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Interference Maxima and Newton's Rings (Please help with trig!)

  1. Apr 21, 2012 #1
    1. The problem statement, all variables and given/known data
    A lens with radius of curvature R sits on a flat glass plate and is illuminated from above by light with wavelength λ (see picture below). Circular interference patterns, Newton's Rings, are seen when viewed from above. They are associated with variable thickness d of the air film between the lens and the plate. Find the radii r of the interference maxima assuming r/R <<1.


    2. Relevant equations

    2L = (m + 1/2) λ

    3. The attempt at a solution

    I understand that we will use 2L = (m + 1/2) λ here. However, I can't figure out how we relate d to R or r. I have my professor's answer key, and he defines θ as the top angle in the picture (formed by R and the normal line to the glass surfaces). He then says that r/R = sinθ which equals θ. Then, he says d = R(1-cosθ) and uses the expansion of cosθ = 1 - θ^2/2 + θ^4/4!, etc.

    I can't wrap my head around how he found d = R(1-cosθ). If someone can help me see it, I would be greatly appreciative.

    Thanks!
     

    Attached Files:

  2. jcsd
  3. Apr 21, 2012 #2

    tiny-tim

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    hi mm2424! :smile:
    shift that little arrow to the middle of the picture …

    it's the gap between the arc (of the circle) and the triangle …

    = R - Rcosθ :wink:
     
  4. Apr 21, 2012 #3
    I'm still missing something, haha. I'm not sure what you mean by shift the arrow to the middle of the picture, and I'm not clear on where Rcosθ comes from. Is it some type of trig relationship involving arcs?
     
  5. Apr 21, 2012 #4

    tiny-tim

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    you have a right-angled triangle with hypotenuse R and top angle θ …

    so the vertical side has length Rcosθ :smile:
     
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