Distance between adjacent planes in a crystal This question has been asked before, but there doesn't seem to be a decent answer.
Many sources state that " For cubic crystals with lattice constant a, the spacing d between adjacent (ℓmn) lattice planes is:
$$  {\displaystyle d_{\ell mn}={\frac {a}{\sqrt {\ell ^{2}+m^{2}+n^{2}}}}}$$
"
https://en.wikipedia.org/wiki/Crystal_structure
Could someone please explain what "adjacent" means in this case (Is it planes that share the same side, is it parallel planes, are these panes in the same unit cell or neighbouring cells etc)? Better yet, does anyone know of a sketch explaining this ?
I am really at a loss here and this has been driving me nuts the whole day
 A: 
I could only find this poor-quality picture. It should give you an idea, anyway. For example, consider the first picture in the first row: $(l,m,n)=(1,0,0)$ in that case, and it is easy to verify that the distance between the grey planes is
$$d=a$$
In the second case, $(l,m,n)=(1,1,0)$, and you can see that
$$d=\frac a {\sqrt 2}$$
etc.
A: In answer to the question: adjacent planes are planes that are closest to one another when distance is measured along the normal to the plane. It is important to understand that every lattice point has exactly one of the infinite set of planes described by the Miller indices $(h k \ell)$ passing through it. (I will use $(h k \ell)$ instead of $(\ell m n)$ like the OP.) I agree that many explanations out there seem to lack some important information, so here is a somewhat rigorous treatment.
Disregarding some special cases, Miller indices are defined as follows. First, find the intersections of the plane in question along the three crystal axes $\pmb{a}, \pmb{b}, \pmb{c}$ in terms of multiples of the lattice constants, i.e., $m a, n b, o c$ for integers $m, n, o$. Then take the reciprocals of $m, n, o$ and find three integers $h, k, \ell$ having the same ratio, and whose greatest common divisor is 1. As an example, consider the plane that intersects the $\pmb{a}$ axis at the second lattice site, the $\pmb{b}$ axis at the third lattice site, and the $\pmb{c}$ axis at the first lattice site. The reciprocals of $2, 3, 1$ are $\frac{1}{2}, \frac{1}{3}, 1$, which have the same ratio as $3, 2, 6$. The plane is thus called $(h k \ell) = (326)$.
To find the distance between adjacent planes, it helps to use the ``reciprocal lattice vectors'', which may be defined as
$$ \pmb{a^*} = V^{-1}\pmb{b} \times \pmb{c} \, , \quad  \pmb{b^*} = V^{-1}\pmb{c} \times \pmb{a} \, , \quad \pmb{c^*} = V^{-1}\pmb{a} \times \pmb{b} \, $$
where $V = \pmb{a} \cdot ( \pmb{b} \times \pmb{c})$ is the volume of the unit cell. By construction, these have the convenient property that, for example, $\pmb{a} \cdot \pmb{a^*} = 1$, while $\pmb{a} \cdot \pmb{b^*} = 0$, and so on. It turns out that the vector $ \pmb{H} = h \pmb{a^*} + k \pmb{b^*} + \ell \pmb{c^*} $ is normal to the $(h k \ell)$ plane. This can be demonstrated by showing that the dot products of $\pmb{H}$ with two non-colinear vectors in the $(h k \ell)$-plane, for example, $n \pmb{b} - m \pmb{a}$ and $o \pmb{c} - n \pmb{b}$, are zero.
Consider now the plane $P_0$ that passes through the lattice point at the origin and is defined by $ \pmb{H} \cdot \pmb{r} = 0 \, , $ where
$ \pmb{r} = x \pmb{a} + y \pmb{b} + z \pmb{c} $ for coordinates $x, y, z$. Because of the convenient properties of the reciprocal lattice vectors described above, we can rewrite $\pmb{H} \cdot \pmb{r} = 0$ as $h x + k y + \ell z = 0$. The lattice points are those $\pmb{r}$ for which $x, y, z$ are integers, call them $p, q, s$, i.e., we have $h p + k q + \ell s = 0$. The origin is the trivial case, where $p = q = s = 0$.
We now wish to find the closest plane, call it $P_1$, by moving from the origin along the positive $\pmb{H}$ direction. The equation of $P_1$ is $\pmb{H} \cdot \pmb{r} = \delta$, or $h p + k q + \ell s = \delta$ for some delta. The geometrical interpretation of the dot product means that $P_1$ should possess the smallest value of $\delta$ possible. Furthermore, because $h, k, \ell$ and $p, q, s$ are all integers, so too must be $\delta$. The smallest possible integer value of $\delta$ is 1. We are guaranteed to be able to find $p, q, s$ satisfying $h p + k q + \ell s = 1$ because of Bezout's identity, which says that for two integers $a$ and $b$ (not the same $a$ and $b$ as above, but we are running out of variable names) with greatest common factor $f$ (written $\mathrm{gcd}(a,b)=f$), there exist integer $x$ and $y$ (again, not the $x$ and $y$ above) such that $ax + by = f$. This generalizes to more than one pair of integers. Thus, we can always find $p, q, s$ such that $h p + k q + \ell s = 1$ because $\mathrm{gcd}(h, k, \ell) = 1$.
Now that we know $\delta$, we wish to find the distance between $P_0$ and $P_1$ measured along $\pmb{H}$. This can be accomplished first by traveling along $\pmb{a}$ from the origin until we encounter $P_1$, i.e., finding $x$ so that $H \cdot (x \pmb{a}) = 1$. This has solution $x = \frac{1}{h}$, so that the vector $\pmb{v} = \frac{1}{h} \pmb{a}$ reaches from $P_0$ at the origin to $P_1$ along the $\pmb{a}$ direction. Finally then, the planar spacing $d$ is the projection of $\pmb{v}$ along the $\pmb{H}$ direction. That is
$$ d = \pmb{v} \cdot \frac{\pmb{H}}{|\pmb{H}|} = \frac{1}{|\pmb{H}|} \, . $$
For the special case of the primitive cubic lattice, the lattice vectors are all orthogonal with lattice constant $a$, i.e. $\pmb{a} = a \hat{\pmb{x}}$ and so on, and the reciprocal lattice vectors are $\pmb{a^*} = \frac{1}{a} \hat{\pmb{x}}$ and so on. Therefore $|\pmb{H}| = \frac{1}{a} \sqrt{h^2 + k^2 + \ell^2}$, giving
$$ d = \frac{a}{\sqrt{h^2 + k^2 + \ell^2}} \, . $$
