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Path difference and phase difference.

  1. Jun 14, 2004 #1
    Pardon the pun, but what's the difference between path difference and phase difference (when talking about interfering light waves)?
  2. jcsd
  3. Jun 14, 2004 #2
    Path difference is the distance one wave (from a coherent source) has to travel from its source to the observer. ie one observers ear may recieve the crest and the other ear recieve the trough. this would give a path difference of 1/2 wavelength.
    Phase difference is best understood by considering a cosine wave and a sine wave. they are 90degrees out of 'synch' with each other and would produce destructive interference.
    i hope someone with more brains than me can clarify your quandary. I think i am corect but i am sure someone here can be more specific.
  4. Apr 16, 2008 #3
    I believe the equation would be phase difference = 2pi/lambda*path difference
  5. Apr 17, 2008 #4

    Andy Resnick

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    As a beam traverses space, the phase of a ray goes as [tex]\phi (z) = k n z [/tex], where k is the wavevector, n the refractive index, and z the distance. The quantity nz (or nd, where d is the thickness of an object) is referred to as the optical thickness. It's easy to make the sitation more complicated- make the refractive index vary with location, for example. Or make it vary with wavelength. Then the phase delay *relative to another ray*, which is the important thing, is given by more complex versions of the above formula.

    Even so, there is no difference between path and phase difference, when you are able to sensibly speak of rays. This is not always the case- strongly focused light, for example. Simply considering polarization can create difficulties, forcing you to keep track of the propogation axes by hand.
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