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The APS-C sensor in my camera has a 1.62 crop factor, but the sensor is also closer to the lens, so the image projected on the sensor should also should be smaller. Wouldn't the reduced distance (from sensor to lens) cancel out the cropping from the smaller sensor?

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  • \$\begingroup\$ You're missing something in the science, here, which I can't quite figure out. But if this were true, all the new mirrorless cameras would have to have artificially increased lens-sensor gaps to fit the theory. They don't, they're all much shorter than their DSLR counterparts, now there's no need to leave room for a mirror. \$\endgroup\$
    – Tetsujin
    Oct 28, 2022 at 14:30
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    \$\begingroup\$ Yeah, I'm definitely missing something. This nice website lets you play around with light and virtual lenses: photonstophotos.net/GeneralTopics/Lenses/OpticalBench/… This site makes it look like the projected image would be smaller if the sensor were moved closer to the camera \$\endgroup\$
    – iwans
    Oct 28, 2022 at 14:42
  • \$\begingroup\$ Yes, but you're not using the same lens on each camera, nor do you ever move it in relation to the sensor. If you do, it's for macro, which doesn't prevent an image falling, it just changes the focal length at which it will focus. This is why all the new mirrorless cameras have adaptors for your old lenses which were expecting the sensor to be further away, whilst the new ones fit perfectly to the closer format… because that's what they were designed for. BTW, that site you linked may as well be in Greek or Martian for me. \$\endgroup\$
    – Tetsujin
    Oct 28, 2022 at 14:52

4 Answers 4

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A lens is designed to project an image circle of a certain size in order to cover an image area (sensor) of a certain size... and for that image circle to be in focus at that distance (flange distance). If you arbitrarily change the distance you change the crop factor (size relative to the image circle) and adversely affect the lens' ability to focus at all distances.

What you are describing is more commonly known as "bellows factor." If you attach a lens to a bellows, when you move the lens father away from the image area/sensor the projected image circle spreads out more over the increased distance (although the additional area/size created may be masked out). This causes an increase in the effective magnification (increased crop factor); and simultaneously causes a reduction in the light density at the image plane (resulting in a smaller effective aperture). It also prevents the lens from focusing at longer distances (incorrect/long flange distance).

And the opposite is also true... less distance = less projected image spread, less magnification/crop factor, greater light density, and an inability to focus as close.

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  • \$\begingroup\$ Bellows Factor: In close-up photography, the camera lens is detached from the camera body and remounted on a bellows or ring spacer. This procedure invalidates the f-numbers engraved on the lens barrel. This grants higher image magnification. The bellows factor calculates a correction factor that compensates for the f-number error. Formula BF = (M+1) squared Example M (magnification) is life-size = 1 Thus BF = (1+1) squared = 4 4X more = 2 f-stops as each stop is a 2X change. \$\endgroup\$ Oct 29, 2022 at 3:57
  • \$\begingroup\$ @AlanMarcus, Bellows factor is not specific to macro photography. The lens is also mounted on a bellows and focused by changing its' distance with a medium/large format view camera. And the physings of the spread of light over distance (inverse square law) applies at all lens distances... The calculation is (extension÷focal length)²; which is another way of saying Magnification². The +1 only applies to a camera with a normally fixed flange distance... i.e. when the bellows extension equals the FL on a view camera you get the "+1" (the base flange distance for aperture). \$\endgroup\$ Oct 29, 2022 at 13:09
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The format size of the APS-C (Advanced Photo System – Classic format) is approximately 16mm height by 24mm length. This rectangle has a diagonal measure of 28.8mm.

As a rule-of-thumb, the “normal” lens for any format is a lens with a focal length approximately equal to the diagonal measure. For the APS-C, that focal length is the diagonal rounded up to 30mm. In other words, the “normal” focal length for the APS-C is 30mm. By “normal” we are talking not telephoto and not wide angle. Further, the ratio of length to height is 24 ÷ 16 = 1.5. Mount a 30mm lens on this format and holding the camera horizontal (landscape), the angle of view is approximately 45⁰. Further, a “normal” lens approximately replicates the “human perspective”. In other words, an image of a vista taken with a 30mm mounted on an APS-C delivers a perspective that approximates what we humans perceive.

Mount a telephoto on an APS-C, the beginning of this series will be 2X “normal” thus 60mm is the starts the telephoto focal length range.

Mount a wide-angle, starts at 70% of “normal” = 20mm or shorter. By the way, mount a lens equal to the diagonal measure of the format, the resulting image is not likely to be vignetted. Another reason why this rule of thumb is used. Now the distance from lens to sensor (or film) is a measure taken from a point on the lens called the rear model. The lens maker takes this into account when designing a lens for a specific camera body.

Now the lens projects an image of the outside world on the surface of the sensor. The size of object projected is a function of the focal length of the lens. The crop factor is obtained by dividing the diagonal measures. A full frame camera sports a frame that is 24mm height by 36mm long. The diagonal of this rectangle is 43.3mm. The diagonal measure of the APS-C is 28.8. We divide to get the crop factor thus 43.3 ÷ 28.8 = 1.5 (this is the average, some variation between models is reasonable).

Thus, the distance lens to sensitized surface is a function of “normal” focal length. Lenses for a specific format are chosen to allow the lens to focus at the infinity position. The smaller the format, the shorter the lenses must be. Thus a “normal” on a full frame delivers an angle of view of about 45⁰. The “normal” on an APS-C delivers and angle of view of about 45⁰. The crop factor should be called a “magnification” factor. The 1.5 factor tells me that if I want to make a final display image from an APS-C, I must enlarge the APS-C image 1.5X more than a comparable image made on a full frame.

No canceling out - we must enlarge the image made by a smaller format to force it to be the same size as its larger cousin.

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  • \$\begingroup\$ I'm not quite sure at any point does this address the actual question asked. \$\endgroup\$
    – osullic
    Oct 28, 2022 at 20:57
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The distance of the sensor from the lens is irrelevant (when talking about crop factors), because lenses and bodies are designed in tandem, or rather, lenses are designed for particular bodies, particular flange focal distances.

Think – a Nikon DSLR (for example), with its reflex "mirror box", has a much longer flange focal distance than a Leica rangefinder. But they're both full-frame cameras (let's say), and a 35mm lens on either will result in essentially the same angle of view. The fact that the sensor in the Leica is much closer to the lens has no bearing on the angle of view – because the lens has been designed for this flange focal distance. The crop factor only comes into play when considering a lens being used on two camera bodies that differ in their sensor size (i.e. how much of the image circle they capture), but which are designed for mounting the same lens (and consequently with the same flange focal distance).

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I assume you assume you want to keep the same lens. My understanding is that while the sensor is closer, you need to refocus the lens, essentially moving the lens further away from the sensor, to get a sharp image.

So I think, for a 50mm lens, the sensor will always end up 50mm away (for focusing at infinity). This results in the different field of view.

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