A new way of thinking and calculating about DOF

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ChrisR
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Post by ChrisR »

Thanks _ while I'm pondering,
Quote:
At lower mags, the wave method is pessimistic about DOF, though I'm not sure why. (At 1:1, f/2.8, it's about half.)
I suspect you're looking at a situation where the Airy disk is small compared to the classic circle of confusion and/or sensor resolution. In that case the 1/4-lambda criterion doesn't apply because you've decided to accept lots more degradation than just pushing down the MTF curve a bit.

If you show me the analysis or describe the experiment, maybe I could tell better.
That's easy:
Image

--
By the way the column in Rik's last-posted tables marked "Step Size" should be "DOF" I think, looking at the numbers.

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Post by ChrisR »

Gotta go out, but, sort of thinking aloud, I'm not sure what I can take from these apertures that's useful.
The f numbers in the chart all seem a good stop too big.

I must admit I also allow a large overlap if I'm being fussy, too, like 50%,and that's on the classic dof calulations, which on objectives are too tight already.

One thing I hadn't realised is that if using microscope objectives, (ie about 2x NA 0.1, up) then all that determines the DOF is the NA. Forget the magnification. The Wave calculation rules so a 10x 0.45 gives the same dof as a 50x 0.45. (Need to check that!)

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Post by rjlittlefield »

At effective aperture f/5.6, an aberration-free lens would be way out-resolving the sensor. So this situation is exactly what I described, "the Airy disk is small compared to the classic circle of confusion and/or sensor resolution."
By the way the column in Rik's last-posted tables marked "Step Size" should be "DOF" I think, looking at the numbers.
The questions I get all refer to step size. It's a matter of taste whether the step size is equal to the 1/4-lambda DOF, versus somewhat larger or smaller. I've used "equal" because that seems to be a reasonable compromise. Chris S. would run smaller, seta666 would run larger.
The f numbers in the chart all seem a good stop too big.
The f numbers in the chart correspond to effective f/10.7 on a 15 megapixel APS-C sensor and f/14.2 on a 23 megapixel full-frame sensor. Those correspond pretty closely to "Diffraction Limits Standard Grayscale Resolution" at CambridgeInColour's diffraction calculator. A user who wants sharper images at the cost of shooting more frames will prefer wider apertures, corresponding to fewer pixels across the circle of confusion.
One thing I hadn't realised is that if using microscope objectives, (ie about 2x NA 0.1, up) then all that determines the DOF is the NA. Forget the magnification. The Wave calculation rules so a 10x 0.45 gives the same dof as a 50x 0.45. (Need to check that!)
The wave calculation rules only if the captured images are diffraction-limited. 10X NA 0.45 is effective f/11. CambridgeInColour shows that an image at f/11 is affected by diffraction on a 15-18 megapixel APS-C sensor, but we know from the demo HERE that f/11 is also sensor-limited by almost a full stop. In contrast, 50X NA 0.45 is effective f/56, which is strongly diffraction limited by everybody's evaluation. The 1/4-lambda DOF will be the same for 10X NA 0.45 and 50X NA 0.45, but it will be a lot harder to tell that with the 10X setup, using current sensors.

--Rik

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Post by rjlittlefield »

Switching focus a bit, I should point out that this same 1/4-lambda approach also works well for long distance applications like landscape photography. In that case what we do is to turn around the model and figure out what defocus distances are acceptable back on the sensor, then for the user's convenience convert those lens-to-sensor distances into subject-to-lens distances that correspond to something you'd use a measuring tape on. In this case we also need to know the lens focal length so that we can do the conversion to subject distances.

For example, if we've decided to use 50 mm and f/8 because it gives the sharpness we want, and we want to stack between 2 meters and 10 meters, we can calculate as follows:

effective f-number = f/8
NA = 1/(2*8 ) = 0.0625
1/4-lambda DOF = 0.00055/NA^2 = 0.1408 mm (This is now Depth Of Focus, not Depth Of Field)
far target, back focal distance = 1/(1/50-1/(10*1000)) = 50.251 mm
near target, back focal distance = 1/(1/50-1/(2*1000)) = 51.282 mm
first back focus point = 50.251 + 0.1408/2 = 50.321 mm (one-half step, catching the far target at 1/4-lambda defocus)
second back focus point = 50.321 + 0.1408 = 50.462 mm (one full step)
third back focus point. etc, at full step increments
last back focus point = 51.282 - 0.1408/2 = 51.212 mm (catching the near target at 1/4-lambda defocus)

Converting these back to subject distances, we have:

first focus distance = 1/(1/50-1/50.321) = 7.8 meters
second focus distance = 1/(1/50-1/50.462) = 5.5 meters
etc
last focus distance = 1/(1/50-1/51.212) = 2.1 meters

I'm not sure what's the best way to package this for users. I can imagine users being uncomfortable with the idea that they said 2 to 10 meters and the calculation is giving them 2.1 to 7.8. Maybe it would work better to just have them shoot exactly at front and back, with full steps in between. That would only be one more frame at worst.

Edit: Oops, silly oversight in the example. I neglected to notice that f/8 is probably going to be sensor limited, not diffraction limited. When the system is sensor-limited, then just like in the macro/micro case we should use the classic geometry model based on circle-of-confusion. That gives a larger depth of focus. The remainder of the calculation has the same structure: calculate focus points at the sensor and transform those back to subject distances.

--Rik

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