## 2 Some Mathematics on Spatial Resolution  
### Spatial Resolution for Unpolarized Neutrons  
Spatial resolution principally depends on the incident beam divergence \(\phi\), in neutron radiography and tomography it is called L/D ratio, the inverse number of \(\sim\phi\). There are different definitions of the L/D ratio (see [4]); however, here we use the most common accepted one, depicted in Figure 1. Conventional neutron imaging instruments use a pinhole aperture (size D) as source, which together with the distance L from the object determines the divergence \(\phi\) of the incident beam. Depending on the size of the source D, or on the mosaic spread of a crystal source D\({}_{\textit{mosaic}}\) a point in a sample is magnified on a screen as d = l\({}_{d}\)D/L (Figure 1). In the case of a crystal monochromator, D must be replaced by d\({}_{\textit{mosaic}}\), d\({}_{\textit{mosaic}}\) = mean grain size of the crystal monochromator and L/D then becomes L/d\({}_{\textit{mosaic}}\)[41]. One can  
Figure 1: To distinguish two points in a sample, L, D, l\({}_{d}\) must be matched to each other. pay attention to the refraction of the beam at edges that causes additional tiny blurring in the image, but refraction is indeed used as well as an imaging signal. This is not to be confused with the additional blur caused by refraction. Here, edge diffraction can be neglected, details on this subject are given in [42; 43; 44; 45; 46; 47]. A finite L/D ratio is always linked with beam divergence \(\phi\) and \(\Delta\lambda/\lambda>0\). Furthermore, a finite precision of scanning and detection (pixel size) contribute as well to smearing effects. To briefly recapitulate the mathematics of spatial resolution, divergence and blur, one considers a perfect straight edge, which is represented by the Heaviside step function \(h(x)\). If an edge is illuminated by divergent radiation, given by \(r=r(x)\), the geometrical image (shadow) is the convolution of \(h(x)\otimes r(x)\). The function \(r(x)\) includes divergence and all other blurring effects and can be best represented by a Gauss function \(g(x)\),  
\[g(x)=\exp\left(-\left[\frac{x-x_{0}}{2\cdot\sigma}\right]^{2}\right) \tag{1}\]  
The convolution \(h(x)\otimes g(x)\) is given as  
\[h(x)\otimes g(x)=\int\limits_{-\infty}^{\infty}\left(\frac{1}{2}+\frac{1}{2} \cdot\text{sgn}(x)\right)\exp\left(-\left[\frac{(x-x^{\prime})-x_{0}}{2\cdot \sigma}\right]^{2}\right)dx^{\prime} \tag{2}\]  
The question to be answered is the \"correct\" scanning (sampling) of the image of the edge, which requires that the sampling frequency f\({}_{s}\) must be twice as high as the highest occurring frequency f\({}_{max}\) in the corresponding frequency spectrum, i.e., f\({}_{s}\) = 2 f\({}_{max}\) (Shannon theorem [48]). The scanning frequency is usually the number of detector pixels per unit length. The highest occurring frequency f\({}_{max}\) in the corresponding frequency spectrum is obtained from the Fourier Transform \(\mathcal{FT}\) of the convolution \(h(x)\otimes g(x)\), C(f) = \(\mathcal{FT}\{h(x)\otimes g(x)\}\). If \(H(f)\) and \(G(f)\) are the Fourier Transforms of \(h(x)\) and \(g(x)\) and using the convolution theorem  
\[\mathcal{FT}\{h(x)\otimes g(x)\}=H(f)\cdot G(f),\quad H(f)=\mathcal{FT}\{h(x) \},\quad G(f)=\mathcal{FT}\{g(x)\} \tag{3}\]  
and with  
\[\mathcal{FT}\{h(x)\} = \frac{1}{\pi}\cdot\delta(f)-\frac{1i}{2\pi f}, \tag{4}\] \[\mathcal{FT}\{g(x)\} = \sqrt{2\cdot\pi}\cdot\sigma\cdot\exp\left(-\frac{f\cdot\left(f \cdot\sigma^{2}+f_{k}\cdot 2i\right)}{2}\right)\]  
one obtains the Fourier transform of the convolution \(H(f)\cdot G(f)\). Please note that \(H(f)\simeq\delta(f)\) function. The normalized absolute value of \(\|G(f)\|\) yields the modulation transfer function (MTF) or frequency contrast function \(C(f)\). The frequency at 10% contrast gives the spatial resolution in the image. Please note that vice versa the integration of the Gauss function yields the error function erf(x) which perfectly can fit to an image of an edge.