# Separation distance of bright fringes in double slit experiment

so I'm working on homework problem that seems like it should be pretty straight forward but the computer won't accept my answer. The question is below.

A double slit aperture with slit separation 0.186 mm is illuminated by light of wavelength 633 ⁢ nm. Fringes are observed on a screen 4.17 m away. Find the spacing between the $m=-100$ and $m=+100$ bright fringes. Express your answer in meters.

In class we showed the distance between bright fringes is $\Delta y=\lambda \frac{L}{d}$. It seemed like the answer should be just be $200 \Delta y$ but that's not working. My mathematica input is shown below just to make sure it's clear that there's no unit errors or typos involved here.

$$200\frac{633 \times10^{-9} m\times 4.17 m}{0.186\times10^{-3}m}=2.8382m.$$

I imagine that maybe the problem is that that formula is only accurate for low order fringes since it's constructed from a small angle approximation and there's supposed to be a way to get around the approximation somehow? Any thoughts?

• Yes, the distance between the maxima -100, and +100 is of the same order of magnitude as L. So, L should be replaced. But in this case L won't be anymore a constant, one and the same for all the 201 maxima, it will depend on y, i.e. will be different for each maximum. – Sofia Nov 24 '14 at 10:37 You can calculate the path lengths $A$ and $B$ using Pythagoras' theorem. The 100th bright fringe is positioned where the different in path length is 100$\lambda$ i.e.
$$B - A = 100\lambda$$
The resulting equation is messy but easily solved numerically. I got $s \approx 1.51$m at the 100th fringe, so the spacing between the 100th and -100th fringes would be about $3.02$m.