All the DOF calculators I looked at were - and probably still are - based on the laws of optics of an ideal, infinitely thin lens. Within this theoretical model and the cases we’re talking about in this part of the forum, the distance between sensor and film should be 400mm when a 100mm lens is used to capture e negative with a scale of 1:1.
The usual camera lens is everything but infinitely short, which suggests that calculations based on theoretical models might still apply, but in a completely different way. Let’s take a look at this lens:
At 1:1, I measured the distance between sensor and negative to be 300mm, which - according to “ideal optics” - makes the focal length of this lens shrink to 75mm and move the “ideal lens” from position A to position B. With additional measurements, I get this:
Does any of the above matter in respect to DOF? I don’t know yet, I’ll post results as soon as I get them. Meanwhile, you can check DOF here, it’s the only site I found so far that even considers the macro situation.
Anyone any ideas or manufacturer info regarding DOF @ macro?
What is it you are interested in? Do you want to predict the DOF @macro such that it would be easier to predict the performance of different lenses? Or do you want to understand/discuss DOF just for the sake of curiosity?
They way I see DOF (as a microscopist) is just a relation between the angle of incident rays and the lateral resolution (smallest distance I can distinguish). If we consider resolution to be constant (e.g. pixel size or diffraction limit), then the DOF can by computed by some basic trig:
DOF = resolution / tan(theta),
where theta is the angle of the incident rays. This angle depends on the distance from the “virtual” lens to the object and on the aperture.
Well, the DOF calculators gave me values that simply felt too optimistic, although I think that the macro DOF calculator (see link above) is closer to what I see.
Let me have a look at your formula: DOF = Pixelsize / tan(theta).
According to my math leftovers, we could also write DOF = 2D*Pixelsize/d
with D being the distance between the virtual lens and the negative
and d being the diameter of the entry pupil at operating aperture and distance settings.
while applied to an object point located on the optical axis.
Exactly, instead of tan(theta) you can also see it as two congruent triangles (like in your formula). In your example in the OP you show that the virtual lens moves forward thereby increasing the angle and decreasing the DOF.
Maybe this is interesting for you: https://www.microscopyu.com/microscopy-basics/depth-of-field-and-depth-of-focus
Just to make applying the formulas extra-difficult, a lot of modern macro lenses do change focal length as you focus. Instead of simply extending the entire lens farther from the image plane (the old way to focus closer) they shift elements inside the lens much as a zoom lens does, varying the focusing distance by varying the effective focal length.
indeed, and this also makes the DOF calculations difficult. As for focal length at macro distances, Here is what I get with my Canon EF 100 f/2.8 Macro USM:
The measurement at 305mm (between sensor and object) is out of spec but provides acceptable sharpness for negatives that have emotional rather than technical values. The horizontal axis represents a scale of 1:1 up to approx. 1:2.5.
Would you mind detailing a bit more how you get/compute the values of focal distance?
Clear explanation, got it, thank you for indulging
Did another measurement with manual focusing and focus peeking with an EOS M6
Here’s the curve:
The pink circles mark data when scanning FF, 645 and 6x6 negatives with a FF sensor.
I also found that both cameras I used refused to autofocus at minimal distance, even when liveview showed perfect focus. I noticed no focus shift on the DSLR and the M6 can’t have any. This can only mean that the lens has a FL that is just a tad longer than 100mm (so what, all data on a spec sheet are rounded anyway)