Spherical aberration entity vs. distance from the lens? Consider a positive lens and spherical aberration, which causes the formation of that region of intersection of different rays as in picture.

Named $q$ the distance of the image from the (idealized) lens, how do the dimension of this region of overlapping of rays change with $q$?
My guess is that this region is more limited if $q$ is small, while if $q$ is big,then the region becomes longer and longer, is this correct? 
 A: It's a bit complicated. There are two competing effects here as one varies the focal length of a spherical lens. Assuming that the beamwidth $w$ is constant, the $f$-number $\frac{q}{w}$ is then proportional to $q$, and the numerical aperture $\frac{w}{2\,q}$ inversely proportional to $q$. Then the effects are:
Aberration Free Scaling
In the low aberration case (i.e. for $q > 20\,w$, roughly), if one plots the transverse electric field as a function of position around the focus (maximal intensity point), then the contours of constant intensity are approximately ellipsoids whose axial diameter is proportional to $q^2$ and whose lateral (sideways, transverse) diameter is proportional to $q$. The peak intensity is proportional to $q^{-2}$. Another way of looking at this effect is to witness that, in the aberration free case, there is a "prototypical plot" of the electromagnetic field vectors that is always the same, and the co-ordinates on this plot are in normalized, optical units. So this prototypical plot gets stretched by a factor proportional to $q^2$ along the axial direction and stretched by a factor proportional to $q$ in the lateral (sideways) direction. The prototypical plot depends on the apodization; if you light the lens with a Gaussian beam, the transverse electric field vector varies as:
$$E(r,\,z) \approx \frac{e^{-i\,k\,z}}{z+i\,z_R}\,\exp\left(-i\,\frac{k}{2}\,\frac{r^2}{z+i\,z_R}\right)$$
where $z_R = \frac{\lambda\,w^2}{4\,\pi\,q^2}$ is the Rayleigh length and $w$ is measured as the $1/e^2$ intensity diameter of the beam. If, however, the input beam is of uniform intensity, then the electric field  pattern in the focal plane varies like 
$$E(r,\,0) \approx \frac{2\,\lambda\,q\,J_1\left(\frac{\pi\,w\,r}{\lambda\,q}\right)}{\pi\,w\,r}$$
whereas the variation along the optical axis is:
$$E(0,\,z) \approx e^{-i\,k\,z}\,\operatorname{sinc}\left(\frac{\pi\,w^2\,z}{8\,\lambda\,q^2}\right)$$
Increasing Spherical Aberration
The above will give you an excellent idea of the field distributions for
$q>20\,w$. However, as $q$ becomes smaller than this value, the spherical aberration increases very swiftly, and the shrinking prototypical plot effect competes with the spreading of the plot by increasing wavefront aberration. A rough rule of thumb is that the RMS wavefront error $\sigma$ varies roughly in proportion to the fourth power of the numerical aperture, and the peak intensity varies roughly like $I \approx I_0\,\exp(-4\,\pi^2\,\sigma^2)$, where $I_0$ is the peak aberration free intensity. You can see that the focal region suddenly spreads out massively as $q$ decreases; the details generally need to be worked out by simulation.
