![]() ![]() Without it, we would not understand the structure of matter. It is justifiably famous, because this very simple little equation launched the whole field of X-ray diffraction, and thereby the study of the solid state. Which is the very, very famous Bragg Law. Reiterating, strong scattering occurs when Clearly, when all the atoms are scattering phase threads which have exactly the same phase in the diffraction plane, then at this particular scattering angle, we get a huge scattered amplitude (intensity peak). at 2d, 3d, 4d, etc), will scatter in phase to the diffraction pattern. When this condition is satisfied, all atoms in either plane, or indeed any other plane which is positioned an integral number of distances d away from the first plane (i.e. From any one of the little right-angle triangles drawn above, we see that this distance is 2d sin θ B, where d is the separation of the planes. For these threads to end up in phase, the extra path length experienced by the lower thread (the red line in the diagram above) must be an integral number of wavelengths long: that is, nλ, where n is integer. To work out the answer, we now draw two phase threads, one scattered by the top blue plane, the other scattered by the next blue plane down. We now consider the following problem: for a particular separation of adjacent blue planes (Bragg planes), at what scattering angle will atoms positioned in either plane also scatter with identical phase into the Fraunhofer plane? We’ve already said that any atom positioned anywhere on a particular blue plane (the planes extend in and out of the paper/computer screen) will scatter into the Fraunhofer plane with the same phase. Be warned that in crystallography and most (but not all) contexts in electron microscopy, θ is assumed to be the Bragg angle, the subscript being left out. It is half the total scattering angle which until this point we have called θ (shown coloured purple above). The green angle is called the ‘Bragg angle’ which I will call θ B. In most of textbooks, only the blue and brown lines are drawn, like this:įurthermore, the picture is usually rotated to look like this:Īnd then if we zoom in to area I have circled, the following geometric construction is made: Note that this vector is perpendicular to the blue planes. The orange vector is the scattering vector, as defined before. I have chosen to draw these lines passing through the gaps in the original diffraction grating. ![]() The blue parallel lines are the planes in which, for any particular phase thread satisfying this particular scattering angle (see above), the position of any actual scattering point in such a plane does not alter the resultant phase of the wave scattered into the Fraunhofer diffraction pattern. The dark brown vectors are the k-vectors corresponding to the incident and scattered wave. The dotted black line is a diffraction grating. In this rather cluttered picture, the green lines represent points of constant phase in the incident and scattered waves. Now let’s rearrange the angles involved so that once again our incident beam is coming from the left of the diagram and is scattering off a diffraction grating. In fact, we can move the scattering point out of the paper (computer screen), so the blue line is really a blue plane of points, all of which add in the same phase in the detector plane. If the eye of the needle moves anywhere along the blue line above, the phase thread remains the same length: any atom or scattering point positioned anywhere on the blue line will give a phase thread that adds with exactly the same phase in the detector plane. Zooming into this, we see that in this symmetrical set up, we can move the needle left or right without changing the length of the thread: That means we can just concentrate on positions close to the red thread above. That means the allowable positions of the needle describe an ellipse, as below:īut of course we are only interested in Fraunhofer diffraction, which means the distances of the needle from the source and the detector are very large. Let’s draw the thread and the needle on a larger scale, and more symmetrically, as follows: The question I want to ask is: how am I allowed to move the atom so that the phase relationship between the incident beam and the scattered beam remains constant? This is equivalent to asking: how can I move the needle keeping the thread which extends from the source to the detector taught and of constant length? The eye of the needle is, in this analogy, the position of the atom. In my flippant introduction to crystallography and diffraction theory, I suggested that scattering from a single atom could be represented by a phase thread passing through the eye of needle, like this: ![]()
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