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B Distance between fringes in an interference pattern?

  1. Apr 6, 2016 #1
    For the sake of simplicity let's suppose I'm talking about the diffraction of light through one slit.

    fraungeo.gif

    Are the distances between any two consecutive fringes (apart from the central maximum) exactly equal? Assuming the wavelength, slit width, and distance between slit and screen do not change, that is. Or is it one of those cases of small angle approximation where the difference is so small it's negligible?

    Why is the central maximum twice as wide?

    Also, is the "distance between 2 consecutive fringes" equal to the width of a fringe? When talking about distance between fringes I assume we are talking about the distance between their centres (since there aren't really any gaps between them), which is the distance from crest to crest, which should be the same as the distance from trough to trough, which is the width of a fringe. Am I right?

    Another question I have is why the image in the spoiler above and many others I have found say that the condition for a minimum is when a⋅sinθ = mλ. Assuming m is an integer value, if one ray of light is offset compared to another by a length equal to a multiple of the wavelength, then surely they're in phase, which means they interfere constructively? That should lead to a maximum, not a minimum. Where have I gone wrong in this thought process?
     
    Last edited: Apr 6, 2016
  2. jcsd
  3. Apr 6, 2016 #2

    blue_leaf77

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    No, they are not. The diffraction pattern from a single slit has the form
    $$
    I(y) = I_0 \left( \frac{\sin ((a\pi/\lambda) \sin \theta)}{(a\pi/\lambda) \sin \theta} \right)^2
    $$
    The above equation implies that the maxima of ##I(y)## do not coincide those of ##\sin ((a\pi/\lambda) \sin \theta)## which are equally separated for small ##\theta##.
    Twice as wide as what?
    Although the maxima of ##I(y)## do not coincide those of ##\sin ((a\pi/\lambda) \sin \theta)##, the minima do. The minima of the latter occur when
    $$
    (a\pi/\lambda) \sin \theta = m\pi
    $$
    which trivially reduces to
    $$
    a \sin \theta = m\lambda
    $$
     
  4. Apr 6, 2016 #3
    How does it imply that? Good lord sorry I feel slow today

    My bad, I meant to say that if the widths of the fringes are constant, then why is it that only the central maximum is twice as wide as every other fringe? But since my premise is wrong this doesn't need answering

    I understand the mathematical explanation of this, but I'm having a hard time following it intuitively. If the path difference, denoted by ##a \sin \theta## as shown in the diagram in the first post, is a multiple of ##\lambda## then logically speaking they must interfere constructively? What's wrong in the way that I'm thinking about this?

    Or is it because ##a \sin \theta## isn't exactly the path difference as that assumes that the rays of light are parallel, which of course they cannot be if they meet at a point?

    Thanks a million by the way
     
    Last edited: Apr 6, 2016
  5. Apr 6, 2016 #4

    blue_leaf77

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    I should also add that ##m## can only take values ##1,2,3,\ldots##.
    I will rewrite the diffraction pattern as
    $$
    I(y) = I_0 (\sin^2 ay) f(y)
    $$
    with ## f(y) = \left(\frac{1}{ay}\right)^2##. As far as I know, ##I(y)## can have maxima that occur at the same places as ##\sin^2 ay## if ##f(y)## is equal to either a constant or another periodic function with the same period and also whose maxima occur at the same places as ##\sin^2 ay##.
    I guess you are applying the reasoning about the phase difference between the rays in the case of diffraction grating and double slit interference. Single slit interference is different from them, i.e. you cannot apply the usual ray analysis to calculate the phase differences.
     
  6. Apr 6, 2016 #5

    Charles Link

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    The reason a single slit diffraction pattern has zeros at what is the same angle as the interference maximum for point sources located at the edges of the slit is that the slit can be split in two (or some multiple of two segments), and each infinitesimal Huygens source can be paired up with a corresponding point somewhere on the slit that is exactly ## \pi ## out of phase with that location. e.g. one edge will be ## \pi ## out of phase with the Huygens source in the center of the slit (when the Huygens sources located at the edges of the slit are ## 2 \pi ## out of phase and would constructively interfere (for ## m=1 ##)). This happens for ##m \lambda=b \sin(\theta) ## for all integer ## m ## except for ## m=0 ##, because in the center of the pattern, the rays from all the Huygens sources across the entire slit are in phase with each other.
     
    Last edited: Apr 6, 2016
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