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Phase Difference of a light wave

  1. May 14, 2013 #1
    1. The problem statement, all variables and given/known data
    Light with wavelength 540 nm is split into two beams that travel along two paths. The difference between the path lengths is 3600 nm. What is the effective phase difference when the light recombines?


    2. Relevant equations
    [itex]\phi[/itex] = (2pi/λ)*ΔL


    3. The attempt at a solution
    It seems like a simple problem, I plug in 540 for λ and 3600 for ΔL but I'm not getting the right answer.
     
  2. jcsd
  3. May 14, 2013 #2

    cepheid

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    Show us your calculations. What answer are you getting?
     
  4. May 14, 2013 #3
    [itex]\phi[/itex] = (2pi/(540x10^-9) * 3600x10^-9 = 41.88

    answer is 120°
     
  5. May 15, 2013 #4

    cepheid

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    Sure, the phase difference is 41.88, which is a huge number, but remember that phase is cyclical. If the phase difference were 4pi, would this be any different from the case where the phase difference were 2pi, or 0? No, because all of these correspond to a shift by a whole number of cycles. A full cycle causes you to increase in phase by 2pi RADIANS.

    Note: the word in all caps above also hints at a second error you were making, involving units. So, the question you have to answer is, how far out of phase are they two waves actually, given that 41.88 radians corresponds to a whole number of cycles + some excess? It's this excess (which is less than a full integer of cycles) that you're interested in, because it tells you how out of phase the waves are.
     
  6. May 15, 2013 #5
    Got it, thanks.
     
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