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About interferometer

  1. Dec 21, 2008 #1

    KFC

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    I am reading a scientific essay about a experiment with interferometer. They use so called 50/50 beam splitter and perfect reflector. Well they have the input wave be the plane wave [tex]exp(ikx)[/tex], what interesting is: whenever the wave reflected by the splitter or perfect reflector, it add half of the wavelength to the wave, but if the wave was transmitted through the splitter, no wavelength was added to it. It is so confusing. What's the rule behind this?

    1) When will the half wavelength being added to the wave? Only happen in reflection?
    2) If the reflector is not perfect and the splitter is not 50/50 (30/70 or 80/20), will it still be a half-wavelength added? or any other portion of wavelength being added?
     
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  3. Dec 22, 2008 #2

    jtbell

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    Only in reflection, and only when the index of refraction (n) of the material "after" the surface is greater than the index of refraction "before" the surface. E.g. when the light is traveling in air and hits a glass surface, the reflected wave has the half-wavelength added, but when the light is traveling in glass and hits a surface with air on the other side, this does not happen.

    The portion of the wave that is reflected has the half-wavelength added (provided the index of refraction on the two sides is as described above), the transmitted portion of the wave does not (regardless of how the index of refraction goes).
     
  4. Dec 22, 2008 #3

    Andy Resnick

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    It's not exactly adding a half-wavelength, it's more correctly a reversal of the wavefront- equivalently, the phase is advanced 1/2 of a period when reflections occur from some interfaces, while the phase is not advanced when reflection soccur from other inteerfaces:

    http://www.kettering.edu/~drussell/Demos/reflect/reflect.html
     
  5. Dec 22, 2008 #4

    jtbell

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    Yeah, some books handle it as a phase shift of [itex]\pi[/itex] radians (which is what it "really" is), and some books handle it by adding [itex]\lambda / 2[/itex] to the optical path length, which is equivalent.
     
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