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Waves-Path and Phase Difference Relationship

  1. Sep 13, 2006 #1
    hi all.

    i'm a college student doing A-Levels this year and i am studying a subtopic under superposition of waves.

    i have trouble understanding a table of relationship between path difference and phase difference.i haved asked my lecturer but i still dont get the table.

    could someone actually explains the table below for me?thanks alot :blushing:

    [​IMG]
     
  2. jcsd
  3. Sep 13, 2006 #2

    jtbell

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    Staff: Mentor

    What is it that you don't understand about the table?
     
  4. Sep 13, 2006 #3
    erm..i dont get the whole thing

    its like why when the source in antiphase is pie,its phase difference is 0 wavelength?

    why when the source in antiphase is 2 pie,its phase difference is 0.5 wavelength?

    i just need to understand one in order to get the whole list.

    and when the source in phase is constructive,why the source in antiphase is destructive?

    hope someone can help me out.been trying to figure out by drawing but really have no idea how to start.
     
  5. Sep 13, 2006 #4

    Doc Al

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    Do you understand that one wavelength is equivalent to 2 pi radians? (Or 360 degrees.) So being out of phase by a half wavelength is the same as being out of phase by pi radians or 180 degrees.

    Read this: Interference and Phase
     
  6. Sep 13, 2006 #5
    yeap.i do know that 1 wavelength is equal to 2 pi rad.

    i dont get the part where when sources of antiphase is 2pi,why is the path difference is 0.5wavelength instead of 1 wavelength?
     
  7. Sep 13, 2006 #6

    jtbell

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    You're referring to the second line in the table, right? Try reading it like this: if the path difference is [itex]0.5 \lambda[/itex] and the sources are in antiphase (in English we usually say "out of phase", I guess your instructor's native language isn't English), then the net phase difference when the waves meet is [itex]2 \pi[/itex] radians.

    When the waves start out, from two different sources, they already have a phase difference of [itex]\pi[/itex] radians because the sources are out of phase. They travel different paths to the point where they meet, and those paths are different in length by [itex]0.5 \lambda[/itex]. This introduces another [itex]\pi[/itex] radians of phase difference, for a net phase difference of [itex]2 \pi[/itex] radians.
     
  8. Sep 13, 2006 #7
    owh..ok..i got what you mean.i think i understand it by theory that sources out of phase started out with 1pi and end with another pi thats its why it is always 1pi more than source in phase in the list.

    but i still couldnt picture the diagram of the situation that you just explained to me.

    i hope someone can draw a diagram for that situation where sources out of phase started out with 1pi and end with another pi.cause i really dont know how to start.

    and thanks alot for explaining the situation for me.really appreciate it.

    thx..
     
  9. Sep 13, 2006 #8

    Doc Al

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    Staff: Mentor

  10. Sep 13, 2006 #9
    thanks so much! :smile:
     
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