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Michelson and Energy Conservation

  1. May 22, 2012 #1


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    I'm wondering about the Michelson interferometer and energy conservation. It seems like if one arm has an optical path difference that results in a pi phase shift relative to the other arm, then there will be complete destructive interference at the output port. Talking to some people, I gather that the solution to the problem of energy conservation is that the beams in the interferometer are spatially extended, and hence what you get at the output port is actually a central dark fringe, followed by a series of bright and dark fringes as a function of angle away from the optical axis. In other words, this is supposed to be no different from the Young double slit.

    My question is this: what happens if you send perfectly collimated (and monochromatic) light through? Shouldn't there be no light at the output port? If so, then energy conservation would lead me to expect that the light rays that go from the beam splitter and end up coming out back the way they came (i.e. through the input port) should interfere constructively in order to conserve energy. But I can't see how this is possible given the geometry. The two outgoing rays that meet back at the input port also have a pi phase shift relative to each other, just like the two rays that meet at the output port.

    I should know the answer to this, but it's been a while for me since taking Optics.
  2. jcsd
  3. May 22, 2012 #2


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    It's not possible to have perfectly collimated light with a finite sized beam. You hit the diffraction limit. There's always a fringe.

    At small distances, Fraunhofer diffraction breaks down, and you can have situations where the light is reflected back to the source. This is how interference filters or thin film coatings work.
    Last edited: May 22, 2012
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