(1) = 0
Ix = cos[45]*(1) = 0.25

Now because the intensity of the light is of both components the final intensity is 0.25/2 or 1/8 of the original. If you look at it as being a case of 2 pairs of polarizers being 45degrees to eachother and not 2 polarizers at 90degrees with one at 45 between the whole problem loses its paradoxical nature. (on paper)

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