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We know that two wave sources can interfere if

  1. two sources have the same frequency
  2. two sources must have a constant phase difference over time.

But I think that if condition 1 is satisfied, condition 2 is too. So is the second condition unnecessary? There must be some counter-example, isn't there?

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When we say two sources have the "same" frequency, there is often a limit on what "same" means. Specifically, we usually recognize coherence as an essential attribute for creating interference.

Let's look at the example of a laser. To create interference, you can split a laser into two beams, then let these beams interfere with each other. With the right setup, you get fringes. But if you add an additional (constant) path to one "leg" of the laser light, you will see that as the total path difference gets bigger, the fringes become less well defined.

In essence, what is happening in that example is that you start to measure the limits of coherence of your laser - at what time/path difference is the phase relationship between the two beams still well-defined? Tiny fluctuations in the frequency eventually build up and cause "decoherence" at sufficiently long path lengths. This mechanism can be used to measure just how much a frequency source fluctuates - the width of the autocorrelation of the signal tells you about the bandwidth.

The key here is that there is no such thing as "monochromatic" light, or a source of "constant" frequency. Every signal made in nature has a degree of incoherence - whether it be Doppler shifting due to thermal motion, or any one of a myriad of other causes. Great care is taken to make some very stable oscillators - the Cesium clock standard is at an error rate of less than a second in 100 million years - but that is still not "constant". So even two Cesium clocks will eventually drift; and oscillators derived from these clock will not guarantee a constant interference pattern.

The way this is usually solved is by making sure that path lengths from the source to the point of interference are kept constant. This way, any phase error in one arm is replicated in the other arm, and the interference pattern is preserved.

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    $\begingroup$ Thta's doing things the hard way... $\endgroup$ Aug 30 '15 at 12:56
  • $\begingroup$ @CarlWitthoft - rewritten. Better? $\endgroup$
    – Floris
    Aug 30 '15 at 13:07
  • $\begingroup$ Newton's rings are quite visible in white light -- it just requires reasonably careful setup to stay within the coherence length of an incandescent source. $\endgroup$ Aug 30 '15 at 13:12
  • $\begingroup$ Exactly. Newton's rings is an example of decoherence - the first couple of fringes are visible as all frequencies have zero phase difference initially, but then the rings explode in color as different fringes have different spacings. $\endgroup$
    – Floris
    Aug 30 '15 at 13:57
  • $\begingroup$ Apologies, Floris, for sounding accusatory. I got dragged off to other stuff for a while. Your answer looks good, although my interpretation of coherence length is not so much a change or drift in laser wavelength as a simple discontinuous "hop" in output phase that occurs inside the resonant cavity. $\endgroup$ Aug 30 '15 at 15:03
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Most light sources do not emit a constant phase, unlike radar or audio sources. So while all the light may be of a given wavelength, the random (in time) phase shifts can kill interference.

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  • $\begingroup$ Strictly speaking if the frequency is constant the two sources must be coherent. What you are describing is incoherent light "of the same frequency" - which is not really of the same frequency (except in short bursts). Perhaps a detour on the topic of coherence is in order to fully answer this question. $\endgroup$
    – Floris
    Aug 30 '15 at 12:58
  • $\begingroup$ Floris, see my comment to your post. There really is a difference between frequency matching and phase matching. $\endgroup$ Aug 30 '15 at 15:04
  • $\begingroup$ This could become a lengthy argument... but bottom line is that any kind of "phase jitter" (whether continuous or discrete) results in frequency spreading: if you take the Fourier transform of a jittery signal you see peak broadening. Which is tantamount to saying it consists "not solely of one frequency" - although over a short interval it looks a lot like it. $\endgroup$
    – Floris
    Aug 30 '15 at 18:50

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