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Homework Help: Anti reflective coating, finding the 'correct thickness'

  1. Apr 13, 2010 #1
    1. The problem statement, all variables and given/known data
    To reduce reflections from glass lenses (n ≈ 1.5), the glass surfaces
    are coated with a thin layer of magnesium fluoride (n ≈ 1.39).
    What is the correct thickness of the coating for green light (510 nm
    vacuum wavelength)?

    2. Relevant equations
    [tex]\lambda[/tex] = 2nd/(m+1/2)

    3. The attempt at a solution
    I don't really know where to start. It says "correct thickness"...what defines the correct thickness? Is it when a certain value is 0? Or when a certain amount of light gets through? I'm confused as to where to start because I'm not sure what this means...if anyone could give me a hint, that would be awesome. Thanks.
  2. jcsd
  3. Apr 13, 2010 #2
    I've been studying this a bit - is the correct thickness they're looking for one that gives off destructive interference? Normally, to get destructive interference...I would do this:

    r1 has a phase shift of pi when its reflected off of the coating (in my notes that is also written as lambda/2...not sure why that is). r2 continues on through the coating and reflects off the other edge, travels back through the coating and out into the air where it interferes with r1.

    In a previous example I did, there was air on the other side of the thin film and so there was no phase shift when r2 reflects off of the other edge. In that case, I would want the phase of r2 to remain unchanged or changed only by k[tex]\lambda[/tex]/2n. Right?

    But in this case, there is an additional phase shift off the outer edge because the light is going from the coating to the glass (n=1.39 to n=1.5). I'm confused as how to fit that into my solution.

    Any help would be appreciated.
  4. Apr 13, 2010 #3
    Okay. Did a bit more thinking on this:

    r1 is shifted by [tex]\lambda[/tex]/2 (or pi, still not entirely sure how those two equate. The only formula I have concerning them is k=2pi/[tex]\lambda[/tex]...where does the k go?)

    Anyway, since there is this issue going to the coating --> glass (low to high again) there is ANOTHER phase shift in r2 by [tex]\lambda[/tex]/2. Normally there wouldn't be the same shift so we could just have a thickness that leaves the phase of r2 untouched...but in this case they have the same shift, so if we have a thickness that leaves the phase of r2 untouched...they will be the same and therefore constructive interference will occur. Right?

    So then we want to have a thickness that causes a slight shift in the phase of r2. The smallest shift we can do is by 1/2 a wavelength...so:

    2L = [tex]\lambda[/tex](1/2) <-- the [tex]\lambda[/tex] we want is the wavelength of the light once its going through the coating though, so:
    2L = [tex]\lambda[/tex]0/n (1/2)
    L = t = [tex]\lambda[/tex]0/2n (1/2)
    t = [tex]\lambda[/tex]0/4n
    = 501nm/4(1.39)
    = 92µm (the correct answer)

    That is the right answer, but there are a few questions I put in there along the way. I want to make sure I'm understanding this pretty comprehensively so if there is anything anyone wants to clarify for me I would be grateful.
  5. Apr 13, 2010 #4
    It's because if you phase shift any sine or cosine function by pi radians you move some integer multiple of lambda/2.

    If k is 2pi/lambda, then that's not the correct equation. If k is some integer then that's correct.

    Again you shouldn't be using k like this, probably some notation confusion going on here.

    Yes. If you shift anything by a integer number of wavelengths the waves will still line up like they were the same wave (different from shifting by integer number of half wavelengths).

    To remember this:

    Integer number of wavelengths --> Doesn't change the initial phase difference
    Integer number of half-wavelengths --> Switches the initial phase difference (in phase to out of phase, vice versa)
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