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Polarizers kicking my behind! |
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| Jan17-13, 01:25 PM | #1 |
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Polarizers kicking my behind!
Cross two polarizers and all light is blocked. Inserting a third polarizer in between two crossed polarizers and light again reemerges from the polarizers. Why is this? I'm trying to wrap my mind around this.
I noticed that certain transparent plastics (cellophane, e.g.,) also allow some light to pass through the crossed polarizers because the plastic acts as a sort of 1/2 wave plate. Could it be that the third polarizer is also acting as a 1/2 wave plate, allowing light to transmit through the crossed polarizers? Again, how can light be transmitted through crossed polarizers when a third polarizer is introduced in between the crossed polarizers? |
| Jan17-13, 01:54 PM | #2 |
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The easiest way to think about this polarizer experiment is in terms of vector projections.
The first polarizer allows only the component of the E-Field parallel to the polarizer axis through to the other side. Call this direction the 'x-axis.' The second polarizer is rotated 90degrees relative to the first. Thus, only the component of the light's E-Field parallel to the 'y-axis' will get through. Since the first polarizer forces the light to be polarized along the x-axis, the light's field has no projection along y. Thus, no light gets through the second polarizer. When we add a third polarizer at, say, a 45 degree angle relative to the first (call this the x'-direction) and in between the first two polarizers, the situation changes. After the first polarizer, the light is still polarized along the x-axis. Next, the light hits the new polarizer, which only allows a component of the light parallel to the x'-axis through. Light polarized along the x-axis does have a projection along the x' axis. Therefore some light gets through, polarized along x'. When the x' polarized light hits the last polarizer some now get's through, because light polarized along x' has a projection along the y direction, unlike light polarized along x. This is hard to explain without pictures. For some more info check out: http://web.physics.ucsb.edu/~lecture...0polarizer.htm |
| Jan17-13, 02:12 PM | #3 |
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hi 2112rush2112! welcome to pf!
![]() i suspect you're comparing polarizers with colour filters if you have a red filter and a green filter, nothing will pass through both filters, and if you put another filter between them, it makes no difference this is because a colour filter doesn't change anything … it may block out some light, but any light that does get through is unchanged a polarizer, however, does change the light that gets through
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| Jan17-13, 02:44 PM | #4 |
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Polarizers kicking my behind!And if it is, then it would make sense that the x' polarizer (the polarizer we sandwiched in between the crossed polarizers) would transmit 1/2 the light that fell on it from the first, "x" polarizer (you called this, "x' polarized light"). And since the 'x' polarized light' is rotated 45 degrees from the y polarizer (the last polarizer in the system), then the y polarizer will pass 1/2 of the light from the x' polarizer. Is this correct? You also said, "Since the first polarizer forces the light to be polarized along the x-axis..." I thought polarizers were passive devices; devices that can't "force" anything, but passively allows certain components (x or y components) to pass through, no? |
| Jan17-13, 02:59 PM | #5 |
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| Jan17-13, 04:44 PM | #6 |
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However, the polarizers do modify the state of the photons passing through, so it's not entirely passive. The polarizer acts as a measurement device, and you cannot measure something without changing it. That's why adding a second polarizer can increase the transmission through the third polarizer.
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| Jan17-13, 05:24 PM | #7 |
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Rush is using passive and active in the engineering sense, which makes sense in this context. A passive device is one that doesn't require energy input from to function. An active device is one that does require energy input to function. In this sense, both a lossy optical medium (one that absorbs light) and a gain medium inside a laser (amplifies light) alter light by absorbing and emitting photons, but the first is "passive" and the second is "active." |
| Jan17-13, 10:59 PM | #8 |
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What we are seeing therefore, and pardon my redundancy, is one-fourth the number of photons on the movie screen that originated from the initial light source. This one-fourth composition of photons consists of not (emphasis added) linearly-polarized photons in the x-plane, nor linearly-polarized photons in the y-plane. Rather, we see the photons on the movie screen that were somewhere in-between the x-plane and y-plane (the x-plane photons were blocked by the y-polarizer and the y-photons were blocked by the x-polarizer. The photons in-between ultimately made their way to the movie screen as a spot of light). For rhetoric's sake, I'll refer to these photons as the "in-between photons"--photons that were neither polarized in the x-plane nor the y-plane, but somewhere "in between" the x and the y. An 'in-between' photon, from our previous posts on this thread, has a 50/50 chance of passing through a second polarizer that's rotated at 45 degrees from the first, then a 50/50 chance of being passed through the third polarizer. Suppose we consider only the 'in-between' photon that passed through all three polarizers. The second polarizer did nothing to the photon; the photon merely passed straight through it! And at the end of the day a photon successfully navigated crossed polarizers. <sigh> To crystallize my thought: The first polarizer will pass some in-between photons emitted by a light source. By the aforementioned 'law', any polarizer orthogonal to the first must likewise pass some of the in-between photons. But we don't see this in nature; crossed polarizers do not pass any light from their native light source!!! Can you see my problem here? Thank God this forum doesn't have an IQ requirement, otherwise I never would have made it past forum registration... |
| Jan17-13, 11:06 PM | #9 |
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| Jan18-13, 04:48 AM | #10 |
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The stuff about movie screens would be OK if linear polarisation were used. Actually, to allow people to tilt their heads without messing up the picture, circular polarisations are used in stereo Cinema. Do you really want to talk about circularly polarised photons - or just stick with waves and fields, where the answer comes out with much more ease? Either model is valid.
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| Jan18-13, 09:38 AM | #11 |
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It was a bad idea for me to use it tho, since polarizers are used in actual stereoscopic cinema... :( This discussion is confined to linear polarizers. And since a polarizer can pass photons that aren't entirely in the polarizer's axis (a polarizer that blocks x-axis photons can still pass photons that are in between the x-and y-axis), then why do crossed polarizers block 100% of light? |
| Jan18-13, 10:07 AM | #12 |
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sophiecentaur is right that you should stick to the wave model for light. You are confusing yourself by talking about photons. Polarizers can, of course, be described with a photon model of light, but doing so requires study of quantum mechanics and my guess is that you have not gotten there yet. One step at a time. |
| Jan18-13, 10:19 AM | #13 |
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@ rush
OK - just avoiding the uninformed grabbing hold of the wrong end of the stick when they read what you wrote. As you competently demonstrated, an explanation of polarisation in 'particle terms' is very convoluted, compared with one in which we just resolve field vectors. As both interpretations of EM are valid then why not go for the more straightforward explanation? People seem to feel the need to reach for 'photons' so often - as if it really implies a greater understanding. To my mind, it's like making love standing up in a hammock to do it that way |
| Jan18-13, 10:38 AM | #14 |
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| Jan20-13, 05:14 PM | #15 |
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| Jan20-13, 05:21 PM | #16 |
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Sophiewcentaur, do you agree that the polarizer alters the orientation of some of the polarized light that passes through it? I thought polarizers just pass/block light. ...And if we want to discuss this with QM formalism, I'm hip with that. I don't see photons as sub-microscopic 'baseballs' being sent through the polarizers; no. They're probability amplitudes... |
| Jan20-13, 05:44 PM | #17 |
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I don't see much point in discussing the phenomenon in terms of photons but, in terms of E and H fields in the EM wave, a linear polariser selects the component of the field in one particular direction and rejects / absorbs components the other plane. If you take plane polarised light and pass it through a plane polariser at 45 degrees then you will end up with a plane polarised wave with half power (1/(root two) amplitude) in the plane of the polariser. You 'could' say it had effectively altered the polarisation of the original wave but it has just selected a component of the E field. This new wave could, of course, be passed through a second polariser, at a further 45 degrees to the first which would then produce a wave with polarisation at right angles to the original wave and 1/4 power (1/2 amplitude). You could carry on doing this, losing half the poser each time you rotate the polarisation by 45 degrees. People seem to think that this is, somehow, unthinkable and violates something fundamental. Why? I think the notion of a polariser being "active" (something that was mentioned further up the thread) is nonsense because "active" implies a power source. |
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