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Thin Film

  1. May 5, 2015 #1
    1. The problem statement, all variables and given/known data
    http://lon-capa.bd.psu.edu/res/psuerie/gwb6/physics/optics/e1p4.png
    A thin 200nm film of oil (n = 2.0) floats on water (n = 1.33). To a normal human, what wavelength will the film appear when viewed from above (nearly perpendicular to the film)?

    2. Relevant equations

    for constructive interference
    2 * pi * m = ( 2 * pi * n * 2 * d) / lambda + pi (in this case)

    3. The attempt at a solution

    This was a question I missed on the midterm. I want to know where I went wrong for the final. Here is what I did.

    There is a poem our prof taught us; high to low, phase no. Low to high, phase change pi.
    Based on the picture, which hopefully shows up, above I can see there is 1 phase change of pi where air meets oil. my equation is then

    2 * pi * m = ( 2 * pi * n * 2 * d) / lambda + pi
    where m = 1, n = oil = 2, d = 200 nm

    I solve for lambda and get the answer of 800 nm.

    This is not the correct answer, and I know this right away because light of this wavelength is not in the visible spectrum.

    I know the answer is 533 nm but I don't know why. I feel like I am missing something here conceptually.

    Please help.


    So, it looks like the picture didn't make it through to the post it looks something like this the lines separate the different mediums.

    air
    _________________________

    oil
    _________________________

    water
    _________________________
     
    Last edited: May 5, 2015
  2. jcsd
  3. May 5, 2015 #2
    Ok, I figured it out.

    First off my math was bad when I came up with the 800 nm. I only multiplied by d rather than 2 d had I done this I would have found a wavelength of 1600 nm. Still not the right answer but it is the right equation.

    It then occurred to me that I have a variable that I can play with which is m. I had assumed that m = 1. Bad assumption because there really isn't any reason m cant be any integer I want. If I make m = 2 the problem is solved.
     
  4. May 5, 2015 #3

    Doc Al

    User Avatar

    Staff: Mentor

    I like that!

    I don't see how you got that answer. Using the same basic approach, I get the given answer.

    Ah... I see you've got it now. (Too late! But good for you.)
     
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