In digital photography, there's the principle of lens equivalency, according to which, when you shoot with a camera which has a smaller sensor than a full frame sensor (which is the same dimension as classic photography film), you effectively reduce your field of view (relative to shooting with the same optics and a full frame sensor). The resulting crop of the optical image is associated with saying that you are shooting a full frame sensor, but a lens with a greater focal length.

So in terms of the resulting image, one should think of the lens as having a focal length of $f$ (the real focal length) times the crop factor $C$ associated with the sensor being used ($C \geq 1$). this is all well and good intuitively but then comes the 'weird' part.

It is said that if you're to think of your lens to have a focal length $fC$, then you must also think of the camera's f-number as being multiplied by $C$ as well. This means that if you increase your f-number (which is proportional to the numerical aperture), then you also increase the depth of field of the camera - and surely enough, for a smaller sensor, you yield a greater depth of field (for the same field of view) - even when your actual physical f-number is the same in both cases.

How can two optical systems with the same f-number\NA, have different depths of field?

I realize that some of you might say it's related to the pixel size of the cameras, but it goes the other way around: the system with smaller pixels (which means less tolerance for the spread of the PSF) has the greater depth of field. This is best demonstrated in modern days smartphones, where for the same f-number as full frame cameras, the depths of field are worlds apart.

What is the physical explanation for this phenomenon? My best guess is that for systems with smaller sensors, one must position the camera further away from the object to capture the same field of view, such that the light rays travel a greater distance and thus become more parallel before being collected by the optics, but I'd like to know for sure.

  • $\begingroup$ When I refer to the f#, I refer to the actual physical f# (calculated as f/D), which in my example is the same as the full frame camera. Photographers claim that in order to calculate the DOF, you should use the effective f#, and it seems to hold water :). About pixel size: it matters because DOF (physical or otherwise) is calculated with respect to a condition that the image is agreeably sharp. So if your PSF spans over 1 pixel or less, it would be agreeably sharp, while if it spans over 2 or more pixels, it would be dimmed blurred. $\endgroup$ – Yuval Weissler Aug 6 '17 at 12:39
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    $\begingroup$ Sorry I deleted the comment that asks the questions that you answered. After thinking, I decided that while I know something about optics, I don't know enough about photography. $\endgroup$ – garyp Aug 6 '17 at 13:10
  • $\begingroup$ I think this is more about geometrical optics rather than physics. The people at photo.SE will be more familiar with your terminology. I think you can only get greater depth of field if you are only using the center portion of the lens, i.e., light entering the edges of the lens does not hit the sensor. $\endgroup$ – Keith McClary Aug 8 '17 at 4:29

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