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Wavelength difference

  1. Feb 13, 2016 #1
    1. The problem statement, all variables and given/known data

    2vlnwxj.jpg

    2. Relevant equations


    3. The attempt at a solution

    (Sorry my poor English). I just don't understand why should the minimum path difference be 0.75 wavelengths for angles below the centerline instead 0.25 wavelenghts as it is for angles above the centerline.
     
  2. jcsd
  3. Feb 13, 2016 #2

    TSny

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    For angles below the centerline, is the path length from A to the detector greater or less than the path length from B?
     
  4. Feb 14, 2016 #3
    It's greater, but how would it change the expression for interference?
     
  5. Feb 14, 2016 #4

    TSny

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    Yes, the the path from A is greater when the angle is below the center line.

    The wave from A has a "head start" compared to the wave from B due to the vibration of the speaker at A being 1/4 period ahead of the vibration at B. When the waves arrive at the detector, does the extra distance for A tend to make the phase of the wave from A even farther ahead of B or not as much ahead?

    You might consider the case where the extra distance for A is 1/4 wavelength. A sketch will help.
     
  6. Feb 14, 2016 #5
    (Sorry my poor English). Okay, but instead adding some value, it seems we have to subtracting one so that the right side of the equation equals to 0.75 wavelength

    Xa - (1/2)λ - Xb = λ/4
    Xa - Xb = 3λ/4
     
  7. Feb 15, 2016 #6

    TSny

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    I'm sorry, but I'm not quite following what you did here.

    The extra distance that the wave from A travels will shift the phase of A in a direction that makes the phase of A "behind" the phase of B. But the wave from A started out 1/4 λ ahead of the wave from B. So, if the extra path distance that A has to travel is 1/4 λ, the waves from A and B will end up in phase at the detector.

    You need to find the path difference in order for A and B to end up out of phase at the detector.
     
  8. Feb 15, 2016 #7
    oh thank you TSny I understand it now. For any point above the centerline the distance from A will be always smaler than the distance from B. If waves from B start 1/4λ behind of waves from A, and if the distance from B is 1/4λ greater than distance from A, we add 1/4λ to find the position of the wave from B, wich will be 1/4λ + 1/4λ = 1/2λ out of phase with waves from A. Analogous situation occurs from points below the centerline. Am I right?
     
  9. Feb 15, 2016 #8

    TSny

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    Yes. That's right.
     
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