Predict DoF at low magnifications

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nielsgeode
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Location: Groningen, Netherlands

Predict DoF at low magnifications

Post by nielsgeode »

Just wondering: at f/2n the DoF is twice as the DoF at f/n

How does this work with change in magnifications?

The "correct" aperture is of course calculated from f = f * (1 + x) with x = magnification. However, at magnifications below 1x (e.g. 1:5 and 1:4), the change in aperture is small. Therefore, you can aproximate with x = 0.

Keeping this assumption in mind, what would be the relative new DoF at 1:4 magnification compared to 1:5? Is this 0.2 / 0.25 = 0.8, so a decrease of the DoF by ~20%?

Can someone please make this clear to me?

:D

rjlittlefield
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Post by rjlittlefield »

Can someone please make this clear to me?
Maybe. Bear with me as I step through this...

Under reasonable assumptions, the "exact" formula is the classic one given by Lefkowitz that total depth of field DOF depends on the circle of confusion C, the nominal ("ring") f-number f_r, and the magnification m according to this formula:

DOF(C,f_r,m) = 2*C*f_r*(m+1)/(m^2)

If we hold C and f_r constant, and take the ratio of two DOFs at different m's, then we find that the ratio can be written as

DOF(ma) / DOF(mb) = ((ma+1)/(mb+1)) * (mb/ma)^2

The first fraction on the right side is a ratio of effective f-numbers, which you've chosen to ignore when you say "approximate with x=0". The second fraction, which dominates under the low-mag conditions you've specified, is the square of the ratio of magnifications.

By the simplified formula, going from 1:5 to 1:4 gets you a DOF ratio of 0.2 / 0.25 squared = 0.64 .

The more exact formula would give 0.66667, not significantly different in this case.

Bottom line: if the effective f-number does not change, then DOF varies inversely as the square of the magnification. It turns out that this relationship holds true under a wide range of conditions, including cases where the simple classic formula breaks down.

--Rik

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