I am doing the classic exercise of finding the distance between interplanar planes in a cubic lattice. So i need to demonstrate that the distance d of the (h,k,l) planes is $d = a/(h²+k²+l²)^{1/2}$.

So, using elementary geometry, the distance between parallel planes is $|c-c'|/\sqrt{m²+n²+p²}$, where the plane equations are $0 = xm + y n + z p - c, 0 = x m + y n +z p - c'$.

Now, since (h,k,l) are miller indices, i thought that the right equation for the planes would be $0 = x/h + y/k + z/l - c$, since the miller indices are the reciprocal of these "geometric indices".

Unfortunately, using this equation and the formula, we get $d = a/(1/h²+1/k²+1/l²)^{1/2}$.

I think the main problem here is that i am interpreting the miller indices wrong, even so I don't know where my error is... Could you help me?


$\mathbf{The\;General \;Method}$
Let $\boldsymbol{a_1}$, $\boldsymbol{a_2}$ and $\boldsymbol{a_3}$ be the crystallographic axes. Then the planes with Miller indices $(hkl)$ intercept the crystallographic axes at $n\boldsymbol{a_1}/h,n\boldsymbol{a_2}/k$ and $n\boldsymbol{a_3}/l$, where $n$ is an integer. In general, the vectors $h\boldsymbol{a_1}+k\boldsymbol{a_2}+l\boldsymbol{a_3}$ is not perpendicular to the $(hkl)$ planes. This is why the general method of determination of the interplanar spacing requires reciprocal lattice basis $\boldsymbol{b_1},\boldsymbol{b_2}$ and $\boldsymbol{b_3}$, as the reciprocal lattice vectors $\boldsymbol{G}_{hkl}=h\boldsymbol{b_1}+k\boldsymbol{b_2}+l\boldsymbol{b_3}$ is perpendicular to the family of $(hkl)$ planes. Expoliting this property, the $(hkl)$ interplanar spacing $d_{hkl}$ can be found by projecting the vector joining the separation of two adjacent planes at the $\boldsymbol{a_1}$ axis onto the unit normal, that is $$d_{hkl}=\frac{\boldsymbol{a_1}}{h}\cdot \frac{\boldsymbol{G}_{hkl}}{|\boldsymbol{G}_{hkl}|}. \tag{1}\label{eq1}$$ Since $\boldsymbol{b_i}\cdot \boldsymbol{a_j}=2\pi\delta_{ij},$ Eq.\eqref{eq1} can be expressed as $$d_{hkl}=\frac{2\pi}{|\boldsymbol{G}_{hkl}|}. \tag{2}\label{eq2}$$
$\mathit{Simple\, Cubic \,Lattice}$
For simple cubic lattice, $|\boldsymbol{a_1}|=|\boldsymbol{a_2}|=|\boldsymbol{a_3}|=a.$ The reciprocal lattice vector is $\boldsymbol{G}_{hkl}=\frac{2\pi}{a}(h\hat x+k\hat y+l\hat z),$ where $\hat x, \hat y$ and $\hat z$ are orthogonal unit vectors and by Eq.\eqref{eq2}, the interplanar spacing is $$d_{hkl}=\frac{a}{\sqrt{h^2+k^2+l^2}}. \tag{3}\label{eq3}$$

Having presented the general formalism, now I attempt to use your reasoning to find the interplanar spacing for the simple cubic lattice. The nearest $(hkl)$ plane from the origin is found by setting $n=1,$ that is the plane which intercepts the crystallographic axes at $\boldsymbol{a_1}/h,\boldsymbol{a_2}/k$ and $\boldsymbol{a_3}/l$. Let the plane equation for this plane has the form $Ax/a+By/a+Cz/a=D.$ The x-interception is $a/h$, so we must have $D/A=1/h$. By similar reasoning, we get $D/B=1/k$ and $D/C=1/l$. So the plane equation has the form $$\frac{hx}{a}+\frac{ky}{a}+\frac{lz}{a}=1. \tag{4}\label{eq4}$$ From Eq.\eqref{eq4}, we see that the normal vector to the plane is $\frac{h\hat x}{a}+\frac{k\hat y}{a}+\frac{l\hat z}{a}.$ It can be shown that the adjacent plane to this plane should be $hx/a+ky/a+lz/a=0$ or $hx/a+ky/a+lz/a=2.$ Using the formula for distance $D$ between parallel planes $ax+by+cz=d_0$ and $ax+by+cz=d_1$, $$D=\frac{|d_0-d_1|}{\sqrt{a^2+b^2+c^2}}\tag{5}\label{eq5}$$ and $|d_0-d_1|=1$, the interplanar spacing is found to be $$d_{hkl}=\frac{1}{\sqrt{(h/a)^2+(k/a)^2+(l/a)^2}}=\frac{a}{\sqrt{h^2+k^2+l^2}},\tag{6}\label{eq6}$$ which is identical to Eq.\eqref{eq3}.


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