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Having recently spent some time fighting for light with a high-speed camera having a CS mount and a sensor area under .1 inches2, I was wondering if the f rules apply to cropped sensors?

For example: A full-frame lens is designed to focus an image on a 1.3 square-inch sensor. Suppose I run a full-frame lens at f/2. I can just project that on the crop sensor, which will give me the same amount of light as if I ran a CS lens of identical construction at f/2. But given that we have a fully-resolved image at the full-frame sensor plane, can't we project it onto the crop sensor, thereby producing an f/2 depth of field that's roughly 13 times as bright as what we get from an f/2 CS-mount lens?

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Assuming I understand what you're saying, the answer is no. Or, depending on how you look at it, yes, but. The optics you use to squeeze that full-frame image into the smaller size are effectively reducing the focal length of the lens. Let's say that your crop sensor is ⅔ the (linear) size of your larger sensor, like the difference between APC-S and full-frame.

And, let's say you have a 50mm f/2 lens. Wide open, that means the apparent aperture is 25mm — because 50mm/25mm = 2.

Your projection effectively creates a 33mm lens with a smaller image circle (think of it from a field of view perspective: by shrinking the image circle, you're recording that much more of the scene, so, wider angle). But now, your f-number is 33mm/25mm, or about f/1.3.

So, yeah, you get a lot more light in. But you do it by converting the lens to effectively have a smaller f-number.

This is exactly what the Metabones Speedbooster does. Read more about it here: How can a speedbooster improve the light performance of a lens?

  • Do we get all of the optical effects that go with a larger aperture? I.e., is your example functionally equivalent to running a 33mm at f/1.3? (I guess it might be nice to not have to find an f/1.3 lens, but do we get the reduction in sharpness and depth of field that one would normally get by taking this f/2 image and projecting it down to f/1.3?) – feetwet Aug 26 '17 at 2:20
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    @feetwet Not really, because the DoF for a 50mm lens at f/2 on a FF camera that needs less enlargement for the same display size is not the same as the DoF for a 33mm lens at f/2 that need more enlargement to be viewed at the same display size (when both are focused at the same distance). – Michael C Aug 26 '17 at 3:21
  • @feetwet You get all of the optical effects that go with a larger aperture and that sensor size. If you compare an image taken with this lens and adapter to an image taken in the same situation with a 33mm f/1.3 lens and a full-frame sensor, and print both images at the same size, the image with more enlargement will have deeper apparent depth of field. – mattdm Aug 26 '17 at 3:57
  • Ah ha, well this is great: It means that if you need more light for a given depth of field you can get it by buying a "bigger" lens! – feetwet Aug 26 '17 at 15:24
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I can just project that on the crop sensor, which will give me the same amount of light as if I ran a CS lens of identical construction at f/2. But given that we have a fully-resolved image at the full-frame sensor plane, can't we project it onto the crop sensor, thereby producing an f/2 depth of field that's roughly 13 times as bright as what we get from an f/2 CS-mount lens?

If you modify the same lens and reduce the size of the image circle cast so that all of the light collected is now projected in an image circle only about 10mm in diameter (instead of 44mm), you have changed the focal length of the lens by a factor of 1/4.4X. Thus you have also changed the f-number by the same factor. You've concentrated the same total amount of light in a smaller area, thus increasing the field density. This is true whether applying it to FF vs. APS-C sized light circles or 7.5mm sensor chips vs. 5mm sensor chips in phones.

  • I would like to be able to work out the generalized relationship as you appear to be doing. But continuing my example: we've taken a 50mm full-frame lens at f/2 and converted it into ... an 11mm CS lens at f/0.45? That doesn't sound right: The angle of view is the same as 50mm at full-frame, and I thought focal length exactly determines (uncropped) angle of view? – feetwet Aug 26 '17 at 15:38
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    @feetwet Focal length and media size (or "crop") together determine angle of view. – mattdm Aug 26 '17 at 16:42
  • A 50mm lens on a Micro Four thirds or even APS-C camera is mildly telephoto. A 50mm lens on a FF/35mm camera is normal. A 50mm lens on a Medium or Large format camera is wide angle. The size of the image circle projected by a lens is independent of its focal length. The focal length is the distance behind the lens at which collimated light falling on the front of the lens converges behind the lens. For a 50mm lens to project a larger image circle, it must collect light from a wider angle of view than a 50mm lens that projects a smaller image circle. (cont) – Michael C Aug 26 '17 at 17:36
  • (cont) The projection at the center of both images will be identical, it's just that the image circle of the lens with the larger image circle will include things on the edges that are past the edge of the smaller image circle. – Michael C Aug 26 '17 at 17:37
  • For more, please see this answer to Shooting 50mm EF vs EF-S – Michael C Aug 26 '17 at 17:40
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This is basically what astrophotgraphers do (with telescopes) to decrease exposure times for long exposure astrophotography. By adding a focal reducer, they reduce the image scale and decrease the effective focal length, increasing the effective speed and giving a wider field of view (due to the smaller image scale, limited by the original image circle and the reducer design).

Think of it as the opposite of a teleconverter.

Note, though, that your final image quality depends on the quality of both the original lens, and that of the converter. For astronomical use, around 0.6x to 0.8x is fairly common (and for small sensors maybe 0.5 to 0.33x)

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