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Young's Experiment - (problem with equation)

  1. Sep 16, 2006 #1
    So my book has the following expression for Young's double slit experiment.

    (maxima - bright fringes) [itex] d \sin \theta = m \lambda [/itex]
    for [itex] m = 0, \, 1 \, 2 \, \ldots [/itex].

    So what if you solve this for wavelength.

    [tex] \lambda = \frac{d \sin \theta}{m} [/tex]

    How is this valid when [itex] m = 0 [/itex]

    Is this because if [itex] m = 0 [/itex], [itex] \theta [/itex] HAS to equal 0?

    by the way
    d is the distance between the slits
    theta is the angle from the central axis
    lambda is the wavelength
    m is the index of where the maxima occur
  2. jcsd
  3. Sep 16, 2006 #2

    Andrew Mason

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    The first maximum is always at 0 angle regardless of wavelength. So you can't determine the wavelength from the position of the central maximum.

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