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Interference fringes with reflected light

  1. Dec 3, 2006 #1
    1. The problem statement, all variables and given/known data

    A point source of light (λ = 589 nm) is placed 0.4 mm above the surface of a glass mirror. Interference fringes are observed on a screen 6m away, and the interference is coming from light reflecting off the mirror and light coming directly from the source. Find the spacing of the fringes

    2. Relevant equations

    reflection from the mirror will cause a phase shift of pi

    the phase is also shifted due to the path difference:
    (Δr)(2pi)/λ

    so total phase shift is pi + (Δr)(2pi)/λ

    3. The attempt at a solution

    I can calculate interference at the point on the screen directly in front of the point source:

    pi + (Δr)(2pi)/λ = pi + (0.8 mm)(2 pi) / 589 nm

    but i'm not sure where to begin finding the fringe spaces on the screen. I can't come up with a function to relate interference on the screen to the path difference caused by the mirror. Any hints would be appreciated!
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Dec 3, 2006 #2

    OlderDan

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    By "directly in front of" do you mean the direct ray is horizontal so you have two right triangles to work with? If that is the case you know how to do, then probalby what you need is the law of sines or law of cosines to tackle the non-horizontal cases.
     
  4. Dec 3, 2006 #3
    Well, I made a diagram:

    [​IMG]

    I thought it might help, but I still haven't even figured out where to start.
     
  5. Dec 4, 2006 #4

    OlderDan

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    I think your diagram is not correct. I interpret the problem to mean the mirror is laying horizontally with the source slightly above it. The direct and reflected rays are nearly horizontal propegating to the wall. The angles of incidence and reflection at the mirror are equal. Drawing the diagram with well spaced lines and larger than expected angles is fine, but I think the .4mm is a vertical distance while the 6m is a horizontal distance.

    So, more or less, if you put your source at the left wall and the mirror where you have the source, make the vertical distance from the source to the mirror .4mm and the horizontal distance from the source to the right wall 6m, and make equal angles at the mirror, you have it. Then you need to compare the direct path length to the total (two legs) reflected path length and account for any phase reversals.

    Before you get too carried away with the calculations, see if you can think of any connection between this and a double slit problem. Where is the image of the source in the mirror?
     
    Last edited: Dec 4, 2006
  6. Dec 4, 2006 #5
    You just made me slap my forehead! Thank you kindly for pointing out the obvious :)
     
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