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Superposition, interference and mathematical description

  1. Apr 6, 2008 #1

    Air

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    I understand that this concept is about 2 (or more) waves meeting in constructive (when waves are added to form larger amplitude wave) or destructive (when waves are subtracted to form smaller amplitude waves) inteference but I don't really understand what path difference is about?

    I've read up and it shows me many numerical interpretations but I can't understand how I can relate it to superposition.

    It would be helpful if someone can explain the concept of path difference. Thank you in advance. :smile:
     
  2. jcsd
  3. Apr 6, 2008 #2

    Doc Al

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    Staff: Mentor

    Say you have two waves that converge at a singe point. To determine the resulting superposition, you need to know their relative phase. If you know their phase at some earlier point (their origin, perhaps) and how far they traveled to get to the destination, you can figure out their phase. If both waves started out with the same phase, you can determine their phase difference by comparing the difference in the paths that each took. (Think of the path length as measured in wavelengths.) The "path difference" will tell you their relative phase. If they started out in phase, but one traveled an extra quarter wavelength (say) the path difference would be [itex]\lambda/4[/itex], which corresponds to a phase difference of 90 degrees.

    So if two waves started out in phase and had a path difference of [itex]\lambda/2[/itex] when they arrived at the destination, they would be 180 degrees out of phase and would demonstrate destructive interference.

    Make sense?
     
    Last edited: Apr 6, 2008
  4. Apr 6, 2008 #3

    Air

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    Yes, It does. Thank you.

    But, I have one more question...
    If two waves are out out phase by [itex]90^{\circ}[/itex], would this cause constructive interference? I thought constructive interference is caused when the waves meet in phase so would this have the same affect but their amplitude would not be as large?
     
  5. Apr 6, 2008 #4

    Doc Al

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    Completely constructive interference would occur when the phase difference is an integral multiple of [itex]2 \pi[/itex] (360 degrees); Completely destructive interference would occur when the phase difference is [itex] \pi[/itex] (180 degrees). Any other phase difference--such as 90 degrees--would give you something in between.

    But you are correct that two waves out of phase by 90 degrees would have a resultant amplitude greater than either one but less than their sum.
     
  6. Apr 11, 2008 #5

    Air

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    Sorry, for bringing back this thread back but my question is related to path difference.

    Q: The path difference is [itex]2.22 \times 10^{-7}m[/itex] in the soap film. The wavelength of the yellow sodium light in the soap film is [itex]4.44 \times 10^{-7}[/itex]. Explain why the part of the film appears dark to the student.
    A: The answer said that the path difference is [itex]\frac{\lambda}{2}[/itex] so waves will meet in antiphase.

    Isn't [itex]\frac{\lambda}{2}[/itex] equal to [itex]90^{\circ}[/itex] out of phase so it wouldn't meet in antiphase? I agree there may be some destructive interference but I thought waves wouldn't meet in antiphase as it is [itex]180^{\circ}[/itex] out of phase.
     
  7. Apr 11, 2008 #6

    Doc Al

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    No, a complete wavelength equals 360 degrees of phase difference (and thus in phase), so [itex]\lambda/2 = 180[/itex] degrees (and thus perfectly out of phase).
     
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