Why does the Debye-Scherrer procedure work (powder diffractometer) In the Debye-Scherrer procedure a sample of crystalline powder is hit by a beam of monochromatic photons. The diffracted photons are measured with a detector. We have constructive interference of the photons if the Bragg conditions are met.
In my lecture notes it says that there are always crystalline planes oriented in a way that the Bragg conditions are fulfilled, thus leading to the typical rings observed with powder diffractometer.

(Graphic taken from Wikipedia)
I have a hard time seeing why there are discrete rings, because from the reason above I would expect that for every angle there is constructive interference and hence we would have a continuous intensity distribution for all angles.
Question: Why are there discrete rings and not a continuous distribution?
 A: 1. X-ray diffraction takes pictures of Fourier space
As briefly described on pg. 34 of "Introduction to Solid State Physics" 7th edition by Kittel, the scattering amplitude for an arbitrarily-shaped object is $$F(\mathbf{k}_i,\mathbf{k}_o)=\widehat{N}(\mathbf{k}_i-\mathbf{k}_o)$$
where $\mathbf{k}_i$ is the incident wavevector, $\mathbf{k}_o$ is the outgoing (or scattered) wavevector, and $\widehat{N}$ is the Fourier transform of the material's scattering density $N$. Since it's often electrons which scatter light, $N$ is often assumed to be the electron density as a function of location in the material.
This is a surprisingly simple equation; to illustrate it's visual meaning, imagine an object is placed inside a hollow sphere whose walls are lined with photographic paper, and through a small hole the object is bombarded with X-rays of fixed wavevector $\mathbf{k}_i$, which are scattered by the object and strike the interior of the imaging sphere, forming an image. What is this image? Imagine a bubble of radius $|\mathbf{k}_i|$ centered at $\mathbf{k}_i$ in Fourier space. The image formed on the photographic paper is the image of $\widehat{N}$ on the surface of that bubble, a fact which is simple to deduce by examining the set of points of the form $\mathbf{k}_i-\mathbf{k}_o$ for a fixed value of $\mathbf{k}_i$, and noting that $|\mathbf{k}_i|=|\mathbf{k}_o|$. That is what X-ray diffraction examines; it literally takes a bubble-shaped slice of an object in Fourier space. This bubble is sometimes call the "Ewald bubble", and it is what diffraction takes a picture of.
2. Fourier space of a crystal
Now, let's insert the fact that we're looking at a crystal. A very simple approximation of a crystal is a 3D Dirac Comb (ie, a Dirac Delta placed at each unit cell), whereupon $$N(\mathbf{r})\approx \mbox{DiracComb}(\mathbf{Ar})$$
where $\mathbf{A}$ is the inverse of the $3\times3$ matrix whose columns are the 3 lattice basis vectors for the crystal. One then Fourier transforms to obtain
$$\widehat{N}(\mathbf{k})\approx\mbox{Det}(A)^{-1}\mbox{DiracComb}\left(\frac{\mathbf{A}^{-1}\mathbf{k}}{2\pi}\right)$$
which essentially tells you that the Fourier transform of the lattice $N$ is also another lattice. This lattice, residing in Fourier space, is often referred to as the "reciprocal lattice" of a crystal.
3. Crystal diffraction
Now let's combine the previous two results. If a reciprocal lattice point of $\widehat{N}$ happens to reside on the surface of the Ewald bubble, then light will be diffracted in that direction in real space. But how do we image the rest of reciprocal space, not just the points that happen to lie on that one bubble-shaped slice? Suppose we rotate the crystal through Euler angles $\alpha,\beta,\gamma$. Then $N(\mathbf{r})$ will become $N(\mathbf{R}(\alpha,\beta,\gamma)\cdot\mathbf{r})$ where $\mathbf{R}(\alpha,\beta,\gamma)$ is the rotation matrix for the crystal rotation. Since $$\mathbf{R}(\alpha,\beta,\gamma)^{-1}=\mathbf{R}(-\gamma,-\beta,-\alpha),$$ we see that $\widehat{N}(\mathbf{r})$ becomes $\widehat{N}(\mathbf{R}(-\gamma,-\beta,-\alpha)\cdot\mathbf{r})$ (rotations are unitary so the determinant is 1), and hence rotation of a crystal in real space correspond to a reversed rotation of the crystal's reciprocal space.
Thus, by rotating the crystal and imaging it after each rotation, we can sweep out a spherical region of radius $2|\mathbf{k}|$ in reciprocal space. Visually, we are rotating the Ewald bubble, whose surface is attached to the origin, to sweep out a spherical region whose radius is the diameter of the imaging bubble (which is $2|\mathbf{k}|$).
4. Powder diffraction
With those preliminaries out of the way, powder diffraction is quite simple. A powder is a large number of very small crystals oriented in random directions. If there are a large number of crystals all oriented randomly (isotropic fine powder) then we can average over all orientations to get $$\widehat{N}_{avg}(\mathbf{k})\propto\int_0^{2\pi}d\alpha\int_0^{\pi}d\beta\mbox{sin}(\beta)\int_0^{2\pi}d\gamma\widehat{N}(\mathbf{R}(-\gamma,-\beta,-\alpha)\cdot\mathbf{k})$$
but you don't need to worry about integrating it, because it's visually obvious that each Delta function located at a reciprocal lattice point $\mathbf{G}$ will be "rotationally smeared" out to form a series of concentric bubbles of radius $|\mathbf{G}|$ centered at the origin.
Where these concentric bubbles intersect the Ewald bubble, diffraction will occur in that direction. And it is a simple fact of geometry that when one bubble intersects another, the intersection region is a circular ring common to both. As a result, there will be ring-shaped regions on the Ewald bubble where constructive interference can occur, and thus, the sample will emit cone-shaped beams of light. And that's how the cone-shaped beams of light in the picture above come to fruition.
