Improving the Quality of X-Ray Images Formed by a Diffraction Lens from a Two-Block Crystalline System

Abstract

The possibility of improving the quality of X-ray images formed using diffraction on a two-block crystalline system is considered. It is shown that the low spatial coherence of the initial radiation improves imaging quality. Conventional wide-focus X-ray tubes can be used as a source for such radiation. The feasibility of using a scanning scheme to reduce the background level in the formed image was also considered by numerical simulation.

APPENDIX 1

The purpose of this Appendix is to estimate the signal-to-background energy flux ratio of the image for the initial δ-shaped wave. With an incident narrow beam under the exact Bragg direction, according to [1], the twice-reflected electric field in the second crystal block is represented by the expression

$$E\left( {x,{{z}_{1}},{{z}_{2}}} \right) \sim \int\limits_{ - \infty }^\infty {\frac{{\sin \Omega {{z}_{1}}\sin \Omega {{z}_{2}}}}{{{{\Omega }^{2}}}}\exp \left( {i\omega x} \right)} d\omega ,$$

(1)

where $\Omega = \sqrt {{{\omega }^{2}} + {{{{\chi }^{2}}} \mathord{\left/ {\vphantom {{{{\chi }^{2}}} 4}} \right. \kern-0em} 4}}$. Here we use a dimensionless coordinate system (x, z) in the scattering plane with axes x parallel and z perpendicular to the surfaces of the crystal blocks, as defined in [10]. In this case, z₁ is the thickness of the first crystalline block, and z₂ is the depth of the point in question in the second crystalline block.

Converting the sines in (1) to exponentials, up to an insignificant constant factor, we obtain

$$E\left( {x,{{z}_{1}},{{z}_{2}}} \right) = {{E}_{1}}\left( {x,{{z}_{1}},{{z}_{2}}} \right) + {{E}_{2}}\left( {x,{{z}_{1}},{{z}_{2}}} \right),$$

(2)

where

$$\begin{gathered} {{E}_{1}}\left( {x,{{z}_{1}},{{z}_{2}}} \right) = \frac{1}{2}\sum\limits_{\nu = \pm 1} {\int\limits_{ - \infty }^{ + \infty } {\exp \left\{ {i\left( {\nu \Omega \delta + \omega x} \right)} \right\}\frac{{d\omega }}{{{{\Omega }^{2}}}}} } = \int\limits_{ - \infty }^{ + \infty } {\frac{{\cos \left( {\Omega \delta } \right)}}{{{{\Omega }^{2}}}}{{e}^{{i\omega x}}}d\omega } , \\ {{E}_{2}}\left( {x,{{z}_{1}},{{z}_{2}}} \right) = - \frac{1}{2}\sum\limits_{\nu = \pm 1} {\int\limits_{ - \infty }^{ + \infty } {\exp \left\{ {i\left( {\nu \Omega Z + \omega x} \right)} \right\}\frac{{d\omega }}{{{{\Omega }^{2}}}}} } = - \int\limits_{ - \infty }^{ + \infty } {\frac{{\cos \left( {\Omega Z} \right)}}{{{{\Omega }^{2}}}}{{e}^{{i\omega x}}}d\omega } , \\ \end{gathered} $$

(3)

$Z = {{z}_{2}} + {{z}_{1}}$, $\delta = {{z}_{2}} - {{z}_{1}}$. The field ${{E}_{1}}\left( {x,{{z}_{1}},{{z}_{2}}} \right)$ corresponds to the above-mentioned rays, which undergo interbranch scattering when transitioning from the first crystalline block to the second and are focused at z₂ = z₁, while ${{E}_{2}}\left( {x,{{z}_{1}},{{z}_{2}}} \right)$—to the rays that do not undergo interbranch scattering, diverge and form a background. This can be confirmed by applying the stationary phase method to the integrals under the summation signs in (3), which in the case $\delta \chi \gg 1$ will result in the ray trajectories in the second block

$$x = \mp \frac{\omega }{{\sqrt {{{\omega }^{2}} + {{{{\chi }^{2}}} \mathord{\left/ {\vphantom {{{{\chi }^{2}}} 4}} \right. \kern-0em} 4}} }}\delta ,\,\,\,\,\left( { - \infty < \omega < + \infty } \right)$$

for the terms of the field ${{E}_{1}}$ and

$$x = \mp \frac{\omega }{{\sqrt {{{\omega }^{2}} + {{{{\chi }^{2}}} \mathord{\left/ {\vphantom {{{{\chi }^{2}}} 4}} \right. \kern-0em} 4}} }}Z,\,\,\,\,\left( { - \infty < \omega < + \infty } \right)$$

for the terms of ${{E}_{2}}$.

From (3), using Parseval’s theorem for the energy fluxes of fields ${{E}_{1}}$, and ${{E}_{2}}$, at z₂ = z₁, we obtain, respectively

$$\begin{gathered} {{\Phi }_{1}} \equiv \int\limits_{ - \infty }^{ + \infty } {{{{\left| {{{E}_{1}}\left( x \right)} \right|}}^{2}}dx = \int\limits_{ - \infty }^{ + \infty } {\frac{{d\omega }}{{{{\Omega }^{4}}}}} = \frac{8}{{{{\chi }^{3}}}}} \int\limits_{ - \infty }^{ + \infty } {\frac{{dp}}{{{{{\left( {{{p}^{2}} + 1} \right)}}^{2}}}}} , \\ {{\Phi }_{2}} \equiv \int\limits_{ - \infty }^{ + \infty } {{{{\left| {{{E}_{2}}\left( x \right)} \right|}}^{2}}dx = \int\limits_{ - \infty }^{ + \infty } {\frac{{{{{\cos }}^{2}}\left( {\Omega Z} \right)}}{{{{\Omega }^{4}}}}d\omega } } = \frac{1}{2}\int\limits_{ - \infty }^{ + \infty } {\frac{{1 + \cos \left( {2\Omega Z} \right)}}{{{{\Omega }^{4}}}}d\omega } \\ = \frac{4}{{{{\chi }^{3}}}}\left[ {\int\limits_{ - \infty }^{ + \infty } {\frac{{dp}}{{{{{\left( {{{p}^{2}} + 1} \right)}}^{2}}}}} + \int\limits_{ - \infty }^{ + \infty } {\frac{{\cos \left( {\sqrt {{{p}^{2}} + 1} Z\chi } \right)}}{{{{{\left( {{{p}^{2}} + 1} \right)}}^{2}}}}dp} } \right]. \\ \end{gathered} $$

Using table integral $\int_{ - \infty }^{ + \infty } {{{{\left( {{{p}^{2}} + 1} \right)}}^{{ - 2}}}dp = {\pi \mathord{\left/ {\vphantom {\pi 2}} \right. \kern-0em} 2}}$ and computing the last integral in the expression of ${{\Phi }_{2}}$ using the stationary phase method, in the case of $Z\chi \gg 1$, we obtain

$${{\Phi }_{1}} = \frac{{4\pi }}{{{{\chi }^{3}}}},\,\,\,\,{{\Phi }_{2}} \approx {{\Phi }_{1}}\left[ {\frac{1}{2} + \sqrt {\frac{2}{{\pi Z\chi }}} \;\cos \left( {\frac{\pi }{4} + Z\chi } \right)} \right] \approx \frac{{{{\Phi }_{1}}}}{2}.$$

Thus, when $\delta = 0$ and $Z\chi \gg 1$, the signal-to-background energy flux ratio is close to 2.

Note that in transitioning from dimensionless coordinates to normal ones, the condition $Z\chi \gg 1$ is transformed to ${{2\pi \left( {{{t}_{1}} + {{t}_{2}}} \right)} \mathord{\left/ {\vphantom {{2\pi \left( {{{t}_{1}} + {{t}_{2}}} \right)} {\Lambda \gg 1}}} \right. \kern-0em} {\Lambda \gg 1}}$, where ${{t}_{1}}$ and ${{t}_{2}}$ are the thicknesses of the first and second crystalline blocks in normal dimensions, respectively.