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Interference and Defraction

  1. Aug 6, 2008 #1
    1. The problem statement, all variables and given/known data
    7: a: A thin film of oil (n2=1.2 and t(thickness) = 600nm) lies on top of some water (n3=1.33) after spilling from an oil tanker. What wavelengths of visible light would be bright in reflection? use n1 = 1 (air)


    2. Relevant equations
    CI (constructive interference): 2t = (m+1/2)Lambda
    DI (destructive interference): 2t = mLamda
    Lamda = Lamda/n

    Layers of film
    air= n1
    oil = n2
    water = n3

    3. The attempt at a solution

    The problem asks for the bright areas so I used the constructive interference equation. my set it up as one part of the wave being completely reflected while another part of the wave refracted into the oil and then is reflected by the water and refracted into the air. I can't figure out what to do next.

    The answer is: (720nm, 480nm)
     
  2. jcsd
  3. Aug 7, 2008 #2
    I think dsin(theta) = m(lambda) would be more useful since there are two answers.

    lambda = (dsin(theta))/m

    d= (1/6000)x10^(-9)

    Im not sure what the n's are in your question, but the two answers and most probably due to +/- theta.

    maybe 'm' in the equation has something to do with the 'n's' from the question.

    Hope this helps.
     
  4. Aug 7, 2008 #3

    alphysicist

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    Homework Helper

    Hi UAPhys03,

    I don't believe these equations are correct for this case, which would explain why you are not getting the right answer. Do you see why they are not right?

     
  5. Aug 7, 2008 #4
    Ok i figured it out. I had my equations mixed up like you suggested.

    So for bright reflection that would be constructive interference and since both rays undergo a pi phase change, they are in phase.

    CI: 2nt = mLamda , where n =1.2 for oil

    I used m = 3

    which gave 480nm

    and m = 2

    and that gave 720nm

    Thanks!
     
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