Spacing between adjacent planes of a given set in a simple cubic crystal

Question

Today I learned about Miller indices in a cubic crystal, and I learned that adjacent planes $(hkl)$ in a simple cubic crystal are spaced a distance $d_{hkl}$ from each other, with $d_{hkl}$ given by:

$$d_{hkl} = \frac{a}{\sqrt{h^2 + k^2 + l^2}}$$

For example, take the origin any of the below vertices of the cube with z axis directed upwards, the $(0 0 2)$ plane cuts horizontally the middle of the cube, and for this set of planes, $d_{hkl} = \frac{a}{2}$.

May be a silly question, but if we followed the rule of spacing, we would get that the next $(0 0 2)$ plane is the very upper horizontal plane in the cube, which - i guess - is not true since the latter plane has $(0 0 1)$ and not in any case $(0 0 2)$ miller indices. However, if we considered that $(0 0 2)$ planes are only those passing horizontally through the middle of unit cells, then the distance between them will equal $a$ and not $\frac{a}{2}$ (using imagination and not rules).

I must be missing something very trivial here, but it is only today that i started learning these stuff and i am really confused. Thanks for help.

Answer 1

When Miller indices are different only in a common factor, the planes are in the same orientation. The higher order reflections share some planes with the lower order reflections. In the case mentioned, every second plane of the (2 0 0) planes coincides with the (1 0 0) planes, whose spacing is double that of the former.