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Optics and a thin wedge

  1. Aug 3, 2010 #1
    I'm having trouble understanding this problem. I think I'm missing something or misunderstanding the question:

    A wedge-shaped film of air is made by placing a small slip of paper between the edges of two pieces of glass as shown below. Light of wave-length 600nm is incident normally on the glass, and interference fringes are observed by reection. If the angle Theta made by the plates is 3x10^-4 rad, how many interference fringes per centimetre are observed? (Hint: Use the small-angle approximation Theta = t/x.)

    My thinking is that there is a path difference of 3x10^-4 rad and it is similar to a double slit experiment (or single slit experiment) causing interference on a screen some distance L away. However we haven't been given any such distance.

    Here is a diagram: http://i.imgur.com/KX1C4.png
  2. jcsd
  3. Aug 4, 2010 #2


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    Homework Helper

    Ηι Malby, welcome to PF.

    The condition for the constructive interference is

    2*t = m*λ, where t is the thickness of the air wedge at a distance x from the point of contact of the glass plates.

    t/x = θ. So 2*x*θ = m*λ.

    Hence the number of fringes per unit length = m = 2*x*θ/λ.
  4. Aug 4, 2010 #3
    In this case isn't m the fringe number and not the fringe number per unit length? I stumbled upon another answer that said m/x was 2*theta/Lambda which gives you a value. I'm not sure about that though because surely you need to know the distance the "viewing screen" (in this case the glass) is away from the source.
  5. Aug 4, 2010 #4


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    2*x*θ = m*λ.

    If β is the fringe width ( distance between successive bright or dark fringes) then

    2*(x+β)*θ = (m+1)λ

    So β = λ/(2θ)

    And number of fringes per unit length is

    1/β = 2θ/λ
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