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Homework Help: Diffraction patterns

  1. Aug 26, 2011 #1
    1. The problem statement, all variables and given/known data

    A beam of monochromatic light is incident on a single slit of width 0.560 mm. A diffraction pattern forms on a wall 1.35 m beyond the slit. The distance between the positions of zero intensity on both sides of the central maximum is 2.32 mm. Calculate the wavelength of the light.

    2. Relevant equations

    [tex]y=L \ sin \theta_{dark}[/tex]

    [tex]sin \theta_{dark} = m \frac{\lambda}{a}[/tex]

    3. The attempt at a solution

    [tex]sin \theta_{dark} = \frac{\lambda}{a}[/tex]

    And since

    [tex]sin \theta_{dark} = \frac{y}{L}[/tex]

    We have [tex]\lambda = \frac{ya}{L} = \frac{(2.32 \times 10^{-3})(0.56 \times 10^{-3})}{1.35} = 962.37 \ nm[/tex]

    But why is my answer wrong? :confused:
    Last edited: Aug 27, 2011
  2. jcsd
  3. Aug 27, 2011 #2


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    Homework Helper

    I haven't checked you numbers, just the idea but...

    Have you taken into account the fact that formulas often work with the angle off the axis/normal to the dark fringe, where as the distance was from the dark fringe on the left to the dark fringe on the right?
  4. Aug 27, 2011 #3
    Okay, I tried to do it differently, but the computer still marks me wrong:

    [tex]d \ sin \theta_{min} = \lambda[/tex]

    [tex]tan \theta = (2.32 \times 10^{-3}){1.35} = 0.0017185[/tex]

    [tex]\theta = 0.09846 \ degrees[/tex]

    [tex]\lambda = (0.560 \times 10^{-3}) \times sin 0.09846 = 962.4 nm[/tex]

    What should I do?
  5. Aug 27, 2011 #4
    Look at Peters post again.
  6. Aug 27, 2011 #5
    The angle in the equation is measured from the centre of the pattern.You used a distance of 2.32mm but you should have used half of that distance.
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