In the MIT ME optics OCW slides, it is written that parallel polarized waves do interfere but perpendicularly polarized waves do not interfere. However, isn't circular polarization formed by the interference of 2 perpendicularly polarized waves with some phase difference?

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    $\begingroup$ To help frame the question: define "interference". $\endgroup$ Jul 14 at 8:12
  • $\begingroup$ Yeah, exactly. As far as I know, interference is the superposition of the amplitudes of waves. Then, what can one mean by "perpendicularly polarized waves do not interfere."? $\endgroup$
    – curiouss
    Jul 14 at 8:15

The issue is that "interference" can be taken to mean different things.

On a general footing, you can think of "interference" as any effect that results from the coherent combination of different waves. This covers e.g. the creation of circular polarizations by combining orthogonal linear polarizations at the correct relative phase.

But more specifically, "interference" is often understood in a more restricted sense as referring only to increases or decreases of the total observed intensity as a result of the superposition of two different waves. In this case, the MIT OCW resource is correct: constructive and destructive interference does not occur for perpendicular polarizations but it does occur for parallel polarizations.

The definition you gave in the comments,

interference is the superposition of the amplitudes of waves

is rather vague. Are you thinking of the amplitudes as vectors? Is the superposition required to increase or decrease the total (norm of the vector) amplitude? The answer to the question (i.e. whether perpendicular polarizations "interfere") will depend on how you specify those details.

(Or, in other words, it's all semantics.)

  • $\begingroup$ Is the circular polarization created, or was it there all along? $\endgroup$
    – user253751
    Jul 14 at 17:21
  • $\begingroup$ @user253751 It depends, doesn't it? If the two polarizations were created separately and then combined (as in e.g. this experiment), can you really argue that it was there all along? $\endgroup$ Jul 14 at 18:39
  • $\begingroup$ you can argue that a circular polarization is just a combination of two linear polarizations. $\endgroup$
    – user253751
    Jul 17 at 18:03

Electromagnetic waves will interfere with others if their direction of polarization is not perpendicular.

Note that the amplitudes of electromagnetic waves have a particular direction in space (perpendicular to the direction of propagation). The interference of such waves depends on their relative polarization directions, and the total intensity$^1$ for two linearly polarized EM waves is given by $$I = I_1 + I_2 + 2(I_1I_2)^{\frac{1}{2}}\cos(\Delta\phi)$$ where $\Delta\phi$ is relative angle of the two waves.

We can see that if $\Delta\phi=90^o$ the third term goes to zero (the intensities simply add up so that there is no interference, as oppose to the case where $\Delta\Phi=180^o$ in which case there will be interference).

And the superposition of two linearly polarized light waves with perpendicular polarization components, can result in linear, elliptical, or a circular polarized wave, but this depends on the amplitudes and the phase difference between polarization components of the two waves. See quarter wave plate.

$^1$ Intensity is proportional to the square of the amplitude.

  • $\begingroup$ If the given formula is for the general case, why do we say "Electromagnetic waves will interfere with others if their direction of polarization is the same." What is the precise definition of interference? Is it that the resulting wave must have the same polarization state as that of the constituting waves? $\endgroup$
    – curiouss
    Jul 14 at 9:43
  • $\begingroup$ Yes. It should technically be read as "Electromagnetic waves will interfere with others if their direction of polarization is not perpendicular". $\endgroup$
    – joseph h
    Jul 14 at 9:56
  • $\begingroup$ @curiouss I'm going to disagree with Joseph here. For water, sound, and other mechanical waves your OCE slides are correct. EM radiation is a very special case, for example all photons (regardless of polarization) passing in the DSE experiment "interfere" but the word interfere is from 1801 (Young's experiment) is way out of date. Photons need to find a path per Feynman theory before they ever travel, they act individually. Photons are created by electrons/atoms and are only absorbed by another electron/atom, they never cancel, that's a violation of conservation of energy. $\endgroup$ Jul 14 at 15:14

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