Focus Tolerance and Format Size

The key variable as far as the tolerances required to position the lens for accurate focus are concerned (at least in a simplified ideal situation) is an appropriate approximate distance between the desired in-focus plane and the actual in-focus plane (which we are assuming is slightly out of focus). It is a distance in the direction perpendicular to the x-y plane normally used to describe position of the image on it, hence the designation delta z, or dz in this post.  The lens’ allowable focus tolerance is therefore  +/- dz, which we will show in this post to vary as the square of the format’s diagonal.

The wave model of defocus makes a number of simplifying assumptions, one of which is that for small out-of-focus situations most of the PSF is due to diffraction and the longer/shorter distance (or Optical Path Difference) that the light incident on the lens must travel in order to reach the slightly out-of-focus sensing plane vs that which would have been traveled by the in-focus light. OPD in wavelengths (\lambda) is therefore an indication of how out of focus the image is, with zero  \lambda perfectly in-focus and more than \frac{1}{2}\lambda unacceptably out-of-focus. Lord Rayleigh’s criterion for in-focus was less than \frac{1}{4}\lambda OPD.

Hopkins (1955) shows that if the exit pupil produces a spherical wave

(1)   \begin{equation*} dz = OPD * 8 * N^2 \end{equation*}

in the same units as wavelength with N the f-number.

For instance for OPD \leq \frac{1}{4}\lambda (let’s call that in-focus) \lambda = 0.55 microns and N = 5.6,  for the image to be considered in-focus dz needs to be less than about 34.6 microns

According to the model such a dz will give rise to a hybrid diffraction/defocus PSF that has the approximate shape of a disk of rough diameter:

(2)   \begin{equation*} k = dz / N \end{equation*}

in the same units as dz.  For the example above, k = about 6.2 microns.

This is all occurring at the optical level, just as diffraction alone does, so a smaller format would of course see the same-sized blur disk of diameter k on the sensor as a larger format all other things being equal, but that same disk would take up a larger portion of its final image than it would on a larger format.

If one substitutes (1) into (2) disk diameter k turns out to be directly proportional to N aotbe

(3)   \begin{equation*} k = OPD * 8 * N \end{equation*}

A smaller format at ‘Equivalent’ apertures would therefore produce a  smaller defocus blur disk, proportionately to the ratio of the two formats’ diagonals, so that the effect of the blur would be the same on equivalent images viewed at the same size, all else equal.

Let’s now take a look at dz on its own, which is proportional to the square of N.  If we use \leq \frac{1}{4}\lambda optical path difference as the in-focus criterion

(4)   \begin{equation*} dz = 2 * \lambda * N^2 \end{equation*}

In a simplified ‘Equivalent’ situation at final image size, a smaller format will need better in-focus tolerance by the square of the ratio of the sensor diagonals.

Using as an example FF vs mFT at f/5.6, if we assume the same perfect lens on both systems, \lambda = 0.55 microns, Full Frame’s focus positioning servos and mechanics can afford to have a play of +/- 34.5 microns (dz) while still staying within the in-focus criterion, as calculated above.

However at an Equivalent f/2.8, mFT will need focus positioning tolerances of only +/- 8.6 microns to meet the same in-focus criterion for the same relatively sized disk on the final image because of the N^2 term – a much tighter precision required of the mechanics and electronics of the smaller format .  In practice this may not be an issue.