Vertical focussing.

Next: Crystal and source dimensions. Up: Some basic facts. Previous: Horizontal focussing. Contents

Vertical focussing.

Source and detector are mounted on a cylinder axis with the Bragg crystal on a cylindrical surface [89]. The crystal is bent with a curvature $R_{V}$ . The distance of source to detector is given by

$\begin{displaymath} x=2\cdot R_{V}\cdot cot\Theta _{B}. \end{displaymath}$

(18)

By invoking Bragg's law the energy dispersion is given by

$\begin{displaymath} \frac{dx}{dE}=-\frac{2\pi \hbar c}{E^{2}}\frac{R_{V}}{d\cdot cos\Theta _{B}\cdot sin^{2}\Theta _{B}} \end{displaymath}$

(19)

$\begin{displaymath} \frac{dx}{dE}=-\frac{2\cdot d\cdot R_{V}}{\pi \cdot \hbar \c... ...c{1}{\sqrt{1-(\frac{\pi \cdot \hbar \cdot c}{d\cdot E})^{2}}}. \end{displaymath}$

(20)

As from the horizontal focussing condition the distance source $\leftrightarrow$ detector is required to be $2R_{c}\cdot sin\Theta _{B}\cdot cos\Theta _{B}$ and the focal length is given by $R_{c}\cdot sin\Theta _{B}$ . As these two quantities form a rectangular triangle together with $R_{V}$ ( $R^{2}_{V}=R_{c}^{2}sin^{2}\Theta _{B}-\frac{L^{2}}{4}$ , the relation between $R_{V}$ and $R_{C}$ is given by

$\begin{displaymath} R_{V}=R_{c}\cdot sin^{2}\Theta _{B}. \end{displaymath}$

(21)

Therefore the expression for the energy dispersion has the form

$\begin{displaymath} \frac{dx}{dE}=\frac{2\pi \hbar c}{E^{2}}\frac{R_{c}}{d\cdot cos\Theta _{B}} \end{displaymath}$

(22)

In practice $R_{V}=2\cdot R=R_{c}$ is chosen because the otherwise necessary ellipsoidal curvature faces major technical difficulties. The loss in vertical focussing is still tolerable.

Next: Crystal and source dimensions. Up: Some basic facts. Previous: Horizontal focussing. Contents

Pionic Hydrogen Collaboration
1998