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"I can't help you"

No, you can't, because you are painfully ignorant on this topic.

Literally, spend 30 minutes with an iPhone and an SLR and you'd be illuminated. Instead you seriously argue that I need to look at the "physics" (which is farcical when you ignore the most important part of a camera, which is the focusing from the lens to the sensor. Dismissing that betrays a complete misunderstanding of optics).

This conversation is clearly futile, but again - spend 30 minutes and actually test your theories. Or, you know, read any single article on the tubes.

Or how about simply ask yourself "why does the iPhone need to do computational bokeh"? 65mm equivalent lens, f/2.2...should be the easiest thing in the world. In SLR world that is bokeh gold.

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I have a phone with a macro lens. I have a mirrorless camera. As I told you, what matters for bokeh is the distance to the object and the diameter of the aperture. The iPhone needs computational bokeh because the aperture is 2.4mm wide, whereas one of my lenses has a 40mm aperture. That's why my camera produces more bokeh - the aperture has a wider diameter while the distance to the object is the same.

That is literally the one and only thing that matters. The diameter of the lens, and the distance from the object. Take a piece of paper, draw the lens as a slit, draw the object as a point, and make a line from the two edges of the slit to the point, that continues furhter back. You'll get two triangles. Everything that is contained in those two triangles will be focused to the same point on the sensor. That's why the ratio between the two is what matters. That's why closer objects produce a more out of focus background than objects farther appart. That's what I'm trying to explain to you.

The DoF formula that photographers use does not work for comparisons across two different film sizes.

You understand that cameras don't use a slit, right? Do you understand the optics in a modern camera?

Further my 70mm lens has a smaller aperture than my 35mm f1.4 lens. Yet it has a much smaller depth of field for a given distance. Weird! Lens makers must not know your remarkable "slit lens" trick.

At this point I'm convinced you are either trolling, or have dug so far into the depths of wrongness that you're dedicated to sticking with it. So good luck with that. I'm out of this conversation.

Revolve the entire setup around the axis perpendicular to the slit and you will have a very accurate representation of how a camera-lens system works.

The ratio between distance and focal length only works if the focal lengths are equivalent across the two cameras. Otherwise it doesn't work. That's to say, a 70mm f/2.8 has the same depth of field as a 35mm 1.4 lens if the second is on a camera with 2x crop factor.

Try it out, crop the image of your 70mm lens at f/2.8 and compare it to the image of your 35mm f/1.4 lens and you will get exactly the same image with the same blur (assuming the lenses are exactly 70mm and 35mm at the focus setting, which is not guaranteed due to focus breathing and manufacturers rounding off their focal lengths)

I think you are talking past each other, depth-of-field is dependent on the physical aperture not "F-Stops", which are often also called "aperture". Yes, afaik it's derived from single-element lenses but so are most other measures, and I'd be surprised if a real lens behaved different (at least in the center).

Your 35mm f/1.4 lens has a physical aperture of 35mm/1.4 = 25mm, so the equivalent 70mm lens with a 25mm aperture would have an F-stop of f/2.8. Hmm, can't think of many modern 70mm lenses besides Sigma's 70mm/2.8 macro which should have the same DoF, or if it's a standard zoom they should have equivalent DoF as well (unless it's Canon's f/2 zoom).

The (acceptable) depth-of-field is derived from blur-disk diameter, and the circle-of-confusion, for an object at a certain distance from subject ("point of focus") and relies only on physical aperture and distance to subject as stated (or alternatively, f-stop _and_ focal-length, because "phys. aperture = focal-length / f-stop").

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