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Ideal destructive interference of two beams - Where is the energy?

  1. Dec 5, 2009 #1
    Two perfectly coherent laser beams point at two different positions on a screen. The energy transmitted by each of the beams is dissipated at the screen - it's either reflected (and thus visible to an observer) or becomes a potential energy of some sort.

    Now both beams are pointed at a single spot so that they interfere complete destructively. Where does the energy which is inserted at the one end of the laser go now?
     
  2. jcsd
  3. Dec 5, 2009 #2

    Vanadium 50

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    This doesn't happen the way you describe it. You have an interference pattern with regions of constructive interference and regions of destructive interference.
     
  4. Dec 5, 2009 #3
    We are not talking about some slit-something experiment. Why would you expect an interference pattern?
     
  5. Dec 5, 2009 #4

    Vanadium 50

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    You're the one who is claiming there is interference. Where there is interference, there is an interference pattern, no?
     
  6. Dec 5, 2009 #5
    Vanadium 50 is correct. When the experiment described by ManDay is performed, you get rings of constructive and destructive interference. Actually the experiment is usually performed with a single laser beam that is split by a partially reflective mirror. This is how an interferometer works.
     
  7. Dec 5, 2009 #6
    I'd agree with this, but he could get the laser 'spot' completely cancelled if he had a reference beam aimed at the screen from the other side (ie, holograpic principle).
     
  8. Dec 5, 2009 #7

    Vanadium 50

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    If the screen is in the x-y plane, this gives you a standing wave in z. If you move the screen a quarter wavelength in z, you get a spot that's 4 times as bright as a single laser. And that's where the energy goes.
     
  9. Dec 6, 2009 #8
    I think I know where you are coming from. Bottom line, the laser power does not disappear, it is never generated in the first place!!!!

    If you change this to one laser, passing through a Mach-Zhender interferometer, sending two beams at the same point (and on the same vector) on the paper screen, then (with ideal optics), it is possible to adjust the delay of one of the arms to cycle through a single "bright fringe" and a single "dark fringe". In the case of a dark fringe, the laser must stop emitting!! The caveat of "ideal optics" is important. Stopping the laser from emitting isn't easy!

    Experiments in the 1990's proved this with a single atom located in a cavity null (dark fringe location). The atom stayed in its excited state for much longer than it's "natural" lifetime. It couldn't emit the (narrowband) photon since it would not satisfy the cavity requirement that there be an integer number of optical wavelengths (for this particular photon) in a round-trip of the cavity.

    It was explained to me that the vacuum fluctuations are "responsible" for atomic transition rates but in this case, there was a node for the vacuum fluctuations also.

    I just did a search on this topic and found what looks to be a nice survey of experiments and theory of the new field of "Cavity Quantum Electrodynamics".
    http://pdfserve.informaworld.com/192957__713825121.pdf
     
    Last edited by a moderator: Apr 24, 2017
  10. Jan 11, 2010 #9
    thanks TakeTwo, very constructive post and a good point for me to start research from (pdf link dead tho). Vanadium, no offense, but no: If there is interference there is not necessarily an interference pattern. The rings in differen interferometers emerge from simple classical optics, because to the outer perimeter of the ring its a longer distance than to the center. take a laser for instance, which is perfectly aligned and parallel: if you adjust the mirrors at both ends to be at a distance of .. 1/2 the wavelength of the wave in the medium, the laser will output no light! no interference pattern at all - simply no light due to interference.
     
  11. Jan 16, 2010 #10
    For a gain medium with a bandwidth smaller than the longitudinal mode separation (delta_nu), ManDay is correct. For example, a CO2 laser with collisional broadening of 5-6 MHz per Torr, at 20 Torr, the bandwidth is ~ 100 MHz. Put this medium into a 1m long cavity, delta_nu = c/2L = 150 MHz. With a piezo-driven mirror, the laser output will oscillate from max to near zero (curved mirrors needed for cavity "stability" => c/2L is not a delta function) as the cavity moves 1/2 lambda_0 =5.3 microns, in the CO2-laser case with lambda_0 = 10.6 microns. Near lasing threshold, the output will drop more than we can measure with the dynamic range of the detector (> 30 X or so) when the "round-trip condition" is violated by having n+1/2 lambda_0 wavelengths within the cavity round-trip.

    Why do we care that one longitudinal mode exists? With one mode, we get a nice, smooth output. If we tune 1/4 wavelength away, we get a sinusoidal output (squared) as two longitudinal modes of difference frequency c/2L interfere. This is useless for our application, so we always try to keep the cavity length L tuned to (N+1/2) of the the laser wavelength where N ~ 1 meter / 10.6 microns ~ 10^5.

    Back to "Cavity Quantum Electrodynamics", here is a currently working link...
    http://www.rp-photonics.com/spotlight_2006_08_01.html
    I take it that these effects show up for N ~ 10^2, so these are small cavities but not as small as the Casimir-effect cavities ; ).
     
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