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Tricky relfection/absorbtion optics question

  1. Sep 19, 2006 #1
    I have the following question:

    A disabled tanker leaks kerosene (n = 1.20) into the Persian
    Gulf creating a large slick on the top of the water (n = 1.30).

    a) If you looking straight down from an aeroplane, while the sun is overhead, at a region
    of the slick where the thickness is 460 nm, for which wavelength(s) of visible light is
    the reflection brightest because of constructive interference?

    b) If you are scuba diving directly under this same region of the slick, for which
    wavelength(s) of visible light is the transmitted intensity strongest?

    I understand that some of the rays will be reflected and some absorbed by both mediums. I found this formulae somewhere in my lecture notes but im not sure what the 'I's mean: I/I0 = ((n1 - n2)/(n1 + n2))^2

    Ive been trying to slug this question out for ages and its frustrating because i kinda know what is going on just dont know how to quantify it, help is much appreciated - cheers.
  2. jcsd
  3. Sep 20, 2006 #2

    Meir Achuz

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    In your equation, I_0 is the incident intensity and I the reflected ilntensity.
    But that simple equation is for a single interface, not a thin film.
    The derivation for a film is done in some grad EM texts.
    (It is given as a problem in Jackson.) The results are:
    a) The reflection is a maximum when \lambda=n_2 d=1.2X460 nm.
    b) The transmission is a maximum when \lambda is twice that in (a).
  4. Sep 20, 2006 #3
    ok thanks mate that really helps - cheers
  5. Sep 20, 2006 #4


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    You don't need to go to that depth to solve this problem. There are two standard simple equations for the wavelengths that produce maximum and minimum intensity in thin-film interference:

    [tex]m \lambda = 2nt[/tex]

    [tex](m + 1/2) \lambda = 2nt[/tex]

    where m is an integer and t is the thickness of the film. One of these gives the maximum and the other one gives the minimum, but which one does which depends on the situation. I'm too lazy tonight to write a lecture about it on the spot, so look in your textbook for these formulas and read the discussion surrounding them (along with any examples). Or try here:


    That page uses d instead of t, and includes a [itex]\cos \beta[/itex] to cover situations where the light isn't perpendicular (normal) to the film. In your problem, [itex]\cos \beta = 1[/itex].

    After reading that stuff, if you have questions, ask away!
    Last edited: Sep 20, 2006
  6. Sep 22, 2006 #5

    Meir Achuz

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    Yes, those were the simple formulas I used. The complicated derivation I referred to is to get the reflection coefficient as a function of wavelength.
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