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If I'd use a 30mm lens on a crop sensor, would it be like a fullframesensor with a 50mm on it? 30*1.6=48~50

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We calculate a crop or magnification factor by finding the corner to corner measure of both formats and then divide. The full frame measures 24mm by 36mm. The diagonal measure of this rectangle is 43.3mm. The crop sensor measures 16mm by 24mm. The diagonal measure is 28.8mm. We divide 43.3/28.8 = 1.5. This is the factor that we will use. A 30mm on a crop sensor is the equivalent of 30 X 1.5 = 45mm on a full frame. Also 1/1.5 X 100 = 66%. This tells us that the crop sensor is 66% of the size of a full frame.

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  • \$\begingroup\$ Saying the crop sensor is 66% the size of the FF sensor is a bit misleading, even if it is true in terms of linear dimensions. In terms of area, which is probably the more applicable metric, it's only 44.4% the size of a FF sensor. \$\endgroup\$
    – Michael C
    Dec 27, 2017 at 14:32
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There's no such thing as true equivalence.

[All of the following assume the camera is at the same position shooting a subject at the same position so that the subject distance is maintained as a constant.]

We can compensate for differences in sensor size by using different focal length lenses. Using a 30-35mm lens on an APS-C camera will give the same field of view (FoV) as a 50mm lens on a full frame camera. Canon uses sensors roughly 22.23x14.7 mm to 22.7x15.1 mm for an ≈1.6X crop factor, everyone else uses sensors slightly smaller than 24x16 mm for an ≈ 1.5X crop factor in their APS-C camera model lines.

We can adjust the aperture to get the same depth of field (DoF). If we use f/2 on the 50mm lens with the FF camera, we need to use f/1.33 (1.5X) or 1.25 (1.6X) with an APS-C crop body. Therein lies one of the problems. To replicate a 50mm f/1.4 lens on a FF camera, we need a 30mm f/0.875 lens with a 1.6X crop body. But there aren't many, if any 30mm f/0.875 lenses available for APS-C cameras!

Even if we can find a lens that gives us the same FoV and DoF, the difference in aperture means the exposure value (Ev) has changed. The image taken using the lower f-number (to get the same DoF with the same FoV at the same shooting distance with the cropped camera) will be brighter. To compensate for the difference in exposure, we must either shorten the shutter time or use a lower ISO. Either change can materially affect how the image looks (i.e. subject motion or image noise). We're also, again, bumping against the limits of available gear at times. If we shoot with a FF camera using a 50mm lens, ISO 100, f/1.4, and 1/8000 second (It's really bright in here!) with a FF camera, to take more or less the same photo with and APS-C camera would require a 30mm lens, f/0.875, and either ISO 50 @ 1/8000 second or ISO 100 @ 1/16000 second. There are many FF cameras and 50mm lenses that can take that photo. To the best of my knowledge there are no mass produced APS-C cameras/lenses that can take the same photo. On the other side of the coin, there are photos certain combinations of APS-C cameras and available lenses can take that cannot be replicated with currently available FF cameras/lenses.

There's a saying that has been around a long time: Cheap, fast, or good. You can get any two at once, but you can't have all three. The same is true of photographic equivalence:

  • The same Angle of View/Field of View
  • The same Depth of Field
  • The same Exposure Value

You can have any two of the three (if a comparable camera and lens is available), but you can't have all three at the same time with two different sized sensors.

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  • \$\begingroup\$ of course, we can't have all 3, but actually, the absolute exposure value (light per area) is artificial and not useful when looking for equivalence between sensors - the amount of light per pixel (sensel) is more appropriate condition, because it's the sensel what finally receives and integrates the signal. under this assumption we can get two different sensors having the same number of pixels, same viewing angle and same aperture (different f/ratio of course) and take equivalent photos where the same amount of light is received by sensels of both sensors. \$\endgroup\$
    – szulat
    Dec 27, 2017 at 13:32
  • \$\begingroup\$ in the ideal world the outcome would be strictly equivalent, in the real world it is not because of technical imperfections, lens characteristics and diffraction effects, but it's the best we can get. that's why it should be called equivalent, even though the (artificially calculated) EV differs. \$\endgroup\$
    – szulat
    Dec 27, 2017 at 13:34
  • \$\begingroup\$ @szulat But even assuming the same technology and identical implementation, smaller sensels have lower full well capacity and therefore lower S/N ratio and dynamic range. It's still not equivalent. The larger sensor could capture more of a high dynamic range scene than the smaller sensor. The only way one could have even theoretical true equivalence is if we could scale the wavelengths of various colors of light (and the energy and mass of photons) by the same factor as we scale everything else. \$\endgroup\$
    – Michael C
    Dec 27, 2017 at 14:27
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    \$\begingroup\$ +1 @szulat should check physics. Want same FoV -- need different focal length. Want same DoF -- need different aperture. Want same exposure -- ??? Want same distortion -- ??? It's not trivial to make same image on different sensor sizes, even if money not the question. \$\endgroup\$
    – aaaaa says reinstate Monica
    Dec 27, 2017 at 17:38
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    \$\begingroup\$ @szulat Point is that "perfect equivalence" is too complicated and doesn't solve any real-world problem to bother. Not a photography but physics/engineering issue \$\endgroup\$
    – aaaaa says reinstate Monica
    Dec 27, 2017 at 18:11

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