## 3 Analysis
### Diffraction-Based Compound System
Therefore, behind the compound lens, with focal length \(f_{c}\) and the diffraction field \(g^{\prime}(x)\), is:  
\[g^{\prime}(x)=\exp\left(j\pi\frac{x_{1}^{2}}{\lambda z_{1}}\right)g(-x_{1},0) \exp\left(-j\pi\frac{x^{2}}{\lambda f_{c}}\right) \tag{14}\]  
Figure 5: Diffraction-based modeling of the system of Figure 4 by the lenses-based approach of Figure 3a. Initial lenses are in dark gray, and the lenses caused by diffraction are in light gray. \[\text{with }x_{1}=\frac{z_{1}}{z_{2}}x \tag{15}\]  
We obtain \(g(-x_{1}\),0), instead of \(g(x_{1}\),0), because of the application of two successive FTs, which is the concept of the telescope setup. Indeed, \(\text{FT}[\text{FT}[g(x)]]=g(-x)\).The scale (zoom) factor of Equation (15) is because the two FTs are performed, with respect to two respective factors: \(z_{1}\) and \(z_{2}\). For the special case of \(z_{1}=z_{2}\), we obtain a 4f-setup. By combining relations (14) and (15), we obtain:  
\[g^{\prime}(x)=\exp\left(j\frac{\pi}{\lambda}\left[\frac{z_{1}}{z_{2}^{2}}- \frac{1}{f_{c}}\right]x^{2}\right)g\left(-\frac{z_{1}}{z_{2}}x,0\right) \tag{16}\]  
For steps 3 and 4, we suppose that the three side-by-side lenses on the right (Figure 5) satisfy:  
\[\frac{1}{f_{2}}=\frac{1}{d-z_{2}}+\frac{1}{z_{3}}=\frac{z_{3}+d-z_{2}}{z_{3}( d-z_{2})} \tag{17}\]  
If we solve Equation (17) for \(z_{3}\), we obtain relation (9). The output diffraction field becomes:  
Figure 4:0  
Figure 4:1  
Leading to:  
Figure 4:2  
or:  
Figure 4:3  
We obtain the radius of divergence \(R\):  
Figure 4:4  
where the scaling factor \(S\):  
Figure 4:5  
By having a close look at Equations (11), (12), (17) and (23) it turns out that: \(S=m\). Moreover, by considering Equation (22), it yields:  
Figure 4:6  
Equations (8), (9) and (24) lead, after little algebra, to (Figure 4):  
Figure 4:7  
The quadratic phase term \(\exp\left(j\frac{\pi}{\lambda R}x^{2}\right)\) of relation (21) represents a spherical wave, focusing on the back focal point F\({}^{\prime}\) (compound system). If the input of the compound system is a plane wave, then the output field is that of a spherical wave converging towards F\({}^{\prime}\). After this point, the wave diverges. At the distance \(z_{3}\), we again obtain: \(\exp\left(j\frac{\pi}{\lambda R}x^{2}\right)\). The term \(g(x/S)\), in relation to (21), becomes the constant 1, which is in full agreement with the framework of geometrical optics. \(z_{3}=h^{\prime}\) presents another interesting situation. This case that the wave is an observed plane at the back principal plane H\({}^{\prime}\). We observe a spherical wave that converges towards (or diverges from) the focal point F', which is in total agreement with the framework of geometrical optics. In Figure 5, we see 14 lenses. Six of them cancel each other out. It is possible to apply a paraxial ray propagator matrix. We then obtain eight matrixes, i.e., A\({}_{1}\) to A\({}_{8}\):  
\[ \begin{split} A_{1}=\left[\begin{array}{cc}1&-1/z_{1}\\ 0&1\end{array}\right],\,A_{2}=\left[\begin{array}{cc}1&1/z_{1}\\ 0&1\end{array}\right],\,A_{3}=\left[\begin{array}{cc}1&1/z_{2}\\ 0&1\end{array}\right],\,A_{4}=\left[\begin{array}{cc}1&-1/z_{2}\\ 0&1\end{array}\right],\\ A_{5}=\left[\begin{array}{cc}1&1/(z_{2}-d)\\ 0&1\end{array}\right],\,A_{6}=\left[\begin{array}{cc}1&1/(d-z_{2})\\ 0&1\end{array}\right],\,A_{7}=\left[\begin{array}{cc}1&1/z_{3}\\ 0&1\end{array}\right],\,A_{8}=\left[\begin{array}{cc}1&-1/z_{3}\\ 0&1\end{array}\right]\end{split}  \tag{26}\]  
The matrixes \(A_{4}\) and \(A_{5}\) could be combined into one matrix \(A_{4,5}=\left[ \begin{array}{cc}1&-1/d\\ 0&1\end{array} \right]\), since the two corresponding lenses are adjacent and can be replaced by a single lens, with focal length -\(d\). We then obtain six transfer matrixes:  
\[