If APS-C and similar crop-sensor digital cameras have a focal length multiplying effect such that a 50mm lens has an apparent focal length closer to the field of view of an 80mm on a full frame camera, and yet at the same time the depth of field for the smaller sensor camera is more like the depth of field a 50mm lens would produce on a full frame camera (using the same aperture), then this would seem to suggest the concept of an "aperture dividing effect."
In other words, a 50mm f/1.8 lens on an APS-C camera would act more like a 80mm f/2.8 (approx. 1.8 * 1.6x) lens in 35mm equivalent — for depth of field, not considering exposure.
Can someone with a better understanding of the physics involved clarify this for me. I've never seen this concept explicitly mentioned anywhere, so I am a bit suspect of it.
Answer
This answer to another question goes into detail on the math behind this. And there's a Wikipedia article with a section specifically about getting the "same picture" with different camera formats. In short, it is approximately true that adjusting both the focal length and aperture by the ratio of the format sizes (the crop factor) will give you the same picture. ¹
But this breaks down if the subject is within the macro range of the larger-format camera (focusing really close). In this case, magnification (and therefore actual sensor size) becomes crucial to the DoF equation, messing up the equivalence.
And, the Wikipedia article casually mentions but does not elaborate on another important point. The assumption is that for the same print size, the acceptable circle of confusion (roughly, the acceptable blur level still considered in focus) will scale exactly with format size. That might not actually hold true, and you might hope (for example) to get greater actual resolution from your full-frame sensor. In that case, the equivalence also isn't valid, but fortunately in a constant way. (You simply have to multiply in your pickiness factor.) ²
You mention "not considering exposure", and now you might be thinking (as I did): wait, hold on. If cropping+enlarging applies to "effective" aperture for depth of field, why doesn't it apply to exposure? It's well known that the basic exposure parameters are universal for all formats, from tiny point and shoots to DSLRs all the way up to large format. If ISO 100, f/5.6, ¹⁄₁₀₀th second gives correct exposure on one camera, it will on any other as well. ³ So, what's going on here?
The secret is: it's because we "cheat" when enlarging. Of course, in all cases the exposure for a given f-number on any area of a sensor is the same. It doesn't matter if you crop or just have a small sensor to start with. But when we enlarge (so that we have, for example, 8×10 prints from that point and shoot to match the large format), we keep the exposure the same, even though the actual photons recorded per area is "stretched". This also has the same correspondence: if you have a 2× crop factor, you have to enlarge 2× in each dimension, and that means each pixel takes 4× the area of the original — or, two stops less actual light recorded. But we don't render it two stops darker, of course.⁴
Footnotes:
[1]: In fact, by changing the f/number, what you are doing is holding the absolute aperture of the lens constant, since the f/number is the focal length over absolute aperture diameter.
[2]: This factor breaks down too, as you approach the hyperfocal distance, because once the smaller format reaches infinity, infinity divided by anything is still infinity.
[3]: Assuming the exact same scene, and minor variations from real-world factors like lens transmission aside.
[4]: Basically, there ain't no such thing as a free lunch. This has the effect of making noise more obvious, and it's a reasonable approximation to say that this increase effectively means the crop factor also applies to noise apparent from ISO amplification.
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