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Interference fringes between rectangular pieces of glass

  1. Mar 21, 2009 #1
    hello again folks, this is the last problem that I haven't gotten in this week's homework. see attached image.

    1. The problem statement, all variables and given/known data
    Two rectangular pieces of glass (see attached image, I painstakingly made it with a very fussy/crash-ey program) are laid on top of one another on a plane surface. A thin strip of paper is inserted between them at one end, so that a wedge of air is formed. The plates are illuminated by perpendicularly incident light of wavelength 595 nm, and 18 interference fringes per centimeter-length of wedge appear. What is the angle of the wedge?


    2. Relevant equations
    x/y = L/d (similar triangles)
    2y = (m + 1/2) * lambda
    Tan theta = (y / x)

    3. The attempt at a solution
    calculating x' from x(m+1) - x(m), I get x'=(L/d) * (lambda/2) . the change in x, x', I think is the space between the fringes. but whether this is 1/ (18/(10^(-2))) or 18/(10^(-2)) I'm not sure. The latter is way too big, and resulted in an incorrect answer. The former is ultimately 1/1800, which seems intuitively right.

    so 1/1800 = (L/d) * (lambda/2).
    L/d = 2/(1800*lambda)
    d/L = (1800*lambda)/2
    theta = Arctan (d/L) =Arctan ((1800*lambda)/2)

    does this look right? I haven't calculated or submitted this answer, as i wanted to wait and think it over before wasting any more of my last 3 chances to get it right.

    Thanks for any tips or suggestions.
    ,Yroyathon
     

    Attached Files:

  2. jcsd
  3. Mar 21, 2009 #2

    Delphi51

    User Avatar
    Homework Helper

    I think you have it right.
    You get a bright fringe when the thickness is nL/2 (using L instead of lambda).
    thickness = x*tan(A) so x = nL/(2*tan(A)) and the horizontal distance between fringes is
    dx = .5*L/tan(A)
    Your dx is .01/18 = .5*L/tan(A)
    tan(A) = 900L
     
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