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B Young's Double Slit Formula

  1. Mar 29, 2017 #1
    In Young's double slit formula, the equation is

    d sin \theta = m \lambda

    Which means that

    sin \theta = m \lambda / d

    But, what if d < \lamda so the sine value becomes greater than 1?

    Will there be any interference?
  2. jcsd
  3. Mar 29, 2017 #2
    I'm finding problem understanding your question, what i know is that the path difference = d sinθ
    and for constructive interference d sinθ =n λ
    where n is an integer.
  4. Mar 29, 2017 #3
    So, what happens if the distance between slits is much smaller than the wavelength?
  5. Mar 29, 2017 #4


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    If d is small, there will be no total cancellation but interference will still cause a drop in the amplitude as the angle increases.
  6. Mar 29, 2017 #5
    So, does the formula still valid for d < lambda?
    Why will there be no total cancellation? I mean, you mean the cancellation caused by the difraction? Or, what? Please explain it.

  7. Mar 29, 2017 #6
    you won't see the fringes no more
  8. Mar 29, 2017 #7


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    It is valid because it only describes the positions of zeros and there is no zero for closely spaced slits.
    The formula is based on a simplified situation with infinitely narrow slits (omnidirectional individual sources)-more like what you get with two co-phased radio frequency dipoles than two slits. If the sources are just greater than λ apart, there will only be a single zero, which will be at nearly 90° off beam. As the spacing gets narrower still, this zero will pass through the 90° direction and disappear. In any case, the beam (the diffraction you are referring to??) pattern of each 'real' single slit will have a zero at 90° or greater. Two dipoles will work fine, though, as a model because they are truly omnidirectional, individually.

    If the slits are too close together, the phase difference between the waves from the two sources can't be as great as π and so you can't have complete cancellation at 90°. Nevertheless, if there is a reflex angle between the two E vectors, the resulting amplitude will be noticeably lower than where they add and you will get a minimum but the zero is, as you say, at an invalid angle for the formula. I tried a brief search for a suitable diagram of the patterns of two omnidirectional sources on Google but I could find nothing. It's the sort of picture you get in books on antenna theory.
    I got to know this stuff from antenna theory, rather than optics and it's actually much more approachable (imo).
  9. Mar 29, 2017 #8
    Will there be any fringes then?
  10. Mar 29, 2017 #9


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    If there fringes then there would have to be zeros, wouldn't there? This is just one very broad maximum.
    We're talking a slit separation of only 600nm or so.
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