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Interference of Light - Thin Film

  1. Nov 30, 2013 #1
    1. The problem statement, all variables and given/known data

    Plan plates of glass are in contact along one side and held apart by a wire 0.05 mm in diameter, parallel to the edge in contact and 20 cm distant. Using filtered green mercury light ([itex]λ = 546 nm[/itex]), directed normally on the air film between plates, interference fringes are seen. Calculate the separation of the dark fringes. How many dark fringes appear between the edge and the wire?

    2. Relevant equations



    3. The attempt at a solution

    I'm trying to follow along with the solutions and I seem to get confused early on

    http://imageshack.us/a/img844/8884/ix2m.png [Broken]

    I understand that

    [itex]Δ = Δ_{p} + Δ_{r} = 2nt = (m + \frac{1}{2})λ[/itex]

    Here [itex]n = 1[/itex] because of air and you use [itex]m + \frac{1}{2}[/itex] instead of just [itex]m[/itex] because you want a dark fringe. I believe the solution is using [itex]Δ_{r} = \frac{λ}{2}[/itex]. I'm however not really sure where this is coming from or why it's included. Apparently the geometrical path difference is [itex]\frac{λ}{2}[/itex] but I'm not really sure why.
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
    Last edited by a moderator: May 6, 2017
  2. jcsd
  3. Nov 30, 2013 #2

    TSny

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    Looks like they should have said "destructive" instead of "constructive" here.

    You should have covered the idea that sometimes a wave "flips over" when it reflects. I think ##\Delta_r## is taking this into account.
     
    Last edited by a moderator: May 6, 2017
  4. Nov 30, 2013 #3

    Simon Bridge

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    Yep - the wave inverts on reflection from one of the interfaces.
    Which one?

    The incoming light reflects of the top surface, but some gets through to reflect off the bottom surface, a distance t away, returns to pass through the top surface again. We are interested in how these two light waves interfere at the top surface.

    They interfere constructively if they are in-phase with respect to each other.
    This depends on the path difference and the wavelength. One wave travels 2t farther than the other, since it traverses the thickness t twice.

    For constructive interference, the path difference is ##m\lambda = 2t## if the wave did not invert.
    Since it does, after a path difference of ##n\lambda##, the returning wave will be pi radians out of phase with the wave reflected from the top surface - pi radians is a half wavelength: so the path difference needs to be an extra half-wavelength long to get them back "in phase" - to get constructive interference.

    Hence ##(m+\frac{1}{2})\lambda=2t## for constructive interference.
     
  5. Nov 30, 2013 #4
    My book doesn't mention this until Michelson interferometer. It doesn't explain why your supposed to use lambda/2. It appeared in my lecture in thin film interference. I can't seem to find though any explanation as to where it comes from. I also have searched the internet for some sort of explanation on it but can't find anything. Do you know of any text on the internet that explains it? I mean I don't necessarily need to know but I need to know when to take it into consideration and when not to at the very least.

    *reading post above*
     
  6. Nov 30, 2013 #5
    This is interesting no one has ever explained this to me. How do I know when they invert and how do I know when they don't?

    [itex]Δ = Δ_{p} + Δ_{r}[/itex]

    The inverted wave travels 2t as far since it travels through the thickness twice, so then why do we use 2nt as apposed to just 2t for thin film interference?
     
  7. Nov 30, 2013 #6

    TSny

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    See here for a discussion of thin film interference which states a general rule for how to decide if a wave inverts at reflection.
     
  8. Nov 30, 2013 #7

    Simon Bridge

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    It's the same as for pulses travelling on a string or a slinky spring - when it has to go from one string to another with different properties. If you've never seen this demonstrated, you should.

    Animations showing the detail:

    http://www.acs.psu.edu/drussell/Demos/reflect/reflect.html

    Real life is messier:
    http://www.animations.physics.unsw.edu.au/jw/waves_superposition_reflection.htm

    Where light moves from a high-speed medium to a low-speed medium, the reflected wave is inverted.

    In your case, the light, 1st, goes from glass-to-air, which is slow-to-fast, so the reflected wave is not inverted.
    The transmitted wave goes through air until it meets an air-to-glass (fast-to-slow) interface, where the reflected part of the wave is inverted.
    So the light wave that traverses the gap gets inverted half-way through it's journey.

    It does not have to be air in the gap, and the plates don't have to have the same refractive index (they don't even have to be glass.) So you can get different relations depending on what happens at the different interfaces.
     
    Last edited by a moderator: Sep 25, 2014
  9. Dec 7, 2013 #8
    Thanks for all of the help guys. It make since to me now. I clicked on all of the links, but was shocked by the link in #6 that went to a AP Physics B class website. I took that class four years ago and learned this material then, apparently I have forgotten all about it in such a small amount of time. Makes me wonder if that'll be the case with the things I am currently learning in my undergraduate a couple of years after I graduate.
     
  10. Dec 8, 2013 #9

    Simon Bridge

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    Forgetting that sort of lesson so fast is a sign that you didn't understand it first time around.
    Perhaps you skimmed the material at the time and focussed on memorizing equations for the exam?

    But it is not too unusual for knowledge you don't practice to get demoted to the back of your memory ... one of the advantages of answering questions on PF is that you keep practicing ;)
    Everyone returns to basics sometime.
     
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