Geometrical interpretation of the solution in the zero-absorption case

Authier, A.

doi:10.1107/97809553602060000569

International
Tables for
Crystallography
Volume B
Reciprocal space
Edited by U. Shmueli

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International Tables for Crystallography (2006). Vol. B. ch. 5.1, pp. 540-541 | 1 | 2 |

Section 5.1.3.7. Geometrical interpretation of the solution in the zero-absorption case

A. Authier^a ^*

^a Laboratoire de Minéralogie-Cristallographie, Université P. et M. Curie, 4 Place Jussieu, F-75252 Paris CEDEX 05, France
Correspondence e-mail: authier@lmcp.jussieu.fr

5.1.3.7. Geometrical interpretation of the solution in the zero-absorption case

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5.1.3.7.1. Transmission geometry

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In this case (Fig. 5.1.3.4) $[\hbox{S}(\gamma_{h})]$ is +1 and (5.1.3.10) may be written $[\xi_{j} = -\hbox{S}(C) \left[\eta \pm (\eta^{2} + 1)^{1/2}\right] / \gamma^{1/2}. \eqno(5.1.3.11)]$

Let $[A_{1}]$ and $[A_{2}]$ be the intersections of the normal to the crystal surface drawn from the Lorentz point $[L_{o}]$ with the two branches of the dispersion surface (Fig. 5.1.3.4). From Sections 5.1.3.3 and 5.1.3.4, they are the tie points excited for $[\eta = 0]$ and correspond to the middle of the reflection domain. Let us further consider the tangents to the dispersion surface at $[A_{1}]$ and $[A_{2}]$ and let $[I_{o1}]$ , $[I_{o2}]$ and $[I_{h1}]$ , $[I_{h2}]$ be their intersections with $[T_{o}]$ and $[T_{h}]$ , respectively. It can be shown that $[\overline{I_{o1}I_{h2}}]$ and $[\overline{I_{o2}I_{h1}}]$ intersect the dispersion surface at the tie points excited for $[\eta = -1]$ and $[\eta = +1]$ , respectively, and that the Pendellösung distance $[\Lambda_{L} = 1/\overline{A_{2}A_{1}}]$ , the width of the rocking curve $[2\delta = \overline{I_{o1}I_{o2}}/k]$ and the deviation parameter $[\eta = \overline{M_{o}M_{h}}/\overline{A_{2}A_{1}}]$ , where $[M_{o}]$ and $[M_{h}]$ are the intersections of the normal to the crystal surface drawn from the extremity of any incident wavevector OM with $[T_{o}]$ and $[T_{h}]$ , respectively.

5.1.3.7.2. Reflection geometry

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In this case (Fig. 5.1.3.5) $[\hbox{S}(\gamma_{h})]$ is now −1 and (5.1.3.10) may be written $[\xi_{j} = \hbox{S}(C) \left[\eta \pm (\eta^{2} - 1)^{1/2}\right] / |\gamma|^{1/2}. \eqno(5.1.3.12)]$

Now, let $[I_{1}]$ and $[I_{2}]$ be the points of the dispersion surface where the tangent is parallel to the normal to the crystal surface, and further let $[I_{o1}, I_{h1}, I']$ and $[I_{o2}, I_{h2}, I'']$ be the intersections of these two tangents with $[T_{o}, T_{h}]$ and $[T'_{o}]$ , respectively. For an incident wave of wavevector OM where M lies between I′ and I″, the normal to the crystal surface drawn from M has no real intersection with the dispersion surface and $[\overline{I'I''}]$ defines the total-reflection domain. The tie points $[I_{1}]$ and $[I_{2}]$ correspond to $[\eta = -1]$ and $[\eta = +1]$ , respectively, the extinction distance $[\Lambda_{L} = 1/\overline{I_{o1}I_{h1}}]$ , the width of the total-reflection domain $[2\delta = \overline{I_{o1}I_{o2}}/k = \overline{I'I''}/k]$ and the deviation parameter $[\eta = - \overline{M_{o}M_{h}} / \overline{I_{o1}I_{h1}}]$ , where $[M_{o}]$ and $[M_{h}]$ are the intersections with $[T_{o}]$ and $[T_{h}]$ of the normal to the crystal surface drawn from the extremity of any incident wavevector OM.

References

International Tables for Crystallography (2006). Vol. B. ch. 5.1, pp. 540-541