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Wavelengths / film / reflected light

  1. Feb 11, 2012 #1
    1. The problem statement, all variables and given/known data

    A film of MgF2 (n = 1.38) having thickness 1.44 x 10-5 cm is used to coat a camera lens.
    What are the three longest wavelengths that are intensified in the reflected light? (Enter your answers from smallest to largest.)


    2. Relevant equations
    λn = λ/n

    n = index of refrection

    for constructive interference:
    2t = (m + 1/2)λn
    2nt = (m + 1/2)λ

    destructive interference:
    2nt = mλ

    3. The attempt at a solution
    Honestly have no idea what the question is asking...

    not sure how to combine them, not sure how to find 3 biggest wavelengths

    not sure if it is constructive interference or not soo

    2nt = (m + 1/2)λ

    2(1.38m)(0.000000144m) = (0 + 1/2)λ

    (2)(1.38m)(0.000000144m) / (1/2) = λ

    λ = 795 nano meters (for m = 0)

    ^^ Yet says it is wrong

    All I can think of is to change m to 1, 2, 3 etc but don't think it works.



    Suggestions??
     
  2. jcsd
  3. Feb 11, 2012 #2

    Doc Al

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    Staff: Mentor

    This assumes that the index of refraction of the lens is less than that of the coating. What's the index of refraction of the lens?
     
  4. Feb 11, 2012 #3
    It actually doesn't say it just says it for the coating.

    I don't know how to find it either, any hints??
     
    Last edited: Feb 11, 2012
  5. Feb 12, 2012 #4

    Doc Al

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    I would assume that the index of refraction of the lens (presumably glass) is greater than that of the coating. How would that change your approach?
     
  6. Feb 12, 2012 #5
    I don't really understand what you are trying to imply...

    Would I have to use the index of refraction of the lens instead?

    ( I read that for crown glass, it would be n = 1.52 instead. Is this supposed to be "given" to you ? )
     
  7. Feb 13, 2012 #6

    Doc Al

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    Realize that there is a λ/2 phase shift upon reflection from a surface with higher index of refraction. So it matters. As I said before, just assume that the index of refraction of the lens is greater than that of the coating. That's all you need to know.
     
  8. Feb 13, 2012 #7
    I can't see how that is supposed to help :uhh:

    I seriously can't find any way how to incorporate the phase shift in the problem.


    Also is it right that you DO get each of the big wavelengths with m equal to 0, 1, 2, etc?
     
  9. Feb 13, 2012 #8

    ehild

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    You cited wrong conditions for constructive and destructive interference in the reflected light.
    2nt = (m + 1/2)λ refers to destructive interference (minimum reflectance) and 2nt = mλ for the constructive one (maximum reflectance). The problem asks the wavelengths where constructive interference occurs.
    Calculate lambda for m=1, 2, 3.

    ehild
     
  10. Feb 13, 2012 #9

    Doc Al

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    You need to review how the formula for thin film interference is derived. Whether there is a phase shift on reflection at each surface depends on the relative index of refraction. Assuming that the index of refraction of the coating is less than that of the lens (a reasonable assumption), then your equation for constructive interference is wrong. (As ehild points out, your version describes the condition for destructive interference. But do not simply copy the formula--understand how it is derived and then you'll be good to go no matter what kind of thin film is used.)
    Yes, that's the right idea. But first get the correct formula.
     
  11. Feb 14, 2012 #10
    Thank you, I got it now, and I also got another problem right that I had wrong because of you saying this too !

    Blah, I blame my book for not seeing that. Then again I sort of glanced over it too.
     
    Last edited: Feb 14, 2012
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