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Homework Help: Light wave inteference

  1. Apr 23, 2009 #1
    I feel so silly right now, because this problem is the 1st problem of my physics problem set(and easiest), and I've done all the rest of the harder problems, but I can't get this simple problem!
    1. The problem statement, all variables and given/known data

    two in phase sources of waves are separated by a distance of 4.00m these sources produce identical waves that have a wave length of 5.00 m. on the line between them there are two places at which the same type of interference occurs. a) is it constructive or destructive interference and b) where are the places located

    2. Relevant equations

    l2-l1= m lambda (wavelength)
    l2-l1= (m+1/2) lambda

    3. The attempt at a solution

    l2-l1= 4 meters
    lambda = 5 meters
    m= 0.80 meters

    I'm so lost.
  2. jcsd
  3. Apr 23, 2009 #2
    Aw man, can no one help me?! This sucks!
  4. Apr 23, 2009 #3
    Why don't we do a drawing?

    Position two points, left and right, 4 metres apart.
    Draw a horizontal line between them

    From the left point draw 5 metre sine wave heading right.
    From the right point draw 5 metre sine wave heading left.

    Are there two places where the two waves are equidistant from the horizontal line?
    Can you calculate the distances?
  5. Apr 23, 2009 #4
    I don't see any points parts of the waves that are equidistant from the horizontal line
  6. Apr 23, 2009 #5
    Why don't you post your drawing?
  7. Apr 23, 2009 #6
    It's attached. :)

    Attached Files:

  8. Apr 24, 2009 #7
    I see my explanation was not clear to you. Check this diagram out.

    Attached Files:

  9. Apr 24, 2009 #8


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    This is wrong. Do you see the problem?
  10. Apr 24, 2009 #9
    Think about it this way...
    if they are in phase
    then you already know that the midpoint between them will be maximum constructive interference (since a peak from each wave will take the same amount of time to travel to the midpoint)
    Now, remember that the distance between 2 points of maximum constructive (or the distance between two points of perfect desctrutive) is wavelength/2
    and that a point of perfect descructive is halways between 2 points of maximum constructive, and vice-versa
    use these 2 facts to draw all the spots of maximum constructive and perfect descrtuctive, and just count them to see how many there are , and where they are
  11. May 2, 2009 #10
    Sorry guys, I'm just not getting it. Can someone explain it in numbers?
  12. May 2, 2009 #11


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    Since the point lies in between the two sources, we have

    l2 + l1= 4 m​

    where l1 and l2 are the distances to each source, and are both positive numbers.

    So, what are l1 and l2 when m=0 and

    l2-l1= (m+1/2) lambda​

    Then do the same for m = -1.
  13. May 3, 2009 #12
    l2-l1= 2.5m when m=0
    l2-l1 = -2.5m when m=-1

    ??? still doesn't get me the answer
  14. May 3, 2009 #13


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    Yes, okay. And we also know
    l1 + l2 = 4m​
    You have 2 equations with 2 unknowns here, so it is possible to find l1 and l2.
  15. May 3, 2009 #14
    l1=3.25 and l2=0.75

    Those are the answers! Thanks!

    BUUUT, how did you know l1+l2=4m
    how do i know the wave have destructive interference
    and how was I suppose to know I was looking for l1 and l2
  16. May 3, 2009 #15


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    We are not finished. :smile:

    All we have done, so far, is to find one location, in between the sources, where destructive interference occurs. That location is 3.25m from one source, and 0.75m from the other source.

    To see if there's another location, try the same procedure using m=-1. And remember, l1+l2=4m.

    l1 is the distance to one of the sources.
    l2 is the distance to the other source.
    Since we are considering locations directly in between the two sources, l1+l2 must equal the distance between the two sources. If you don't see that, draw a diagram.

    We didn't know that. We're finding the locations where destructive interference occurs if they exist.

    Huh? You're trying to find locations where there is interference aren't you? Those locations can be specified by the distance to each source, l1 and l2.

    If you haven't yet drawn a diagram indicating both sources, an unspecified location between the sources (where interference might occur), and also showing the distances l1 & l2 in the figure, I urge you to do that. Understanding what's going on starts with having that diagram. Without that diagram, this is pointless.
  17. May 3, 2009 #16
    I've drawn it. It just doesn't make sense to me.
  18. May 3, 2009 #17


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    The diagram looks something like this:
    Code (Text):

       <--- l1 ---> <------ l2 ------>

       :           :                 :
      1st     interference          2nd
    source      location          source
    Is it clearer now why l1+l2 = 4m?
  19. May 4, 2009 #18
    You already have the correct answers.

    In the attached diagram the two wave trains meet and interfere constructively at the dotted line in the centre of the diagram. The other two vertical dotted lines which are one quarter of a wavelength away from the central line represent the locations of destructive interference.

    Wavelength = 5m
    One quarter wavelength = 1.25m
    Central peak is at distance 2m
    2m + 1.25m = 3.25m
    2m - 1.25m = 0.75m

    Attached Files:

  20. May 4, 2009 #19
    Okay I think I get it now! THANKS SO MUCH for sticking with me. You guys are great!
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