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Does polarized light interfere?

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    $\begingroup$ Yes, it does. Could you be more specific, why do you think it would not? $\endgroup$
    – gigacyan
    Feb 9, 2011 at 14:11
  • $\begingroup$ Question is unclear but can be useful. Don't think need to close -- just downvote. $\endgroup$
    – Kostya
    Feb 10, 2011 at 9:36
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    $\begingroup$ @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" $\endgroup$ Feb 10, 2011 at 10:53

5 Answers 5

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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 $E_{2y} = E_0\,\cos(\omega t+\Delta)$. Here $\Delta$ 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 the average is defined as $\langle f(t)\rangle = \frac{1}{T}\int_0^{T}\,dt\,f(t)$ (so that $\langle \cos^2(\omega t+\Delta)\rangle = \frac{1}{2}$ for any $\Delta$).

Finally we've 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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  • $\begingroup$ Nice way of explaining this, definitely more thorough than my answer. $\endgroup$
    – Colin K
    Feb 9, 2011 at 17:19
  • $\begingroup$ Agreed, +1 for content and style. (tiny typo in the first line though: "not to be unsubstantiated") $\endgroup$
    – qftme
    Feb 9, 2011 at 18:06
  • $\begingroup$ Thank you all for feedback! By my answer should be considered as a continuation of @Colin K. $\endgroup$
    – Kostya
    Feb 9, 2011 at 18:21
  • $\begingroup$ 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! $\endgroup$ Feb 9, 2011 at 20:54
  • $\begingroup$ That is what he said, and he explained why it is the case as well. $\endgroup$
    – Colin K
    Feb 10, 2011 at 4:00
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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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  • $\begingroup$ 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. $\endgroup$ Feb 9, 2011 at 19:35
  • $\begingroup$ you should mention something like coherence and same-wavelength-interference-only $\endgroup$ Feb 10, 2011 at 7:29
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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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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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    $\begingroup$ -2 for what? Didn't I answer the question? $\endgroup$ Feb 9, 2011 at 18:59
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    $\begingroup$ 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 $\endgroup$ Feb 10, 2011 at 7:28
  • $\begingroup$ 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. $\endgroup$ Feb 10, 2011 at 11:02

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