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Thin-Film Interference Question. Quick One!

  1. May 1, 2007 #1
    1. The problem statement, all variables and given/known data

    A film of gasoline (n= 1.40) floats on glass (n = 1.52). Yellow light (wavelength = 580 nm in vacuum) shines perpendicularly on the film. Determine the minimum nonzero thickness of the film, such that the film appears bright yellow due to constructive inteference

    2. Relevant equations

    2t + 1/2*wavelength(film) = wavelength(film)

    3. The attempt at a solution

    I know the constructive interference equation is setup like 2t + 1/2*wavelength(film) = wavelenght(film). However, I am having difficulty determining which n value i should use to get the wavelenghth of the film.

    I know this can be an easy problem for some people so i appreciate all the help i can get. Thanks!
     
  2. jcsd
  3. May 1, 2007 #2
    If the film is made of gasoline you obviously have to use the refractive index of gasoline.
     
  4. May 1, 2007 #3
    ok...does this look correct??

    2t + 1/2*wavelength(film) = wavelength(film)

    t = 1/4*wavelength(film)

    t is for thickness

    I get the wavelength(film) by dividing 580 nm by 1.40 (refractive index of gasoline). THIS IS WHERE MY QUESTION LIES...which n-value should I use? 1.40 or 1.52 or both? arrrgh! :smile:

    ok by dividing 580 nm/1.40 i get 440 nm. Then, i plug into thickness equation t = 1/4*(414 nm) and get 104 nm as the answer.
     
  5. May 1, 2007 #4

    Doc Al

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    This equation does not apply here. Figure out your own equation for the thickness needed to get constructive interference. Hint: What is the phase change upon reflection at each interface? What's the additional path length of the wave reflected from the bottom of the film?

    The film is gasoline--so use the index of refraction of gasoline.

    Redo that calculation (probably just a typo, since you have the correct value below).
    That 1/4 wavelength equation is for destructive interference.
     
  6. May 1, 2007 #5
    ok so should the equation look like this: t = 1/2*wavelength(film) since the extra path traveled is just t???

    t = 414 nm/2 = 207 nm
     
  7. May 1, 2007 #6

    Doc Al

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    The extra path traveled is 2t. But that extra path must equal one wavelength. So your answer is correct anyway (but you'd better understand why).

    Sounds good.
     
  8. May 1, 2007 #7
    Ok so...what about the phase change? do you avoid that?
     
  9. May 1, 2007 #8

    Doc Al

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    You tell me. What's the phase change at each interface?
     
  10. May 1, 2007 #9
    is it 0 in this case? b/c that sorta sounds like something in destructive interference.

    otherwise it would be 1/2*wavelength(film) since half of the wavelength is shifted??

    but for each? hmmm
     
    Last edited: May 1, 2007
  11. May 1, 2007 #10

    Doc Al

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    Just answer the question: Light that reflects off the top of the air/gasoline surface gets a phase change of how much? What about the light that reflects from the gasoline/glass surface?

    What does the phase change depend on?
     
  12. May 1, 2007 #11
    ah that's a clearer picture to me.

    1st question: 1/2 phase shift
    2nd question: nothing - dark fringe?

    phase change depends on the amount of light? true?
     
  13. May 1, 2007 #12

    Doc Al

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    This is correct: light reflecting from the air/gasoline interface gets a 180 degree phase change. Why?
    Why nothing? What's different about the gasoline/glass interface compared to the air/gasoline interface?

    Any talk of fringes requires comparing two reflections, not just one interface.

    Nope. :wink: (This should be in your text. Look up the derivation of one of the thin film interference formulas.)
     
  14. May 1, 2007 #13
    nevermind. i have no idea what im talking about...btw, you asked why on the first qustion. it's because the incident hits the surface at almost a perpendicular distance (im assuming), which is 180 degrees (or straight down)

    so the phase change is 2 due to 2 of the wavelengths?
     
    Last edited: May 1, 2007
  15. May 1, 2007 #14

    Doc Al

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    That's part of the answer, but not the part that I'm going for. Assuming normal incidence, how can you tell if you get a phase change upon reflection or not?

    Hint: In some cases, the phase change is zero (no phase change); in other cases, it's 180 degrees (half a wavelength). What's the governing rule that tells you which applies?
     
  16. May 1, 2007 #15
    ummm i dont know. the book im using is very brief about this section. it basically derives the equations for you and tells you this is the only way to solve a problem...obviously, that's not the case with this problem.
     
  17. May 2, 2007 #16

    Doc Al

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    Phase change upon reflection

    I suggest that you pick apart the derivation and see what it says about the phase change upon reflection. Whether or not you have a phase change upon reflection depends upon the relative indices of refraction of the two media at the boundary. Read this: Reflection Phase Change

    And while you're at it, you can read more about thin films here: Thin Films

    (FYI: In my opinion, hyperphysics (http://hyperphysics.phy-astr.gsu.edu/hbase/hph.html) is a reliable reference for basic physics; I suggest browsing through it to augment your assigned text.)
     
  18. May 2, 2007 #17
    A phase shift occurs when light enters a more optically dense material (greater index of refraction). Youve got 2 boundaries for consideration, the air-gas boundary and the gas-glass boundary. A phase shift occurs at both boundaries, so in order to constructively interfere, the extra distance must be a full wavelength.

    If the glass was replaced with another layer of air, your equations would apply. But with problems like this, its best to think about the problem and not rely on formulas that only work sometimes
     
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