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Thin Film Interference of oil

  1. Jan 8, 2009 #1
    1. The problem statement, all variables and given/known data
    A transparent oil with an index of refraction of 1.25 spills on the surface of the water (n=1.33), producing a maximum of reflection with incident orange light ([tex]\lambda[/tex] = 602nm in air.) Determine the lowest possible thickness (in nm) of the oil slick.

    2. Relevant equations
    n1[tex]\lambda[/tex]1 = n2[tex]\lambda[/tex]2

    3. The attempt at a solution
    Okay, I know how to get the answer... but I don't quite understand it. The question is asking the lowest possible thickness, and it's producing a maximum of reflection, right? What I thought was, since there's two reflections - one off the oil slick and then another reflection off the water, that would mean 2 phase shifts; creating constructive interference. In that case, wouldn't the lowest possible thickness be 1/4[tex]\lambda[/tex]? I get the answer if I apply the thickness of the film to be 1/2[tex]\lambda[/tex], but I'm not sure why.

    Also, these may be other questions but they're still related to thin films, if that's okay: how do you know when it's trasmitted light or reflected light? It would have to say in the question, right? And all the questions I've done seem to always have constructive or destructive interference that makes the film of either 1/2[tex]\lambda[/tex] or 1/4[tex]\lambda[/tex] thickness... When is it 0 or something like 3/4[tex]\lambda[/tex]? I'm guessing that's when there's more reflections, perhaps?

    Thanks for help.
    Last edited: Jan 8, 2009
  2. jcsd
  3. Jan 8, 2009 #2

    Doc Al

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    Staff: Mentor

    The phase shift upon reflection at both those interfaces (air/oil & oil/water) is 180 degrees. Thus those phase shifts produce no net phase difference between the two reflected beams.
    Don't forget that the light that reflects off the bottom surface makes a round trip through the oil layer. And to get maximum constructive interference, the optical path length difference must be an integral number of wavelengths.

    Realize that each time light is reflected only a small fraction of the incident light is reflected. So you usually ignore multiple bounces, since they will be very weak.
  4. Jan 8, 2009 #3
    What do you mean by an integral number of wavelengths?

    Err.. I'm confused. How exactly do you tell the thickness of the film in terms of [tex]\lambda[/tex], knowing things like in the question above?
  5. Jan 8, 2009 #4

    Doc Al

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    Staff: Mentor

    I mean nλ, where n = 1, 2, 3, ... (an integer)

    I'm unclear as to your exact question here.

    When considering thin film interference (for reflection) you compare:
    (1) The light reflected off the top surface
    (2) The light reflected off the bottom surface

    The light in (2) travels an extra distance equal to twice the thickness of the film. In this problem, you want the phase difference to be zero, thus you want the path difference to be some integral number of wavelengths. The thinnest film will be when 2t = 1λ, thus t = λ/2. (Note that here λ is the wavelength of the light within the film material, not the wavelength in air.)
  6. Jan 8, 2009 #5

    How do you know which interger to use for n?
  7. Jan 8, 2009 #6
    Doc Al is offline so I'll try to help.

    The question is:
    Determine the lowest possible thickness (in nm) of the oil slick.

    Doc Al told you:
    The thinnest film will be when 2t = 1λ, thus t = λ/2.

    The 1 in the above is the value of n.
  8. Jan 8, 2009 #7
    Thanks for helping out as well.

    What I mean is in general, how do you know what n is? So in this example, how did you know n = 1? Sorry if it's obvious, lol.
  9. Jan 9, 2009 #8
    Doc Al told you:
    The thinnest film will be when 2t = 1λ, thus t = λ/2.

    In this case n=1.
    In any other case (n=2,3,...) then this will not be the thinnest film.

    You have an equation.
    The value 2 means the light passes twice through the film. This is a fixed value.
    t is the film thickness. This is a variable.
    λ is the wavelength. For a given monochromatic source of light this is a fixed value.
    n is the number of wavelengths that fit into the double passage through the film. This is a variable.
    So in our equation we have two variables, t and n.
    Set n=1 and we get the smallest possible value of t, and maximum brightness.
    Set n to any higher integer value and you will still get constructive interference, but less brightness due to greater absorption losses.
  10. Jan 9, 2009 #9
    Oh... I see. I get it now. Much more simple than I thought it was. Thanks.
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