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Does polarized light interfere?

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4  
Yes, it does. Could you be more specific, why do you think it would not? –  gigacyan Feb 9 '11 at 14:11
    
Question is unclear but can be useful. Don't think need to close -- just downvote. –  Kostya Feb 10 '11 at 9:36
3  
@Kostya: the problem is, if you down-vote, will you actually return to unvote if the question is improved? I think a closed question is more likely to become "rehabilitated" –  Tobias Kienzler Feb 10 '11 at 10:53

5 Answers 5

Yes, it does and this property is independent from a particular polarization. So non polarized light gives the same interference pattern.

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1  
-2 for what? Didn't I answer the question? –  Vladimir Kalitvianski Feb 9 '11 at 18:59
    
because it's wrong or at least very misleading. While stating "interference always occurs" is kind of valid despite omitting the dependence on coherence and phase difference, your statement non polarized light gives the **same** interference pattern is really wrong –  Tobias Kienzler Feb 10 '11 at 7:28
    
Ah yes, I implied the same frequency, of course, and interferometer conditions. It is especially evident when I speak of interference of photon with itself. –  Vladimir Kalitvianski Feb 10 '11 at 11:02

Yes. In fact, light will only interfere with light of the same polarization. If you take a Mach–Zehnder interferometer, for example, and put a polarization rotating optic (a waveplate) in one of the arms, the interference pattern will lose contrast. If the polarization is rotated 90 degrees, the pattern will vanish completely.

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I can only add that quantum mechanically a photon interferes with itself, i.e., the resulting field is always of the same polarization. If you intervene with the polarization rotation optic, you break the main rule - not to intervene in the interferometer arms. Any such intervention (polarization, intensity, etc.) spoils the patter to this or that extent. –  Vladimir Kalitvianski Feb 9 '11 at 19:35
    
you should mention something like coherence and same-wavelength-interference-only –  Tobias Kienzler Feb 10 '11 at 7:29

Let's do some math in order not to be unsubstantiated.

1. Perpendicular polarizations.

First wave $E_{1x} = E_0\,\cos\omega t$,second wave $E_{2y} = E_0\,\cos(\omega t+\Delta)$.
Where $\Delta$ -- is a phase difference between waves.

Total field: $\vec{E} = E_0\left(\vec{i}\cos\omega t+\vec{j}\cos(\omega t+\Delta)\right)$.

Intensity: $I_\perp\sim \langle|\vec{E}|^2\rangle = E_0^2\langle\cos^2\omega t+\cos^2(\omega t+\Delta)\rangle$
where average means: $\langle f(t)\rangle = \frac{1}{T}\int_0^{T}\,dt\,f(t)$, so that $\langle cos^2(\omega t+\Delta)\rangle = \frac{1}{2}$ for any $\Delta$

Finally we got $I_\perp\sim E_0^2$, which is independent on the phase difference between the waves.

2. Parallel polarizations.

First wave $E_{1x} = E_0\,\cos\omega t$,second wave $E_{2x} = E_0\,\cos(\omega t+\Delta)$.

Total field: $\vec{E} = \vec{i}E_0\left(\cos\omega t+\cos(\omega t+\Delta)\right)$.

Intensity: $I_\parallel\sim E_0^2\langle\cos^2\omega t+2\cos\omega t\cos(\omega t+\Delta)+\cos^2(\omega t+\Delta)\rangle=E_0^2(1+\cos\Delta)$, which nicely depends on the phase shift between the waves.

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Nice way of explaining this, definitely more thorough than my answer. –  Colin K Feb 9 '11 at 17:19
    
Agreed, +1 for content and style. (tiny typo in the first line though: "no**t** to be unsubstantiated") –  qftme Feb 9 '11 at 18:06
    
Thank you all for feedback! By my answer should be considered as a continuation of @Colin K. –  Kostya Feb 9 '11 at 18:21
    
Kostya, when you add two perpendicular fields, do you think you consider an interference? Any interference is by definition made of the same polarization! You consider a one-arm interferometer! –  Vladimir Kalitvianski Feb 9 '11 at 20:54
    
That is what he said, and he explained why it is the case as well. –  Colin K Feb 10 '11 at 4:00

As others have noted, you will not get any intensity modulation from the interference of two linearly polarized light beams with orthogonal polarizations. It's worth noting, though, that this does not mean that beams with perpendicular polarizations don't affect each other. In fact, a counter-propagating pair of beams with orthogonal linear polarizations-- the so-called "lin-perp-lin" configuration-- is the best system for understanding the Sisyphus cooling effect, the explanation of which was a big part of the 1997 Nobel Prize in Physics.

The superposition of two counter-propagating linearly polarized beams with orthogonal polarizations doesn't give you any modulation of intensity, but does create a polarization gradient. For the lin-perp-lin configuration, you get alternating regions of left- and right-circular polarization, and combined with optical pumping this lets you set up a scenario where you can cool atomic vapors to extremely low temepratures. This makes laser cooling vastly more useful than it would be otherwise, and allows all sorts of cool technologies like atomic fountain clocks.

It's not interference in the sense that is usually meant, but it is a cool phenomenon that results from overlapping beams with different polarizations. So you shouldn't think that just because it doesn't produce a pattern of bright and dark spots it's not interesting.

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Your question is quite vague, but in short, the answer is: yes, look it up on wikipedia. But let's be more precise:

As long as the light intensity is low enough to not obtain nonlinear effects, the superposition principle of linear optics is valid. That means, the amplitudes of two electromagnetic (EM) fields sum up, thus yielding interference.

However, since the amplitudes are vectors (while the intensity, being related to the absolute squares of the amplitudes, is a scalar), the interference depends on the relative polarization, the total intensity for two linearly polarized EM waves is

$I = I_1 + I_2 + 2\sqrt{I_1I_2}\cos(\Delta\varphi)$

where $\Delta\varphi$ denotes the angle between the two polarizations. You see that for perpendicular polarization the cosine term vanishes, the intensities just add up and you obtain no interference, while for antiparallel polarization ($\Delta\varphi = 180°$) you obtain destructive interference since the cosine becomes -1. In case you wonder about energy conservation (being proportional to the intensity) keep in mind that only the global energy is conserved while local fluctations are ok.

One final note: The whole thing only works for a well-defined phase relation between two EM waves. That is, only spectral components of the same wavelength can interfere, and the coherence length and time need to be large enough - you won't obtain perfect interference if your light source flickers due to heat for example.

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