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?

  • $\begingroup$ 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. $\endgroup$
    – Sofia
    Nov 24, 2014 at 10:37

1 Answer 1


The experiment looks like:

Young's slits

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.


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