# Interplanar distance given Miller indice of the planes

Say we've a cubic crystal of unity edge length.

A set of planes in such a crystal is specified by their miller indice as $$(3 2 0)$$. One of these planes then has intercepts on the axes as $$(2 ,3 ,\infty)$$. It's well known that the interplanar distance between two adjacent planes in a cubic lattice is given

$$d_{hkl} = \frac{a}{\sqrt{h^2+k^2+l^2}}$$,in our case this gives us

$$d_{320} = 1/\sqrt13$$.

Now if I look at the planes which form the Miller indices of $$3 2 0$$ I find that apart from a plane having intercepts as $$(2 ,3,0)$$ an adjacent parallel plane to this plane having the same indices will pass via origin $$0,0,0$$ and the distance between them is $$d_{320} = 6/\sqrt13$$

Which contradicts the above result! Can anyone please point out what's wrong in my reasoning?

## 1 Answer

Your adjacent plane isn't.

Worth a thousand words, they say.

• Thank you Mr Ivan Neretin. Nov 28, 2020 at 3:58