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Interference of thin films

  1. Mar 13, 2016 #1
    1. The problem statement, all variables and given/known data
    (a) Calculate the minimum thickness of a soap bubble film that results in constructive interference in the reflected light if the film is illuminated with light whose wavelength in free space is λ=600nm. The index of refraction of the soap film is 1.33.

    (b) What is the film is twice as thick? Does this situation produce constructive interference?

    2. Relevant equations
    Constructive interference occurs here, so: 2nt = (m + ½)λn

    3. The attempt at a solution
    (a) Part A is pretty straightforward, as it just involves recognizing that this is a constructive interference case and then plugging in what we know.
    2nt = (m + ½)λn
    t = (0 +½)λ / 2n = λ / 4n = 600nm / 4(1.33) = 113nm

    (b) The answer for Part B shows:
    t = (m+½)λ/2n = (2m + 1) λ/4n
    m = 0, 1, 2...
    "The allowed values of m show that constructive interference occurs for odd multiples of the thickness corresponding to m=0, t=113nm. Therefore, constructive interference doesn't occur for a film that is twice as thickness

    I'm confused as to what exactly they did to arrive at this conclusion. What is meant by "odd multiples of the thickness"? The answers doubled every value in the above equation except for thickness (t). Why wouldn't this be done, as the question is specifically asking "what if the film was twice as thick"?
     
  2. jcsd
  3. Mar 13, 2016 #2

    haruspex

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    They mean odd multiples, (2m+1), of λ/4n. I.e. odd multiples of the minimum thickness for destructive interference.
    They did not double all values except t. They just multiplied the top and bottom of a fraction by 2.
     
  4. Mar 13, 2016 #3
    I'm still confused as to why exactly this means that constructive interference doesn't occur for a film that is twice as thick.
     
  5. Mar 13, 2016 #4

    haruspex

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    Let the minimum thickness for complete destructive interference be t0=λ/4n. So in general it occurs for all odd multiples, (2m+1)t0. 2t0 is an even multiple of t0.
     
  6. Mar 13, 2016 #5
    Sorry just to clarify... isn't the minimum thickness for complete destructive interference 2nt = mλ (m=0), so t = λ/2n?
    And instead, for complete constructive interference be t0=λ/4n. That's what my textbook says, at least.

    Also, why exactly does this only occur for odd multiples of the thickness?



    EDIT: I figured out why it needs to be ODD multiples of the thickness. For anyone else that may be having trouble understanding, I really dumbed it down for myself, but this is how I conceptualized what's going on...

    For us to see constructive interference at the surface of the bubble, the TOTAL phase shift needs to be 2π (ie. 360°)
    If the reflected wave already produces a 180° phase shift, then we know that the wave of light in the film needs to ALSO produce a 180° phase shift -- but from the extra distance traveled (t) in the film, rather than due to the refractive index.

    180° (or π) corresponds to HALF a wavelength. But since the light in the film travels a total distance of 2t, the thickness of the film must correspond to a QUARTER of the wavelength.
    Therefore, in general then, we can say that in order for a shift of additional distance of 180° to be produced, the thickness of the film needs to be t = ¼λfilm. This occurs at t=¼λfilm, t=3/4λfilm, t=5/4λfilm (odd integer multiples of the thickness!)

    Now, if we add up all the phase shifts that have happened:
    Reflected ray -- phase shift of 180°
    Refracted ray in the film -- phase shift of 90°+90° = 180°
    Total shift = 360° or 2π --> our condition for constructive interference
     
    Last edited: Mar 13, 2016
  7. Mar 13, 2016 #6

    haruspex

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    Sorry, I was thinking in terms of the reflected wave. When there's destructive interference in the reflected wave there's constructive interference in the propagated wave. Just swap con- with de- in what I wrote.
     
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