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Interference on a mirror -- waves confusion

  1. Nov 15, 2016 #1
    1. The problem statement, all variables and given/known data
    JZ4fz1F.png
    ##d=0.52 cm,## ##\lambda=431nm##, ##L=119 m##
    2. Relevant equations
    ##dsin\theta=(m+1/2)\lambda## for constructive/bright fringes since there is a pi shift already

    3. The attempt at a solution
    a) $$sin\theta\leq1$$
    $$(.0052/431e-9)-0.5=m=12064$$ 2 \e{3}
    $$ total bright fringes = 2m = 24129$$

    However, I am extremely confused on why my solutions sets ##d=.0052(2)## which gives ##m=24129## from the get go.
    XjYzaxL.png
    Isn't there a mirror underneath? I don't know where this solution is coming from which is messing up my numbers for parts b and c. Is there something I'm missing? The main thing that I don't get is why ##d=.0104m##
     
    Last edited: Nov 15, 2016
  2. jcsd
  3. Nov 15, 2016 #2

    Simon Bridge

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    The point source S is distance L from the screen and distance d above the mirror.
    The interference is between this source and it's image S' in the mirror ... what is the separation between the source and it's image?
     
  4. Nov 15, 2016 #3
    Oh ok. separation between source and image is .0104m. But why do we have to take into account the image in the mirror? I'm not sure how that's relevant in the problem
     
  5. Nov 15, 2016 #4

    Simon Bridge

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    The image in a mirror is the place where light rays appear to radiate from ... this means the image acts like another source.
    So we can replace the physical setup with one that does not have the mirror, but there is a second source identical with the first but with opposite phase... this new setup will produce the exact same interference pattern as the one with the mirror.
    Then do the maths for this setup - remember that there are no fringes in the lower half though.

    You asked why d=0.0104m i the solution - and that is because the d in ##d\sin\theta =(m+\frac{1}{2})\lambda## is not the same as the d in the diagram.
    0.0104m is the separation between S and it's image.

    You don't have to do it that way though - you can just work out the equation for the path difference and derive the result.
     
  6. Nov 15, 2016 #5
    I see. So this is just a slightly different type of double slit problem in which the equations are switched between destructive and constructive because there is a pi shift in one of the sources, providing a relative pi phase shift.
     
  7. Nov 15, 2016 #6

    Simon Bridge

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    Well done ... it's the equivalent to "method of images" in electrostatics.

    The trick to these sorts of things is to be careful about deriving the equations you use.
    If you can replace the setup you have with an equivalent that you already have the answers to, then so much the better.
    Why do more maths than you have to?
     
  8. Nov 16, 2016 #7
    That's true. Thank you so much for helping me out.
     
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