Tech question about DOF - macro environment

[ghammer]

Originally Posted by jaapv

The problem is, the focal length of a lens is fixed at the given value only at infinity and lengthens as it gets focussed closer, creating a smaller aperture of course. So in your formula f is the focal length at infinity and N the aperture at infinity?

No, focal length is an invariant of the lens (by definition). What you mean is extension, which changes by focussing. In the formula the effect of focussing (changing the extension) is represented by m, the aspect ratio. You can of course eliminate m and introduce the extension instead, giving another expression. But usually the ratio is more convenient for macro setups. Accordingly N does not depend on the extension. It is simply the (dimensionless) value you choose by turning the diaphragm.

[luigi bertolotti]

OK, I'm SATISFIED ! Thanks Ghammer, Giordano, Jaap, I see my initial doubt is not significant in practice: even applying the "f related formula" the difference in DOF between 50 and 135 at 1:2 ratio and f11 is in the range of 10^-4 mm ... the CoC value is anyway the driver... the simply act of focusing on the Viso screen makes all this irrilevant, even admitting that the superb Leitz extension rings have surely screwing tolarences UNDER 10^-4 mm ... (not to mention the alternative of Summicron DR with std. RF focusing... )

[jaapv]

Originally Posted by ghammer

No, focal length is an invariant of the lens (by definition). What you mean is extension, which changes by focussing. In the formula the effect of focussing (changing the extension) is represented by m, the aspect ratio. You can of course eliminate m and introduce the extension instead, giving another expression. But usually the ratio is more convenient for macro setups. Accordingly N does not depend on the extension. It is simply the (dimensionless) value you choose by turning the diaphragm.

Thank you; completely clear.

[zapp]

Originally Posted by ghammer

This is an approximation. The exact expression is: 2*c*N*(m+1)/(m^2-(c*N/f)^2)

with f the focal length. So there is really a difference for all ratios m (Luigi is right), but the difference is very small for c approaching 1. For Luigis case: c=0.33, m=0.5, and N=16 it is 6.403mm versus 6.400mm.

magnification ratio already has focal length included since
1/i + 1/o = 1/f and i/o = m
where m is the magnification ratio, i is image distance and o is object distance, and f is the acutal focal length.