Polarizing light

[Hikaru79]

This is a high-school level question, I think. But I'm in high school

 

We are told that a polaroid filter will only allow light to pass if it is polarized on the same axis as the filter. Otherwise, it will not get through. It follows, then, that if you pass light through two polaroid filters, you will get nothing through the other side. Empirically, we can test this, and it confirms what we know.

 

So far so good.

 

Now, say we take a third filter and put it in between the two previous ones. We turn it to a 45-degree angle from both of them. Now, if no light could get through the two filters before, then surely adding a filter would not change this. After all, a filter cannot generate or re-polarize light, right?

 

But if we actually perform this very simple experiment, you will see that as you rotate the third, middle polaroid filter, you do start to see light, contrary to my understanding of the theory.

 

I'm clearly missing something. Can anyone perhaps explain? There was a big argument between my entire class and the teacher over a multiple choice question on the test involving this issue. Her explanations were not satisfactory.

 

Thanks

[timo]

This is a high-school level question, I think. But I'm in high school

I think it can be understood with high-school knowledge, but it´s amazing for any level.

 

We are told that a polaroid filter will only allow light to pass if it is polarized on the same axis as the filter. Otherwise, it will not get through.

Welcome to the wonderful world of simplifications. Above statement is true if you know what it means and pretty useless (or even incorrect) if you don´t know how to interpret it. Let me try to explain what actually happens:

Let´s assume the photon (=light) moves in z-direction. As you hopefully know, it´s polarisation is perpendicular to the movement direction. So if we -for simplicity- rescrict ourselves to linear polarisation, the polarisation of the light can be described as a 2D vector in the x/y-plane (x,y). Vectors within this plane can be descibed by a linear combination of any two perpendicular base vectors e1 and e2. Example: e1 is the x-axis and e2 is the y-axis, then (x,y) = x*e1 + y*e2.

Now what a polarisation filter does is projecting the vector onto a certain direction. Let polarisation filter 1 project on the e1-direction (x, right-left), polarisation filter 2 prohject on the e2-direction (y, up-down) and polarisation filter 3 project on the e1+e2 direction (x+y, right/up - left/down):

 

It follows, then, that if you pass light through two polaroid filters, you will get nothing through the other side. Empirically, we can test this, and it confirms what we know.

Now let´s put this into the language of above projections. The beam of light arrives with any polarisation (x,y) = x*e1 + y*e2. The first filter projects this vector on the e1-axis so after passing this filter the polarisation is (x,0) = x*e1. The 2nd filter then projects the polarisation on the e2-axis, then. So after passing the 2nd filter the polarisation is (0,0) = 0. Now at this point I realize that I should have told you that the magnitude of the polarisation vector is connected to the intensity of light. I don´t want to rewrite all I´ve written so far so just believe me that a zero polarisation vector means "no light" and that a nonzero polarisation vector means "some light" - at least within the context of this post.

 

So far so good.

I hope so.

 

Now, say we take a third filter and put it in between the two previous ones. We turn it to a 45-degree angle from both of them. Now, if no light could get through the two filters before, then surely adding a filter would not change this. After all, a filter cannot generate or re-polarize light, right?

Yes, it is surprising. But let´s do the math. As before, after passing the 1st polarisation filter, the polarisation will be (x,0). Now, instead of arriving at the 2nd filter, the light comes to the filter 3 which projects onto the e1+e2 direction. To calculate the projection onto this direction, let´s rewrite (x,0) as a linear combination of e1+e2 and a perpendicular vector (e1-e2): (x,0) = x*e1 = 0.5*(x*(e1+e2) + x*(e1-e2)). So after projecting on the e1+e2 direction by polarisation filter 3, the polarisation will be (x/2,x/2) = 0.5*x*e1 + 0.5*x*e2.

Already see what will happen now? I guess so but let´s complete the calculation anyways: This polarisation will then arrive at the polarisation filter 2 which projects it on the y-axis. (x/2, x/2) projected on the y-axis simply is (0, x/2) which is not zero (unless x was zero, of course).

[Klaynos]

I quite like the trick of putting 2 square linear polerisers on top of each other so that no light passes through and their sides line up, and then turning one of them over and light passes through, always puzzels people on open days....

[5614]

If you have two polarisers, one vertical and the other at 45° (relative to the first polariser) and you fire vertically polarised light at the system then all of the light will pass through the first polariser and half of the light will pass through the second.

 

So if you had vertical, one at 45° and one horizontal and you fired vertically polarised light at it you'd get all going through polariser 1 (p1) you'd get 50% through p2 and then 50% of that 50% going through p3... so basically you end up with a half of a half (ie. a quarter) of the original light. Note this example assumes original light in all polarised vertically.

 

One more thing to add. When light comes out of a 45° polariser the light will no longer be aligned vertically (like it originally was) it will now be aligned at 45° and so when it arrives at the 3rd polariser the light "sees" it as 45° relative to itself.

 

This is a good example of quantum uncertainty. The photon itself does not have a definite polarisation. The fundemental meaning of uncertainity is that it is not us who can't measure the polarisation exactly, the photon itself does not have a definite polarisation, the concept of a photon with a definite polarisation does not exist in reality.

[timo]

This is a good example of quantum uncertainty. The photon itself does not have a definite polarisation.

Assuming you read my post, I´d like to know what was unclear there. I didn´t use uncertainty, my photons had a definite polarisation (x,y) and I didn´t use QM at any point (although admittedly the base-transformation used for filter 3 was heavily QM-inspired).

[s pepperchin]

If you want to do an interesting experiment with polarization, use corn syrup.

In one of my lab classes we set a polarizing filter on an overhead projecter, then we poured corn syrup into a shallow glass dish placed on top of the filter, then we would hold the other filter above it and look at the light passing through all three as it is projected on the wall. Ask your teacher about it and maybe he will show you.

[5614]

Assuming you read my post, I´d like to know what was unclear there. I didn´t use uncertainty, my photons had a definite polarisation (x,y) and I didn´t use QM at any point (although admittedly the base-transformation used for filter 3 was heavily QM-inspired).

/me confused...

 

There was nothing unclear with your post. And I'm unsure what you didn't like in mine.

 

You might say a photon has a definitive polarisation, lets say vertical. However when you pass your vertically polarised light through a polariser at 45° to the vertical as you know 50% of the light will pass through.

 

The reason for this is that the "vertically" polarised light has an uncertainty about it's polarisation. It has a probability of having a different polarisation. e.g. 0% chance of being horizontal but 50% of being at 45°. Although the photon is vertically polarised it is in a superposition of states and remains so until it interacts with the polariser at which point it takes a definitive polarisation.

 

When I said "This is a good example" I think that "this" referred to the whole experiment. It shows that although you can say "this photon is vertically polarised" due to uncertainty there is a defined probability that the photon will have a different (from vertical) polarisation.

[timo]

/me confused...

 

There was nothing unclear with your post. And I'm unsure what you didn't like in mine.

I do not particularly like to introduce QM where it isn´t nessecary. I don´t know what "high school" is but it doesn´t sounds that someone in high school will be helped by being told that the example is a nice example of QM, especially not when the follow-up description is very vague. Stuff like a polarisation vector "up" and a polarisation vector "right-up" not being perpendicular (and therefore having a non-zero scalar product and a non-zero projection onto each other) is not unique to quantum mechanics but simple linear algebra. Admittedly, those LA properties directly transfer to QM which is based on LA to a large extend.

[swansont]

It's not uncertainty, though, in the normal way that's used in QM. It's about using the proper basis to measure the polarization, which is, as Atheist points out, is basically a linear algebra issue.

[timo]

It's not uncertainty, though, in the normal way that's used in QM.

What 5614 said does have to do with uncertainty in a way. For example, if you write down the projection matrix for a vertical polarisation and the projection matrix for ... say polarisation right-up, these two matrices do not commute which means that you cannot have both observables (quantum numbers) fixed.

[swansont]

What 5614 said does have to do with uncertainty in a way. For example, if you write down the projection matrix for a vertical polarisation and the projection matrix for ... say polarisation right-up, these two matrices do not commute which means that you cannot have both observables (quantum numbers) fixed.

 

But isn't that more a function of choosing one basis that's a superposition of another basis, for angular momentum states (since linear polarization is a itself a superposition of the two circular polarizations)? (Brain atrophy. Don't have my books handy to look it up, and QM was sooo long ago)

[timo]

Well that´s what uncertainty is about. That you have two observables which you cannot find a common base of eigenstates for:

If a system is in an eigenstate of operator 1 (e.g. after measurement of the associated observable) it is not in an eigenstate to op 2. Therefore, the value that will be measured for observable 2 is not clearly determined.

On the level of matrix representation this simply means that both matrices cannot be diagonalized simultaneously.

[swansont]

Yeash, the commutation relation being the uncertainty clicked for me a few minutes after I logged of, after thinking about it as photon states. (Nice to think about physics now and then, instead of budget and having to keep both eyes on a construction project)