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Consider a positive lens and spherical aberration, which causes the formation of that region of intersection of different rays as in picture. enter image description here

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?

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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.

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  • $\begingroup$ Thanks the answer! If I may ask, suppose that I’ve placed the object at a distance p to the left of the lens: in that case what you defined as $w$ is actually similar to the diameter of the lens $D$ (supposing that the beamlight is sufficiently spread)? Therefore the condition to separate the two cases would be $q \leq 20 D$? Also, could you provide book or site references where the two cases (especially the second one) are explained, as I can study this topic a little bit deeper? $\endgroup$ – Sørën May 27 '17 at 14:24
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    $\begingroup$ @Sørën I'll have to think about this one. There are expressions for Siedel spherical aberation and so forth, but that won't tell you what you want. Surprisingly, I don't think there are many expressions around that will do what you want - it's something that is usually explored by simulation. You need to calculate the optical path length as a function of exit pupil co-ordinate, then a Fourier transform will get you the focal plane electric field amplitude, whence you can use a freespace propagation to generate the intensity map. $\endgroup$ – WetSavannaAnimal May 28 '17 at 6:52

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