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Diffraction Transverse Wave Confusion

  1. Oct 3, 2012 #1
    http://en.wikipedia.org/wiki/File:Doubleslit.gif
    As shown in the link, diffraction is shown in the form of these kind of longitudinal waves. The diffraction pattern explains the wave interference pattern observed in the double slit experiment. However in the case of the delayed choice double slit experiment, or if there is only one transverse wave passing through the slit(s) then how can we explain these diffraction patterns with respect to the transverse wave. I understand that in the link, in the straight lines, one color is the crest and the other might be the trough. But how does the transverse wave explain the pattern of the colored bands after passing through the slit(those curves, how are they explained???)? Is it still one transverse wave which is sweeping out curvy areas or is it that the transverse wave is being split into multiple waves which are going in different directions thus totally spreading out and hence cover the entire curve(I know it's not exactly a curve)?

    Please help!!!

    Sorry for being such a physics novice!!!
     
  2. jcsd
  3. Oct 3, 2012 #2

    mfb

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    The type of wave does not matter - those color bands in the animation are just areas of common phase. You can identify blue with "electric field pointing upwards" and black with "electric field pointing downwards", for example, and see it as illustration of electromagnetic waves.

    I don't understand the second part of your post.
     
  4. Oct 4, 2012 #3
    @mfb: so that means that after passing through the slit, would the curve mean that the blue curves would be curved electric fields pointing upwards??? that's my question

    In the second part I ask that do the curves indicate that they are the different paths that the light wave might take or do they represent curved electric fields.
     
  5. Oct 4, 2012 #4

    mfb

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    "curved electric fields"?
    Curved areas where the electric field is pointing upwards? That would be a possible interpretation.

    The light takes all possible paths.
     
  6. Oct 4, 2012 #5

    sophiecentaur

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    What causes the interference is the relative path differences through one slit and the other which is basically a consequence of the geometry of the situation. The nature of the waves doesn't affect the distance travelled by the waves. You could imagine doing a double slit experiment using microwaves (transverse) and ultrasound waves (longitudinal) of the same wavelength and the same slit separation and measuring the wave amplitude at the same distances from the slits. You would get the same interference pattern (same positions of maxima and minima and also the detailed intensity curves. This would be a scaled up version of what you get with light, of course but, as EM waves are the same everywhere in the spectrum, then it's a reasonable thought (or real) experiment to discuss.

    I think the picture you have in your head of waves and curves may not be helping you. The wave front can be looked upon as a line (on a 2D diagram or a surface, in 3D) joining points where the signal has taken the same time to reach (hence the spherical waves from a point).
    If the wavefront happens to be curved, it means that the direction of the E and M fields will be different (not all parallel) at different places.

    However, I am wondering whether you could be thinking in terms of the waves that arrive in a particular place from the two different slits would actually have fields that are not actually parallel to each other, because the fields are at right angles to the 'ray' from each slit to the screen, coming from different directions. That actually will have an effect on the way two transverse waves will interfere as the vectors will not be in the same plane and there will not be complete cancellation. Could that be your problem?
     
  7. Oct 4, 2012 #6
    mfb: When you said light takes all possible paths, do you mean that one light wave covers the entire area or that each transverse wave can take any of the possible paths

    sophiecentaur: yes!!! that's my problem!!! I don't understand the interference in terms of the transverse waves where the fields are perpendicular to the direction of the ray and not parallel.Do you have an explanation for that?
     
  8. Oct 4, 2012 #7

    sophiecentaur

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    Grand - I understood what you were getting at!

    All that happens is that you take the two vectors - not parallel - and you work out the resultant. This will never be exactly zero. Just imagine the situation with two EM waves crossing each other at right angles (easy done for light with an arrangement of half silvered mirrors but it's difficult to invent an easy experiment to demonstrate the situation)

    If you were to take two identical radio waves from two antennae, you could (and people are always doing it) produce an interference pattern with 'nulls' in it. (All directive antenna arrays do this) But at a point in space where the two waves happen to be travelling at right angles to each other, the resultant amplitude can never be zero because the vectors will be at right angles. In the case of radio waves, you can detect them using a simple short dipole, which responds to the component of the field that happens to be parallel to the E field and, by changing its orientation, you can either get good zeros or no zeros at all, depending which components of each of the waves you are picking up.

    If you try the two slits experiment (optical) when the separation of the slits is 'significant', the nulls are just minima and not 'zeros'. They appear at the same places though.
    Of course, there are many other things that can 'spoil' the nulls in the pattern - including the actual width of the slits, the thickness of the slide which they're cut into, the range of wavelength of the light used and how coherent the light is that arrives at the back of the slits.

    PS you may be confusing the idea of a Wave (considering just one direction of propagation) with a 'Wave front' which describes the wave pattern over a region of space.
     
  9. Oct 7, 2012 #8
    Thanks for the answer. I never knew about the concepts of wavefronts till now(not too late since I am still in high school). I read up a few pages about them. Do these wavefronts represent only one wave or is it a collection of the paths of different waves?
    Also are waves(EM) three dimensional? Could they have width?
     
  10. Oct 7, 2012 #9

    sophiecentaur

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    I think this is really a matter of word usage but this is the way things are usually discussed (afaik). A wave is produced by 'something' and spreads out or can be 'directed' mostly in one direction but still spreading out, eventually. Spreading out, a wave front will be spherical but, far enough away, the curvature will be large enough to produce a plane wavefront. Two beams of light (two plane waves) will interfere with each other and the wavefront formed is very compicated (Interference / Diffraction). If you look up Huygen's Principle (sometimes termed Huygen's construction), you will find a way of predicting how a wave will develop as it travels by considering an infinite number of secondary wavelets, constantly forming at a wavefront and these wavelets all interfere to produce the next step - and so on. They mutually cancel in every direction (as it happens) but the direction that other methods will predict what happens to the wave.
     
  11. Oct 8, 2012 #10
    Thanks a lot!!! Could you also answer the second question my previous reply(the width problem)?
     
  12. Oct 8, 2012 #11

    mfb

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    Unless you have very special surface waves, all electromagnetic waves are 3-dimensional. In many systems, one dimension does not matter because the system is the same for every value of that coordinate.
     
  13. Oct 8, 2012 #12
    what about electrons behaving as waves??? are they 3-dimensional waves??? Then wouldn't that imply that the electrons must be at two or more places at a time since if there was only one electron, it would only produce a 2-dimensional wave???
     
  14. Oct 8, 2012 #13

    Drakkith

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    No, the wavefunction is 3 dimensional.
     
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