# Length of the box based on the fringe pattern? I think I'm close.

## Homework Statement

The figure

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?

## Homework Equations

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

Fringe position:
y(m)=(m*lamda*L)/(d)

Fringe spacing:
(delta)y=(lamda*L)/(d)

## 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?

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