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Thickness of AntiReflection Coating for destructive interference

  1. Dec 1, 2012 #1
    1. The problem statement, all variables and given/known data
    Glass, despite being transparent, still reflects a little bit of light. When making lenses and such, you'd rather not waste any light with stray reflections. Fortunately, you can use destructive interference to suppress those reflections. The picture below shows a piece of glass with a coating of magnesium fluoride (MgF2). The arrows represent incoming light. Some of the light will reflect off the air-MgF2 interface. Some of the light will reflect off the MgF2-glass interface. If you can get these two portions of light to interfere destructively, there won't be any visible reflections.

    Use what you know about path length differences and interference to figure out how thick the coating of MgF2 needs to be for an anti-reflective coating for light of wavelength 510 nm. Use the thinnest coating possible. MgF2 has an index of refraction of 1.38 and the glass has an index of 1.56.

    Note: the wavelength given is the wavelength of the light in air.

    http://img811.imageshack.us/img811/5161/arcoating.jpg [Broken]

    Uploaded with ImageShack.us


    2. Relevant equations
    r2-r1=d*sinθ = (m + 1/2)λ
    V(in material) = c/n
    δ = 2*Pi(m + 1/2)


    3. The attempt at a solution
    Eventually I'll need the wavelength in the glass. So: c=λ(in air)*f → f = c/λ(in air)
    V(in glass) = c/n(glass) = λ(in glass) * f →
    λ(in air)/n(in glass) = λ(glass)

    Did some reading, and my physics book says that the phase difference between the beams is only due to the path length difference, which is 2t where t is thickness of film.
    So, basically it said this is true:
    2*Pi(m + 1/2) = 2t*2*pi/λ(glass)
    So, I solved for t, got this: t = λ(glass)/2

    That was wrong.

    Correct answer: t = λ(glass)/4. Where is this factor of 1/2 coming from?
     
    Last edited by a moderator: May 6, 2017
  2. jcsd
  3. Dec 1, 2012 #2

    Doc Al

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    Staff: Mentor

    Good.
    What must the phase difference be for destructive interference.
     
  4. Dec 1, 2012 #3
    Shouldn't it be δ= 2*Pi(m+1/2) as I did? Or should it be simply 2*Pi and I'm confusing my equations?
     
  5. Dec 1, 2012 #4

    Doc Al

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    Staff: Mentor

    That's the correct equation, but what does m equal?
    That is what you did and it's incorrect.
     
  6. Dec 1, 2012 #5
    Well I thought m should equal 1 because we're looking for the first instance of destructive interference to minimize the thickness of the coating.
    Why isn't it 1?
     
  7. Dec 1, 2012 #6

    Doc Al

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    Staff: Mentor

    The first instance is when m = 0.

    The main thing is that for destructive interference the phase difference has to be an odd multiple of pi, not 2pi.
     
  8. Dec 1, 2012 #7
    Oh ok, that is my mistake, though first instance was m = 1.

    And the phase difference being a multiple of pi makes sense, because at 2pi basically there is constructive interference, and half way between 0 and two pi there is destructive, hence the multiples of pi.

    Cool, thanks a lot, makes a lot more sense now!

    Smith
     
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