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I Out of phase light/photons, would it be measurable?

  1. Dec 22, 2016 #1
    With sound you can create out of phase signals and they cancel out, this is not possible with light as it does not interact like sound does.

    If it was possible to create two light sources in that were exactly 180 degrees out of phase with each other and aim them at a common point (eg an isosceles triangle, the two points of identical angle are the light source, the 3rd point is the measurement point) regardless of the distance from the source to the point on which they combine the wavelengths would appear to cancel would they not? If they were of identical frequency and amplitude they would never depart from each other and would continue to appear "cancelled" at every point of measurement despite the photons still being there and containing energy.

    The light would not be coherent until the meeting point, this is the point upon which you would attempt to measure the frequency/amplitude. Would it appear to be zero?

    How would you actually produce an experiment like this, eg how could you lag the phase of a light signal, does it even make sense to attempt something like that?

    The reason for the question is I'm wondering if you could hide the fact a device/experiment is releasing energy by making all measurements appear null.
     
  2. jcsd
  3. Dec 22, 2016 #2

    Dale

    Staff: Mentor

    No, if you have destructive interference at one location then you will have constructive interference at another location.

    The geometry you describe is the standard geometry for a double slit experiment. But with the regions of constructive and destructive interference swapped.
     
  4. Dec 22, 2016 #3
    Interference like you describe is not difficult with light. The very thing you describe: creating co-propagating waves out of phase so that they cancel everywhere is exactly how dielectric mirrors work. Overlapping multiply reflected beams traveling forward cancel everywhere, so the beam has no choice but to reflect. Actually the interference is constructive for the reflected beam. As Dale points out, this is a fundamental principle. The energy can't just disappear. No arrangement of interference can make the beam eliminate itself. What's lost in destructive interference has to show up somewhere in constructive interference.

    The case you describe of bringing beams together at an angle to interfere at an overlap point is how polar illumination in a photolithography machine makes the nm scale features on your computer chip. However due to the angle you don't get cancelation across the focal plane. You get modulation. Technically interference is how all photography and in fact all optics work.

    There are examples of light engineering for interference all around. Iridescent paint, the diffractive optic element on your laser pointer that makes a funny shape, the hologram on your credit cards.
     
  5. Dec 23, 2016 #4

    ZapperZ

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    Er...yes it does!

    Send a beam into an interferometer, adjust one arm of the interferometer so that you get a destructive interference, and voila!

    Zz.
     
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