# Why is the central maximum the brightest?

This is a question directly from a homework sheet, however I can't find any decent answers online so hopefully someone can help!

In single slit diffraction, why is the central maximum double the width of the other bands and the brightest?

I understand vaguely that it is something to do with Fresnel zones, but I'm quite confused. Instead of making another question, I think this question also fits in with the above:

Why does the intensity of light bands decrease away from the central maximum in single slit diffraction?

Thanks!

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A more physical way of restating Ron's answer: in diffraction theory you are adding up Huygens wavelets over the aperture with appropriate phase factors. The central fringe is the only one where the entire aperture is adding in phase, i.e. constructively. The next fringe will have 2/3 adding constructively and 1/3 destructively, then after that 3/5 - 2/5, etc. The farther out you go in fringes, the more of the aperture is just cancelling itself.

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In single slit diffraction, assuming small diffraction angles, the intensity profile is the magnitude-squared of the Fourier transform of the function which is constant between -1 and 1 (up to units of length), and this Fourier transform is $\frac{\sin(x)}{x}$. This has a peak at zero, because of the falloff of $1/x$ but most importantly, and this can be seen qualitatively, the first zero of $\sin(x)$ is absent, the central maximum is twice as wide as all the others.

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One can formulate a relationship between the separation of the slits, s, the wavelength λ, the distance from the slits to the screen D, and the width of the interference bands (the distance between successive bright fringes), x $$λ / s = x / D$$