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Homework Help: Air Wedge

  1. Jun 26, 2010 #1
    1. The problem statement, all variables and given/known data
    A ray of light consisting of 2 monochromatic wavelengths, 550 nm and 672 nm, is incident on an air wedge. It is found that, at the nth bright fringe from the vertex of the air wedge for the 550 nm wave, a bright fringe for the
    672 nm wave is also formed. Determine the smallest value of n.

    2. Relevant equations
    x= λ/(2 tan θ)

    3. The attempt at a solution
    For the wave of 550 nm,
    2t = (n-0.5)x , where t is the thickness at a point of the air wedge

    For the wave of 672 nm,
    2t= (m-0.5)x'

    Since it's the same air wedge i.e. t is the same for both equations,we can equate the 2 equations.

    But I can't find another equation relating m and n.

    The solution given is:

    Distance of nth bright fringe of wavelength λ from the vertex = (n-0.5)x

    Distance of (n-1)th bright fringe of wavelength λ' from the vertex


    My question is, can we consider the fringes of both waves that coincide at a point of the air wedge, nth and (n-1)th fringes from the vertex? Shouldn't it be nth and mth fringes from the vertex since they have different wavelengths?

    Is my approach to the question correct?
  2. jcsd
  3. Jun 27, 2010 #2
    Yes, you're right. And the solution is right, too.They are nth and (n-1)th fringes because n is supposed to be minimum. Think about it :smile:
  4. Jun 27, 2010 #3

    Doc Al

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    Staff: Mentor

    Remember that you are trying to find the smallest value of n for which the bright fringes overlap. So you start with m = n - 1 and see if there's an integral solution. Of course, the problem is set up so there is! :wink: If that didn't work, you'd try m = n - 2, then m = n - 3, and so on.
  5. Jun 27, 2010 #4


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    Science Advisor
    Homework Helper
    Gold Member

    Are you asking why m should differ by one from n and not by some other arbitrary integer? Draw yourself a number of equally spaced arbitrary lines separated by, say, 1.0 cm. With the same starting point, draw underneath them a second set of lines separated by 0.9 cm. Note that when you move to the tenth mark on the bottom set you are 9 cm from the starting point which is the ninth mark on the top set. The marks will also coincide at 18 cm and so on but the smallest value is 9 and 10. If you understand this, you should also understand why m and n must differ by one.

    The situation in this problem is slightly more complicated because the maxima don't start at the same distance from the vertex, but the given solution takes care of that.
  6. Nov 21, 2010 #5
    Is there a mathematical proof for this?
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