It is easy to find equations that quantify this on the internet, but not an explanation as to why...

Also, does this apply to electrons and neutrons? Or just X-ray reflection/diffraction?

  • 2
    $\begingroup$ you know diffraction of light on a grid or double slit? the coarser the grid, or the larger the distance of the slits, the smaller the angle,. This is exactly the same. $\endgroup$
    – trula
    Aug 21, 2020 at 20:32
  • 2
    $\begingroup$ @trula that's an answer $\endgroup$
    – FGSUZ
    Aug 21, 2020 at 22:32
  • $\begingroup$ I noticed that you haven't voted on, commented on, or otherwise interacted with the answer that's there already. If you need an answer with a different approach, what would that be? $\endgroup$
    – uhoh
    Sep 12, 2020 at 3:57

1 Answer 1


It has to do with optical laws, regardless of the wave you use.

Bragg's law gives you the angles in which there is a maximum, but it doesn't say anything about the intensity you will detect on them.

If you perform the calculations, you'll find out that the intensity of the maxima rises with $N^2$, but the width decreases as $1/N$.

You can check this in a simple 1D model. Consider a chain of $N$ equally spaced atoms at a distance $a$. They're indexed from $0$ to $N-1$. Any lattice vector is $R=n\cdot a$

The intensity would be the sum $S$ squared.

$$S=\sum_{n=0}^{N-1} e^{-i \Delta k\cdot R}=\sum_{n=0}^{N-1} e^{-i \Delta k\cdot (na)}=\sum_{n=0}^{N-1} \left(e^{-i \Delta k\cdot a}\right)^n=\dfrac{e^{-i \Delta k\cdot Na}-1}{e^{-i \Delta k\cdot a}-1}$$

$$S=e^{-i \Delta k\cdot (N-1)a/2} \cdot \dfrac{\sin(N\Delta k a/2)}{\sin(\Delta k a/2)}$$

Taylor to First order yields

$$I\propto |S|^2 = 1\cdot \left| \frac{N k a/2}{k a /2 } \right|^2 \propto N^2$$

On the otehr hand, the distance between the two first minima is proportional to $1/N$, so the width of the peak decreasses with $1/N$.


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