Is a polarizer a filter or a converter?

[San K]

Does a polarize simply filter photons
or
does it change the polarization of photons (as well)?

let's consider only one photon being sent, at a time, via the polarizerI think the answer might be that:

for photons polarized at say 45 degree to the horizontal (or vertical) plane of the polarizer
the polarizer is a convertor, i.e. it changes the polarization of the photons into vertical or horizontal (randomly).

for photons polarized at say 0 or 90 degree to the horizontal (or vertical) plane of the polarizer
the polarizer acts as a filter because it is unable to convert the spin of such photons, the angle is too much/great.

maybe I did not understand polarization well

second question:

does a photon have a "definite" polarization prior to measurement/filtering?

 

[y33t]

Polarizer is a mechanical apparatus which only passes the light/wave depending on it's characteristics. For example a CPL (circular polarizer) will only let circularly polarized field components of the wave to pass to other side of it. Part that doesn't pass by is dissipated as heat in the polarizing filter. It's a filter not converter.

Photon it self doesn't have a polarization state, polarization is a behavior of waves. Photons has spins.

 

[Joncon]

Photon it self doesn't have a polarization state, polarization is a behavior of waves. Photons has spins.

But photons show both wave and particle behaviour don't they? If a photon doesn't have a polarization state then what determines how it passes through a polarizer? I thought a photon followed Malus' law when encountering a polarizer.

 

[jfy4]

The polarizer really is an interaction with the light. The light that is polarized at 45 degrees can be thought of as having the possibility of being polarized at 90 degrees, or at 0 degrees. The polarizer simply demands the photon be one or the other. Then either the photon is absorbed or passes through.
This can be written down mathematically as
$$|\psi \rangle = \sin\alpha |\psi\rangle_{||} + \cos\alpha|\psi\rangle_{\perp}$$
for 45 degrees this is
$$=\frac{1}{\sqrt{2}}(|\psi\rangle_{||}+|\psi\rangle_{\perp})$$
But from moving from a single photon to the above is about a large number of photons, not a case-by-case basis.

 

[Drakkith]

I think you can CONVERT the polarization of light using a Wave Plate, but not a polarizer.

http://en.wikipedia.org/wiki/Wave_plate
http://en.wikipedia.org/wiki/Polarizer

 

[Joncon]

But from moving from a single photon to the above is about a large number of photons, not a case-by-case basis.

How can it not be a case by case basis? As I understand, photons don't interact with each other, so they must be acting individually.

I think you can CONVERT the polarization of light using a Wave Plate, but not a polarizer.

OK, but if I hold two linear polarizers at 90 degrees to each other then no light gets through. If I then put one at 45 degrees between these two then some light *does* get through. Doesn't that suggest that the polarization of the photons is being converted by the middle polarizer?

 

[jfy4]

How can it not be a case by case basis? As I understand, photons don't interact with each other, so they must be acting individually.

The individual photons interact with the polarizer, but if you look at the number of photons on the screen behind the polarizer, it is proportional to $\sin^2\alpha$. This is only true after you let a large number of photons upon the polarizer. If you let a single photon upon it, at 45 degrees, it is equally likely to be absorbed or passed through, at random.

 

[Joncon]

Yes but then the photon is polarized at 45 degress. So it has been converted from it's original polarization hasn't it?

 

[Drakkith]

Yes but then the photon is polarized at 45 degress. So it has been converted from it's original polarization hasn't it?

I don't believe so. See here:
http://en.wikipedia.org/wiki/Polarizer#Malus.27_law_and_other_properties

 

[jfy4]

Yes but then the photon is polarized at 45 degress. So it has been converted from it's original polarization hasn't it?

That's an interesting point. I would put the question though, to be more precise I think: What exactly is $\alpha$ in the above?

It seems you could avoid even having to ask your question if you perhaps decided to abandon classical thinking right off the bat, i.e. what does it mean for an individual photon to be polarized at 45 degrees? You could take the math for what it's worth in experiment (a lot) and say that polarization of individual photons is defined as
$$|\psi\rangle=\sin\alpha|\psi\rangle_{||} + \cos\alpha|\psi\rangle_{\perp}$$
But then what does it mean for $\alpha=\pi /4$? Maybe you could say the preparation device for the experiment prepares a photon with the parameter $\alpha=\pi /4$, but that does seem like you are replacing polarization with just something else...

 

[nonequilibrium]

OK, but if I hold two linear polarizers at 90 degrees to each other then no light gets through. If I then put one at 45 degrees between these two then some light *does* get through. Doesn't that suggest that the polarization of the photons is being converted by the middle polarizer?

I understand it intuitively seems to be converting, but upon breaking it down step by step it's clear it's just a filter: (I'll write it down for classical waves)

---
In the first scenario (the one filter), the filter asks "does the polarization have a component parallel to my polarization axis? If "yes", let that part through; if "no", stop everything; answer in this case: "no""

In the second scenario, the first filter asks the same question, sees a parallel component, and let's that component through; the second filter asks the same question, again sees a component parallel and let's that pass.

You see that at each step all a polarizer does is either let a part go through or stop everything; it is obviously a filter, not a converter. The part where your intuition is led astray is because the notion of "having a component parallel to a direction" can change when stopping/deleting certain parts of your wave (as the first filter in the 2nd scenario does). To understand this even more clearly: imagine two sines, in phase, oscillating around the same line, but each in a different plane, and the planes are perpendicular to each other. Call one plane A, the other B. Rotate the whole in your mind such that you're looking at a cross (i.e. the direction of propagation of the waves is pointing away from you). Draw a new plane C that cuts the X you see with a horizonal line through the midpoint. The two sines (i.e. the net sum) does not have a component parallel to C. But if a polarizer deletes the sine in plane A, then the wave suddenly does have a component parallel to C, yet all you did was filter.
---

However, you might say "yes classically I agree it's a filter, but it seems when I think about it quantum mechanically, it looks like it's converting, because once a photon that is polarized by 45° goes through a 90° filter, its polarization has shifted, yet a photon is one particle, so the explanation "the component parallel blabla" doesn't work." (not trying to put words into your mouth, I just think it would be an intuitively acceptable thought)

The thing is --and I might be wrong because my knowledge of quantum mechanics is shabby; someone please correct me if I'm wrong-- the photon isn't really polarized by 45° degrees to begin with. A polarization of 45° degrees simply demonstrates your ignorance about its actual polarization (or in the more popular interpretation of QM: it demonstrates that nature hasn't decided yet what the polarization is); in either interpretation it comes down to: a "45° polarization" means a superposition (in equal amount) of a 0° polarization and a 90° polarization; when you do an experiment to determine its polarization (with a 90° polarizer), there's a 50% chance it'll have a 90° polarization (and thus pass through), and it has a 50% chance it'll have a 0° polarization (and thus be stopped). You see the only "converting" that has been done, is going from ignorance/undeterminateness to knowledge/determinacy (but that can be said about any QM-measurement).[*]

Now the connection between the classical explanation and the QM explanation: say all the photons are polarized at an angle alpha w.r.t. the axis of polarization, this means that the polarization of one photon is the superposition of $\cos \alpha \{ \textrm{axis of polarization} \} + \sin \alpha \{ \textrm{perpendicular direction} \}$, and thus the chance of a photon having the polarization in the "right" direction is (Born's rule) $|\psi|^2 = (\cos \alpha)^2$; now in the classical limit we have a huge number of photons, so the law of large numbers says the fraction of photons that pass through is cos(alpha)², and the numer of photons is proportionate to the classical intensity of a wave (such that the relative intensity of the part of the wave passing through is cos(alpha)²), and as the intensity is the square of the electric field, this means that a fraction cos(alpha) of the original classical wave passed through, which is indeed equivalent to "the component parallel to the axis of polarization".[*] this reasoning does imply that a photon only can be polarized in either the direction of the polarizer or perpendicular to it; the intuitive strangeness is avoided in the usual interpretation because then there is no polarization beforehand (nature is not yet determined); I don't know how deterministic QM interprets this, but this may be off-topic...

EDIT: I started writing my post before jfy4 posted, hence the fact that I'm not referring to his post although I seem to be discussing similar things

 

[ThomasT]

If you put a polarizer in the path of, say, sunlight, then the intensity of the light transmitted by the polarizer is half that of the original sunlight. If you then put a polarizer in the path of the light transmitted by the first polarizer, then the intensity of the light transmitted by the second polarizer varies according to the law of Malus. Both the first and second polarizers are clearly filters. And the second polarizer seems to be a converter as well. But what about the first polarizer?

 

[nonequilibrium]

ThomasT - I think I explained your question "And the second polarizer seems to be a converter as well. But what about the first polarizer?" in my previous post, between the --- symbols (its conclusion is that nothing converts). Unless there is still a reason to call something a converter, but I wouldn't see what.

 

[Drakkith]

If you put a polarizer in the path of, say, sunlight, then the intensity of the light transmitted by the polarizer is half that of the original sunlight. If you then put a polarizer in the path of the light transmitted by the first polarizer, then the intensity of the light transmitted by the second polarizer varies according to the law of Malus. Both the first and second polarizers are clearly filters. And the second polarizer seems to be a converter as well. But what about the first polarizer?

Neither polarizer converts anything.

 

[ThomasT]

Thanks Drakkith and mr. vodka (I must read your explanation more carefully and think about it).
The problem I have in understanding this is that the first polarizer has a particular orientation. So, all of the light transmitted by it can be said to be polarized wrt that orientation. Then the orientation of the second polarizer can be varied and if some light is being transmitted then we can say that that light is polarized wrt the second polarizer's orientation. And if the orientation of the second polarizer is offset from that of the first, then how can it be said that the polarization of the light hasn't been converted by the second polarizer?

 

[San K]

That's an interesting point. I would put the question though, to be more precise I think: What exactly is $\alpha$ in the above?

It seems you could avoid even having to ask your question if you perhaps decided to abandon classical thinking right off the bat, i.e. what does it mean for an individual photon to be polarized at 45 degrees? You could take the math for what it's worth in experiment (a lot) and say that polarization of individual photons is defined as
$$|\psi\rangle=\sin\alpha|\psi\rangle_{||} + \cos\alpha|\psi\rangle_{\perp}$$
But then what does it mean for $\alpha=\pi /4$? Maybe you could say the preparation device for the experiment prepares a photon with the parameter $\alpha=\pi /4$, but that does seem like you are replacing polarization with just something else...

interesting discussion, good points jfy and joncon, still thinking over Mr Vodka's post

just wanted to add that there is no difference between the behavior of the first polarizer versus the second because:

the first one also encounters photons with polarization at say 45%, 50% 30% etc

for 45% it again..."forces" (whatever that means) just like the second polarizer to be V or H.

for the say 30% ...it again "forces", however we know the profitability of it emerging as V or H? The probability of it emerging as H would be higher...?

we once again one of the fundamentals of QM at play...(for a given polarization angle)--you can calculate/know the probabilities but not the individual behavior

on a separate, but related, topic:
also if the photons' polarization prior to encounter with a polarizer/qwp etc is indeterminate (as per a popular QM interpretation) does it remain determinate (forever) after emerging from a polarizer?

 

[nonequilibrium]

@Thomas: Hm, I understand your problem: you're thinking "okay at first I have a collection of sines (i.e. all kinds of polarizations) coming in into the first polarizer, then the polarizer takes a part of it and let's only one sine wave go through, more specifically: a sine with polarization matching the (first) polarizer. Then seeing this "pure" sine wave encountering the second polarizer, I see a new sine wave exiting the 2nd polarizer, rotated with a certain angle (and also a different amplitude, but that's not of the essence here). So basically your problem is: you see one sine wave coming in, and one sine wave coming out, this gives the illusion of conversion, cause what is there to select out of one sine wave, since it is indeed one wave?

Well, the thing is: the concept of "one wave" is faulty. Take a random sine wave, and in a plane perpendicular to the direction of propagation, take two perpendicular axes x and y (make a drawing). Is it not true that this "one" sine wave can be written as the sum of two other (sine) waves? In particular it's the sum of the projections of the original wave onto the two planes formed by the axes and the direction of propagation (again, easier to follow if you made a simple drawing)

My point: the situation at the 2nd polarizer is exactly the same as at the 1st polarizer: what intuitively seems as "one wave" is in fact a sum of waves with different polarizations; it's just a matter of perspective

I'm convinced what I'm trying to say is very simple and convincing, I just find it hard to state in clear terms, but let me know if something is getting through

 

[dlgoff]

Yes but then the photon is polarized at 45 degress. So it has been converted from it's original polarization hasn't it?

Here's a cute little youtube experiment that has been posted here a few times. :)

https://www.youtube.com/watch?v=ZudziPffS9E

 

[ThomasT]

@Thomas: Hm, I understand your problem: you're thinking "okay at first I have a collection of sines (i.e. all kinds of polarizations) coming in into the first polarizer, then the polarizer takes a part of it and let's only one sine wave go through, more specifically: a sine with polarization matching the (first) polarizer. Then seeing this "pure" sine wave encountering the second polarizer, I see a new sine wave exiting the 2nd polarizer, rotated with a certain angle (and also a different amplitude, but that's not of the essence here). So basically your problem is: you see one sine wave coming in, and one sine wave coming out, this gives the illusion of conversion, cause what is there to select out of one sine wave, since it is indeed one wave?

Well, the thing is: the concept of "one wave" is faulty. Take a random sine wave, and in a plane perpendicular to the direction of propagation, take two perpendicular axes x and y (make a drawing). Is it not true that this "one" sine wave can be written as the sum of two other (sine) waves? In particular it's the sum of the projections of the original wave onto the two planes formed by the axes and the direction of propagation (again, easier to follow if you made a simple drawing)

My point: the situation at the 2nd polarizer is exactly the same as at the 1st polarizer: what intuitively seems as "one wave" is in fact a sum of waves with different polarizations; it's just a matter of perspective

I'm convinced what I'm trying to say is very simple and convincing, I just find it hard to state in clear terms, but let me know if something is getting through

Your terms are clear enough. And if I have to say that presently I might not fully understand it, I expect that by tomorrow I will. I should say that my intuitive inclination was to say that polarizers act solely as filters.

I'll either come up with another (what I, imho, think is a good) question tomorrow, or agree with you.

 

[jfy4]

I'm going to have to answer this like this:

The state vector, which seems dependent on $\alpha$, just gives the statistical properties of the photon upon interaction. So when an alpha is chosen, the state vector is chosen, but this can't mean a single result is guaranteed from an interaction. I think it simply means that a statistical distribution of possible results has been set. So specifying the "polarization" of a photon doesn't mean the same thing as it intuitively seems to mean... So a photon with polarization $\alpha=\pi /4$ doesn't mean what it means classically. Instead it would mean that for various classes of interactions with the photon, a particular distribution of results can be expected. That particular distribution of results is what I would call the state vector which is dependent on the parameter we call polarization.

So say we have $\alpha=\pi /4$ which says that for a particular experiment there are two possible results: $\alpha=0$ and $\alpha=\pi$. After an interaction $\alpha=\pi$, this decides a whole other possible distribution of results for the photon for all different classes of interactions.

So did the polarization get converted, sure. But thinking about it this way seems to make it unimportant...

Any thoughts?

 

[y33t]

I'm going to have to answer this like this:

The state vector, which seems dependent on $\alpha$, just gives the statistical properties of the photon upon interaction. So when an alpha is chosen, the state vector is chosen, but this can't mean a single result is guaranteed from an interaction. I think it simply means that a statistical distribution of possible results has been set. So specifying the "polarization" of a photon doesn't mean the same thing as it intuitively seems to mean... So a photon with polarization $\alpha=\pi /4$ doesn't mean what it means classically. Instead it would mean that for various classes of interactions with the photon, a particular distribution of results can be expected. That particular distribution of results is what I would call the state vector which is dependent on the parameter we call polarization.

So say we have $\alpha=\pi /4$ which says that for a particular experiment there are two possible results: $\alpha=0$ and $\alpha=\pi$. After an interaction $\alpha=\pi$, this decides a whole other possible distribution of results for the photon for all different classes of interactions.

So did the polarization get converted, sure. But thinking about it this way seems to make it unimportant...

Any thoughts?

My 1.99$ ;

Polarization is never converted no matter how many you cascade in series or etc. You can not convert polarization after the wave left the source with a polarizer. To change the polarization you need to change the original oscillating field vectors at the source side. What polarizer does is that it simply filters out unwanted portions of the incident wave.

In the video above, the second pcb board exposed to transmission channel acts as a waveguide. This is why some might perceive it as if it converts the polarization but this is not the case.

You can not get a circular polarization from a linearly polarized wave using a polarizer but you can get a linear polarization from circularly polarized wave. First case is possible as others stated but not with polarizers. You can achieve such a conversion using wave plates or waveguides which has far more complicated structure than a polarizer. I am not familiar with the idea that individual photons interact with a polarizer, since it's a mechanical apparatus, polarizer structure should be constructed at the photons scale (in size) to let interaction at that level.

Polarization of the whole wave (say 1 duty cycle) doesn't originate from the spins of individual photons, they are different phenomena.

 

[Joncon]

Hmm, maybe I'm just having a hard time understanding this but I'm not convinced.

I understand that you can view the wave as having both vertical and horizontal components. But if the polarizer was simply a filter then the one aligned at 45 degrees, between two perpendicular polarizers, wouldn't make any difference i.e. the first polarizer (aligned vertically for example) would remove all the the horizontal component, so this photon could never pass a horizontally aligned polarizer. However, the addition of the polarizer in the middle must change some property of the photon, as it now has a 50/50 chance of passing the horizontal polarizer.

 

[harrylin]

Hmm, maybe I'm just having a hard time understanding this but I'm not convinced.

I understand that you can view the wave as having both vertical and horizontal components. But if the polarizer was simply a filter then the one aligned at 45 degrees, between two perpendicular polarizers, wouldn't make any difference i.e. the first polarizer (aligned vertically for example) would remove all the the horizontal component, so this photon could never pass a horizontally aligned polarizer. However, the addition of the polarizer in the middle must change some property of the photon, as it now has a 50/50 chance of passing the horizontal polarizer.

I agree: my reply to the OP's question would be "both".

Only adding a pure filter cannot increase the amount of light that passes through - that's not what a filter does.
It's of course largely a matter of words. :tongue2: A filter takes away stuff, and if it also modifies stuff, then that modification isn't correctly called "filtering".
- http://dictionary.reference.com/browse/filter
When a polarizer filters out a wave component, the resulting wave or photon has its polarization rotated - and rotation is more than just removing something, it's changing one of its properties. If you model light as photons, then it let's pass photons at a modified polarization - else nothing would pass through the system at all.
Now, "converter" is perhaps not the best word choice,
http://dictionary.reference.com/browse/converter
but certainly a polarizer does more than "just filtering".

For the second question, that's the topic of a still ongoing debate related to "Bell's Theorem"
- http://www.desy.de/user/projects/Physics/Quantum/bells_inequality.html

Harald

 

[kith]

Whether the polarization is filtered or converted, is a question of the viewpoint.

If you consider the electromagnetic wave as the fundamental physical object, polarization is filtered because you can always decompose the wave into two orthogonal polarization components.

In the viewpoint of quantum mechanics (not QFT), the single photon is the fundamental physical object. The electromagnetic wave then describes the state of such a photon (which can be occupied by many photons due to their bosonic nature). In this case, polarization is obviously converted, since the photon enters with one fixed polarization state and leaves with another or gets absorbed.

Also note that there is no fundamental difference between polarization and spin. A spin-1 particle has three spin directions which corresponds to the electric field vector having three components. This is analogous to the case of a spin-1/2 particle (like the electron), which is described by an object with 2 components ("Pauli-Spinor"). The Maxwell-Equations now demand that electromagnetic waves are transverse waves, so the number of components is reduced by one. The resulting formalism for the photon ("Jones Calculus") is very similar to the well-known spin-1/2-formalism.

 

[nonequilibrium]

I understand that you can view the wave as having both vertical and horizontal components. But if the polarizer was simply a filter then the one aligned at 45 degrees, between two perpendicular polarizers, wouldn't make any difference i.e. the first polarizer (aligned vertically for example) would remove all the the horizontal component, so this photon could never pass a horizontally aligned polarizer. However, the addition of the polarizer in the middle must change some property of the photon, as it now has a 50/50 chance of passing the horizontal polarizer.

I tried to explain this fallacy on the previous post (in my long post, especially between the --- symbols). Classically, anyway, there is obviously (once you get my post, that is; if my post is unclear, tell me, I'll try again) NO conversion, just filtering.

(As for the QM case: I don't think I know enough)

 

[kith]

I understand that you can view the wave as having both vertical and horizontal components. But if the polarizer was simply a filter then the one aligned at 45 degrees, between two perpendicular polarizers, wouldn't make any difference i.e. the first polarizer (aligned vertically for example) would remove all the the horizontal component, so this photon could never pass a horizontally aligned polarizer. However, the addition of the polarizer in the middle must change some property of the photon, as it now has a 50/50 chance of passing the horizontal polarizer.

Yes, I get your point. So my remark about the classical case is at least incomplete. Now that I think about it, the classical case as well must respect the fact that different components of spin do not commute.

So classically, one polarizer is a filter. Two polarizer in a different angle as 90° are a converter.

 

[jfy4]

I still think that the notion of "a photon polarized at 45 degrees" is a poor concept to hold onto classically. Before even trying to start to answer this question you should ask "What does it mean for a photon to have a polarization of 45 degrees...?"

If you consider a photon passing through a polarizer as a filter; what gets filtered!? There isn't less of a photon, this is a fundamental result of quantum mechanics. But consider it as a "converter"; what is being converted!? I think you will be hard pressed to answer this question any other way (more comprehensively) than: "The possible statistical results of the photon in different classes of interactions."

If this is what you had in mind when you answered the first question, then I think you can walk away and say its a converter following these definitions. If not, then its not so binary.

Note that there is something being "filtered" in a more general sense; the probability of the photon to be absorbed or pass though. If you set up the three polarizers like in the video, after each polarizer the probability to pass through is diminished. So something is getting "converted" in a sense, and "filtered" in a sense, but I think that sense has to be changed from our classical notions (or at least mine).

 

[nonequilibrium]

So classically, one polarizer is a filter. Two polarizer in a different angle as 90° are a converter.

Okay as I seem to be ignored trying to explain it nicely, I'll be blunt: you're wrong. Say system S is a construction of two polarizers at different angle, then S is a filter because whatever comes out was already part of the original wave, there's just something left out.

(Mind you: that is for the classical situation; I would agree with jfy4 that the notions of "filter" and "converter" become a bit murky on a QM level)

 

[DrChinese]

A photon with a known linear polarization at A will have a completely indeterminate polarization at A+45 degrees. That would be consistent with the HUP for a non-commuting observable. Any angle between is a mixture of the two and will be "somewhat" indeterminate.

So to me, a polarizer places a photon in an eigenstate as would be expected from any observation. Even if the photon does not emerge, I guess you would say the same.

 

[nonequilibrium]

How does a polarizer actually work on a QM-level? I can't figure out what it does to one photon. What is the measuring aparatus?

 

[Drakkith]

Obviously one can argue that per QM the idea of polarization of a photon being exactly X is incorrect. We have no idea what it is polarized at and most will say it doesn't have an exact polarization, only the probability of what it can be. To me, looking at the whole situation, X photons go through and less than X photons come out. Since the ones that come out don't appear to be "converted", and instead seem to have already been in the beam to begin with, I would say that a polarizer is a filter, not a converter.

 

[ibysaiyan]

interesting discussion, good points jfy and joncon, still thinking over Mr Vodka's post

just wanted to add that there is no difference between the behavior of the first polarizer versus the second because:

the first one also encounters photons with polarization at say 45%, 50% 30% etc

for 45% it again..."forces" (whatever that means) just like the second polarizer to be V or H.

for the say 30% ...it again "forces", however we know the profitability of it emerging as V or H? The probability of it emerging as H would be higher...?

we once again one of the fundamentals of QM at play...(for a given polarization angle)--you can calculate/know the probabilities but not the individual behavior

on a separate, but related, topic:
also if the photons' polarization prior to encounter with a polarizer/qwp etc is indeterminate (as per a popular QM interpretation) does it remain determinate (forever) after emerging from a polarizer?

I suppose it would return back to wave-function i.e the probability would be indefinite since it's not being observed anymore,just a guess.

 

[San K]

Here's a cute little youtube experiment that has been posted here a few times. :)

https://www.youtube.com/watch?v=ZudziPffS9E

the video says "photon polarization is indeterminate till its measured"

as per one of the popular interpretations of QM

1. is the polarization indeterminate ...before or after ...the photon goes through the 45 degree polarizer? or after it exits the third polarizer?

2. what action/setup constitutes measurement?

3. how do we convert the polarization of a photon
- from determinate to indeterminate -- does it simply happen over time, for one?
- indeterminate to determinate -- just send it through a single polarizer? or through two polarizers? with some angle between their axis so that it constitutes a measurement?

 

[San K]

Here's a cute little youtube experiment that has been posted here a few times. :)

https://www.youtube.com/watch?v=ZudziPffS9E

if we were to replace the 45 degree with 30 degree polarizer, would we still expect light to pass through however it would be a bit dimmer?

 

[agentredlum]

When he put the polarizer in the middle he should have kept rotating it so we could see if it had any effect on the polarizer closer to the camera.

The light that goes through 2 polarizers is distinctly different in color than the light that goes through all 3

With 2 polarizers the intensity changes but the color of the image does not change as he rotates the second polarizer. With 3 polarizers the image in the center looks blue but the image outside the first polarizer AND inside the third polarizer looks the same as originally with only 2 polarizers. Very curious...