Interesting DOF Result: No Matter Focal Length, Ratio is Constant

[Borek]

Perhaps it is known to you, it was something new to me.

After Andy confirmed that reasonable circle of confusion for my camera should be about 0.02mm, I did some calculations - and I was surprised by the result.

Say you have an object that you want to take a picture of (be it flower, book, chair - something like that). Say its size is x cm and you want it to fit whole frame. What focal length lens should you use to get the largest DOF (assuming f-number of 8, as it usually gives best results)?

The answer is... it doesn't matter! Well, almost doesn't matter:

Note how for a given size of the object DOF/base ratio is always the same, no matter what focal length you use. For example - 30 cm object, f-number of 8, DOF/object size is always 0.20.

Picture base is a width of the object, dist is a distance between object and the sensor (for obvious reasons it changes with the focal length), DOF/base is the ratio between DOF and object size - and it turns out this ratio almost doesn't depend on the focal length. At least as long as we are talking about relatively small objects up to 1 meter. Shorter focal lengths give marginally larger DOF - that's what I expected, but I expected the difference to be much larger. Turns out short focal length means you have to get close to the object, and DOF gets smaller - so there is no gain.

 

[Andre]

Is the math available somewhere? I know, it's common assumption, that DOF is supposed to be only a function of magnification, aperture and CoC. However, when I did the math some eons ago, I ended up being unable to eliminate the focal distance. It appears that larger f's decreases DOF very slightly, even with the same magnification and CoC.

I guess I have to redo that math.

 

[Borek]

My spreadsheet was prepared using these http://www.dofmaster.com/equations.html

 

[Andy Resnick]

Perhaps it is known to you, it was something new to me.

This was a very nice result and a very clever piece of work. I hacked at this problem for a while, and although I made some progress, I don't think I completely solved it.

The wiki page (http://en.wikipedia.org/wiki/Depth_of_field) has a lot of useful information, but there's an assumption made very early in the derivation (as does all the other DOF sites I found) that is not always valid. I'm not sure there's value in presenting the detailed work, but I'll present the main results.

Start with Gauss's formula for a thin lens: f/s + f'/s' =1, where f and f' are the front and rear focal lengths, s is the object distance and s' the image distance. Recall that photographic lens focal length specifications are f'. The initial result is (pardon the text formatting):

DOF = 2s/[f/Nc(ff'/(f^2-f's))-Nc/f((f^2-f's)/ff')].

Where N is the f-number, c the CoC, etc. etc.

The problem with the online formulas I found is that the assumption f = f' is made too early. If f = f' then the usual formula is recovered from the above (and using m = f'/(f-s) = f/(f-s)). The issue is determining if f = f' is a valid assumption.

Photographic lenses are not symmetric because s != s', and in order to get good aberration correction, the lenses become asymmetric. My suspicion is that for 'normal' lenses (meaning say 35mm-85mm) f ~ f' because they are generally based on a double-Gauss configuration, but I know that telephoto lenses and microscope objectives have significantly different f and f'.

The key insight was to keep the magnification 'm' constant, or more specifically to keep the ratio s/s' constant as f' is varied. Keeping the approximation f = f' and rearranging the simplified formula gives the result:

DOF = 2Nc(m+1)/(m^2-(Nc/f)^2),

which clearly shows that as the f-number N increases the DOF increases. Now if we make another assumption, that s<<H (the hyperfocal distance) which is usually valid, then the DOF formula further simplifies to

DOF = 2Nc(m+1)/m^2,

which is independent of lens focal length, as Borek found (and is mentioned on the wiki page).

AFAICT, online calculators make use of the assumption that f = f', so the relevant question is the sensitivity of the full result when letting f/f' vary from 1. As a practical matter, since lens manufacturers don't provide data on f, I would have to measure it and that's too time consuming right now. I looked up a few telephoto lens prescriptions and found f/f' can be 3 or larger. My microscope objectives (including the luminars) have f/f' ratios that vary from about 1/1.6 (the 100mm luminar) down to 1/100 (100x objectives).

I had been hoping to generate some sort of simple scaling law for the DOF (DOF/Ns is dimensionless) but I couldn't get very far due to the complicated denominator. Maybe someone else will have better luck.

 

[LudusRex]

Thank you for sharing your findings with us. This is a very interesting result indeed. It seems that as long as the f-number remains constant, the DOF/base ratio remains constant as well, regardless of the focal length. This could be due to the fact that the f-number controls the aperture size, which in turn affects the depth of field. It is also worth noting that this result holds true for objects up to 1 meter in size, which suggests that it may not be applicable for larger objects or longer focal lengths.

It is always exciting to come across new discoveries in the field of photography, and your calculations have shed light on an aspect of DOF that many may have overlooked. This information could be useful for photographers when choosing the appropriate lens for a specific shot, as they can now focus on other factors such as composition and perspective without worrying about the DOF changing based on the focal length.

Further research and experimentation may be needed to fully understand the underlying reasons for this result, but your findings are definitely a valuable contribution to the field. Thank you for sharing your insights with us.