How should depth of field be calculated for a custom 50mm microscopy optic, and why would near DOF exceed far DOF?
Asked 11/16/2022
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I’m trying to calculate depth of field for a custom optical system and got an impossible result where the near DOF distance is larger than the far DOF distance, giving a negative total DOF.
System details:
- Sensor: 7.4 × 5.55 mm, 9.25 mm diagonal
- Pixel size: 0.8 µm
- Lens: 50 mm focal length
- Front/objective diameter: 30 mm
- Marked f-number: f/1.67
- Physical aperture opening used: 10 mm
I assumed the 10 mm opening means the system is effectively f/5. I also used a circle of confusion based on one output pixel, since the images are saved digitally rather than printed.
My questions are:
- Is a negative DOF result physically meaningful?
- For DOF calculations, should I use the physical aperture diameter or the effective entrance pupil / f-number?
- Is it valid to use pixel size as the circle of confusion, or should CoC be based on sensor size instead?
- For microscopy-style imaging, what assumptions should I revisit if I need depth of field greater than about 10 µm?
Originally by Photography Stack Exchange contributor. Source · Licensed CC BY-SA 4.0
Photography Stack Exchange contributor
3y ago
2 Answers
3
A few points of confusion... a 50/1.67 lens has an entrance pupil diameter of 30mm; probably defined by/equal to the diameter of the objective element. This is the relevant/effective aperture used for the calculation; not the physical diameter of the aperture opening (5mm). The entrance pupil is the optical obstruction to light transmission created by the aperture restriction, as magnified optically by the objective elements (as seen by the source/subject).
The circle of confusion (CoC) for any sensor size is calculated as *diagonal÷1500 or 9.25÷1500 in your case... about six times larger than the CoC you used. This gives you the CoC and diffraction limit for an image when viewed normally (distance ~= image diagonal); whether printed or not. And thus the depth of field (DoF) of that image when viewed in that way.
But DoF is not a fixed aspect of an image; when the image is viewed differently (from a different distance/different magnification) the DoF changes as a result. So your use of pixel level diffraction (pixel diagonal as the CoC) wouldn't necessarily be wrong; except that the diagonal of a .8um square pixel is 1.13um... (I'm assuming there is a typo). However, your calculation would result in the DoF for the image when displayed/viewed at 1500x the standard magnification (larger/closer).
*(Kodak used d÷1730 because they defined the limit of normal vision slightly differently)
Originally by user70370. Source · Licensed CC BY-SA 4.0
user70370
3y ago
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A negative DOF is not physically meaningful; it usually means one or more assumptions in the calculation are wrong.
From the answers provided, the main issue is aperture and CoC choice:
- For a 50mm f/1.67 lens, the effective aperture is the entrance pupil, about 30mm. That is the relevant aperture for DOF, not simply the physical stop opening unless you know how it maps to entrance pupil.
- Your CoC assumption is also too small. Using one pixel as CoC is not the standard DOF convention. A common approximation is sensor diagonal ÷ 1500. For a 9.25mm diagonal sensor, that gives about 0.006mm = 6.2µm, much larger than your value.
So yes: your assumptions are likely causing the impossible near/far reversal.
Also note that conventional DOF formulas are based on acceptable blur in the final image, not strict microscopy resolution limits. If you need DOF on the order of 10µm, you may need to consider diffraction and magnification-based microscopy DOF methods rather than standard photographic DOF calculators.
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