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