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Homework Help: Length of the box based on the fringe pattern? I think I'm close.

  1. Nov 30, 2008 #1
    1. The problem statement, all variables and given/known data
    The figure 25.CP42.jpg
    shows a neutron in a one-dimensional box. If the right end of the box is opened, the neutron travels out of the box, impinges on a double slit, and is detected 2.0 m behind the double slit. Repeating the experiment over and over produces the fringe pattern shown in the figure.

    What is the length (in nm) of the box?

    2. Relevant equations

    Variables in equations explained in "The attempt at a solution"

    Fringe position:


    Fringe spacing:

    3. The attempt at a solution
    I thought I could use the fact that the fringe position is y(m)=(m*lamda*L)/(d) where lamda is the wavelength, m is the number of fringes which should be two (m=2) since the central max. is m=0. Then L is the length behind the double slit. d is the fringe spacing. I then plugged that into the formula for the wavelength of a open-closed tube which is lamada(m)=(4*L)/(m) where m in this case is m=4 since there are 4 modes in the wave. The two fours cancel out giving lamda=L so, L=0.0165nm but that is not right. So, I tried using the fact that the fringe spacing (instead of fringe position) was (delta)y=(lamda*L)/(d) and I got an answer for L=0.033nm. I know I am close... I'm not the best at physics so can someone spot where I am going wrong?
  2. jcsd
  3. Dec 1, 2008 #2


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    Staff Emeritus
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    Homework Helper

    1. I wouldn't use L in the diffraction fringe equation, since it could get confused with the length of the box L.

    2. What numbers are you using for Δy and d in the diffraction fringe equation? And what do you get for λ as a result?

    Please check that equation. I'm pretty sure the "4" should be a "2".
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