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Double Slit with air block

  1. May 3, 2013 #1
    1. The problem statement, all variables and given/known data
    2a7sw88.png


    2. Relevant equations



    3. The attempt at a solution

    As air is gradually removed from one box, the fringe pattern shifts towards that box. Hence total number of maximas that pass through the central is calculated.

    vdcn5l.png

    However, according to http://hyperphysics.phy-astr.gsu.edu/hbase/tables/indrf.html the index of refraction of air is 1.0003
     
  2. jcsd
  3. May 4, 2013 #2

    rude man

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    This looks like a difficult problem since the optical path for the box with air varies with the number of the maximum. For the 16th maximum there should be no effect of the boxes (delta optical path = 0, angle close to 90 deg.) wheras for the first few the optical path thru the air-filled box varies appreciably from maximum to maximum. This would seem to hold even if d >> s, l.
     
  4. May 5, 2013 #3
    Hmmm, any ideas on how to solve this then?
     
  5. May 5, 2013 #4

    TSny

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    I think your analysis that leads to 1.0008 is correct, even though the answer does not correspond to the known value of the index of refraction of air at 1 atm.

    I'm not following rude man's comments. The problem states that we are only looking at one spot on the screen (z = 0) and measuring the number of maxima that occur at that fixed spot as air is removed from one of the boxes.
     
  6. May 5, 2013 #5

    rude man

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    I would assume the addition of a converging lens as is done in deriving the baic min-max pattern (fringes). Then, the delta distance between the top & bottom slits, which is basically d sin θ, has to be modified (increased) by the increase in optical path length thru the bottom box. This is primarily an analytic geometry problem, since each bottom ray forming another max. will pass thru different sections of the box. If that delta distance were the same for all rays forming maxima then it would be easy; but that's not the case.

    So instead of winding up with d sin θ = mλ you will have d sin θ + Δop = mλ where Δop = delta optical path for the lower rays & will be different for each ray forming a max.

    Δop = x(n-1.00000) and x is the physical path length of a bottom ray inside the lower cube. Unfortunately, x = x(θ). For the near-central max, θ ~ 0 and x ~ l, the length of a side of the box. For the outermost maxima, θ ~ +/-90 deg and x ~ 0.

    Like I said - difficult analyt geometry problem.
     
  7. May 5, 2013 #6

    rude man

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    I agree, I misinterpreted the problem. So it's far easier than I thought. Thanks for cluing me in, T.

    EDIT: P.S. I also got 1.0008 = 16λ/l - 1.

    PPS Well, maybe the experiment was performed in one of our smoggier areas ... :smile:
     
    Last edited: May 5, 2013
  8. May 5, 2013 #7

    TSny

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    :biggrin:
     
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