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Classic Waves Interference Problem with Wedges

  1. Oct 9, 2014 #1
    1. The problem statement, all variables and given/known data

    http://i.imgur.com/Mp77mDR.png

    2. Relevant equations
    Path length interference equation.

    3. The attempt at a solution
    Why is the answer to (a) a minima? 1. When the light wave hits the top surface of the top glass, a wave will be reflected with phase change of pi. 2. The non-reflected goes to the bottom surface of the top glass to be reflected again with no phase change. 3. The non-reflected part still goes to the top of the down glass and is reflected with phase change pi. 4. We don't care about other reflections as amplitude diminishes too much.

    Now, We have three waves who are reflected towards our eye. 2 and 3 automatically cancel each other. There is only wave 1 left. So, this is not perfectly destructive. How can they call this a minima?
     
  2. jcsd
  3. Oct 9, 2014 #2

    Simon Bridge

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    You mean, if you take into account the reflection from the top and bottom surfaces of the top glass plate?
    There is an assumption that the glass plates are not thin films - the small variations in the glass are enough to cancel out the interference due to having two surfaces leaving the interference due to the wedge to dominate.

    Even so - the "minima" is only a zero amplitude in the first approximation.
    In practice there is some light reflected because materials are not perfect.
    It's like how you see a strong reflection from a window when it is dark on the other side.

    Anyway - the theory you are learning here is incomplete.
    You will get to the rest when your maths catches up.
     
    Last edited: Oct 9, 2014
  4. Oct 9, 2014 #3
    Simon, I understand that the two waves reflected from the bottom surface of the top glass and the top surface of the bottom glass will cancel each other out. But, there will still be the first reflected wave from the top of the top glass. This is not a minima in my head!
    What do you mean "an assumption"?
     
  5. Oct 9, 2014 #4

    Simon Bridge

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    You have to sum all the possible reflections, you cannot just single one out and say that particular one gets left out - why not some combination of the other reflections too? Why not add in reflections from the bottom surface of the bottom glass?

    Something that is assumed.

    [edit]whoops - forgt a bit:
    iirc you do get a reflection off the top surface .. if you are careful doing the experiment, you may see it. But "minima" does not have to mean zero. It just has to be, locally, the lowest point.
     
  6. Oct 9, 2014 #5
    Simon I will be posting a response in two minutes. Please wait if you can.
    [edit] I posted a response. I thought it would have been longer though.
     
    Last edited: Oct 9, 2014
  7. Oct 9, 2014 #6
    How do we know if this is not a local max?
     
  8. Oct 9, 2014 #7

    Simon Bridge

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    Because when you add up the contributions from all the waves, it is a local min.
    Also because you can see that there is a local min there - so the first order relation has proved a good-enough approximation to predict one there.

    [edit]don't get me wrong, it's not a silly question - and leads you on to quantum mechanics and QED... eventually.
     
    Last edited: Oct 9, 2014
  9. Oct 9, 2014 #8
    Hmm. How did you know there was a local min? Sorry if I can't follow.
     
  10. Oct 9, 2014 #9
    Why can't it be a local max if the combination of wave 2 and 3 actually end up having destructive interference with the first wave reflected when the air separation gets larger?
     
  11. Oct 9, 2014 #10

    Simon Bridge

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    To do an interference problem, you have learned to add the waves that reflect from two surfaces right?
    But your question points out that this leaves the waves reflected from other surfaces out.
    I'm saying you should include them in your calculation. Do the math yourself and see.

    Remember to include a reflection of each surface ... there are four in the picture.
    (I will have to step out for a bit.)
     
  12. Oct 9, 2014 #11
    Simon, if I get this right, you are telling me that the wave reflected from the glass is not relevant because glass is not a thin film. But why is it not relevant if glass is not a thin film?
     
  13. Oct 9, 2014 #12

    Simon Bridge

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    The waves from the top and bottom of the same sheet of glass contribute to the interference pattern if the glass is very thin as well as optically flat.
    All this theory you are learning is approximate - for instance, the top plate is angles wrt the bottom one - so waves normal to the top won't, in general, be normal to the bottom ... so the reflected waves are at different angles, so the interference is not a simple addition. But if the angle between the plates is very small, that is a good-enough approximation.

    Have you ever actually got two optically flat sheets of glass and shone monochromatic light through them?

    But I see you were actually making a different point - and what I am saying about that is you should do the actual maths for the situation that includes reflections from all the surfaces in the problem - then you will see why you are taught to just look at the surfaces either side of the air-gap.
    The calculation (there's only four surfaces) is within your abilities so why not do it?
     
  14. Oct 9, 2014 #13
    Sigh. I just don't know how to do it. I've been thinking about it, to no result.
     
  15. Oct 9, 2014 #14
    Simon, could you give me a hint as to how to start? Thanks a lot by the way. Always there when help is needed.
     
  16. Oct 9, 2014 #15

    Simon Bridge

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    Can you not do the calculation for the case where you have only two interfaces? Excuse me I just figured you were doing a course.

    Pick a ray for an exemplar. We are dealing with a whole wave here but the rays make it easy to see the path lengths.

    You need to be able to tak about the various rays - so define a notation like this:
    Number the interfaces 1-2-3-4 from top to bottom.
    Call the incoming ray, ray 0.
    Rays off each interface are either reflected "r" or transmitted "t" ... number the reflected and transmitted rays after the interface ... so at the first surface you have r1 and t1. r1 differs in phase from ray0 (which will be a t0) by pi ... inversion on reflection to a slower medium.
    r1 and r3 invert but r2 and r4 do not.
    We don't care about secondary reflections at this stage.

    You are going to end up with 4 rays emerging from the top, ultimately one from each interface.
    Now pick a situation where r2 and r3 destructively interfere ... so they sum to zero.
    That leaves you with r1 and r4 to add up.
     
  17. Oct 9, 2014 #16
    Yes perfect. I understand to this point. But now, how do I know the path length difference between 1 and 4. The thickness of the glass is not mentionned.
     
  18. Oct 9, 2014 #17
    I found this explanation on this website :

    For this problem, it’s only the interfaces of the thin film that matter. “Thin”
    means of width of the order of a few, or maybe a few tens, of wavelengths of
    light. The inner air layer is thin, so that waves reflecting off top air wedge
    interface and the bottom air wedge interface will be coherently in phase or
    out of phase. A half-wavelength is a large fraction of the thickness of the thin
    inner air layer. The top glass plate is very many wavelengths thick, so that
    a fraction of wavelength is a very tiny fraction of the thickness, and wavesinteracting at the top and bottom of the glass layer will not be coherent– so
    there will be negligible interference due to the glass plate.
    The interfaces that matter for this problem are the top of the air wedge,
    and the bottom of the air wedge.

    http://www.phy.duke.edu/~schol/phy152/faqs/faq14.pdf

    It's still not clearly understandable why this smaller ratio would matter. After all, waves are waves and phases stay constant in a medium.
     
  19. Oct 10, 2014 #18

    Simon Bridge

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    This is true - in ideal situations it should not matter how many times the wavelength the sheet of glass is, the conditions for interference minima and maxima are the same.
    In real life, though, you only get significant interference where the dimensions are of the order of a wavelength.
    Recall - the theory you are learning here is incomplete, it is approximate.

    ... you need to calculate the path difference to surface 1 in every case ... just call the glass thickness "t" and leave it unknown.

    You don't actually need the exact optical path difference - you just need to compare it to the r2 and 3 cases where you know there is destructive interference.

    But all that is if you want to gain a better understanding at this level.
    You could always just take a "those are the rules" approach and wait for the theory to catch up with your questions.
     
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