A mirco four thirds sensor (crop factor 2.0) gathers the same amount of light with a F/2.0 lens compared to a fullframe sensor with a F/4.0 lens since the area of the sensor is four times smaller. If both sensors are having the same pixel count the signal from the photosides of the smaller sensor would be the same as on the other sensor (they get 4 times more light but are 1/4 of the size). If the signal is the same - why would the fullframe camera use four times the ISO of the smaller camera to amplify the same signal four times more?

I could ask this question alternatively: Using two different sized sensors with the same lens - why is the ISO on both cameras the same since they gather a different amount of light because of their size difference?

  • \$\begingroup\$ Half in height and half in width thus 4 times less area. \$\endgroup\$
    – Arji
    Commented Aug 26, 2021 at 21:33
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    \$\begingroup\$ You are assuming that there is some standard that sets ISO. It is a multiplication factor that is applied to the readout of the sensor. I would guess (which is why this is a comment) that for each camera the ISO scale is set so what used to be a proper exposure on film is still a proper exposure on the digital camera. If the sensor on a new camera is twice as efficient at turning light into electrons the manufacturer will adjust the multiplication factor downward by a factor 2. \$\endgroup\$ Commented Aug 27, 2021 at 0:48
  • \$\begingroup\$ I think you are confusing Total Light with Light/unit area. It's the latter that impacts ISO as that is light on a sensor element. \$\endgroup\$ Commented Aug 27, 2021 at 16:31
  • \$\begingroup\$ @BobMacaroniMcStevens Full frame is 24x36, m43 is 13x17.3. \$\endgroup\$
    – Eric S
    Commented Aug 27, 2021 at 18:49
  • \$\begingroup\$ Possible duplicate: photo.stackexchange.com/questions/55885/… \$\endgroup\$
    – Rafael
    Commented Aug 28, 2021 at 20:30

2 Answers 2


I'll try to answer based on your second alternative.

ISO values were defined using the chemical film model, defining which film exposure (in lumen-seconds per square meter) gives a decent gray or color level. The benefit of ISO values is that you don't need object distance, focal length, sensor size or pixel size to determine the correct exposure, only the scene lighting, the aperture value, the exposure time and the ISO value.

Comparing with film, an m43 sensor corresponds to cutting a 13x17mm part out of a bigger 24x36mm film. Of course, this film cutout still has the same sensitivity.

You're right that a smaller sensor gets less total light, but that is because it only captures a smaller part of the scene (as long as you use the same lens). The scene parts that fall onto the cutout still get the same illuminance through the lens.

If you use digital sensors with the same pixel size in both cameras, it's clear that both cameras behave the same, need the same signal amplification and give the same ISO values.

Same pixel size means half the linear resolution in the m43 camera, and this isn't state of the art. Typically, the m43 camera might have the same resolution as the full-size one, meaning that the sensor elements have only a quarter of the area, compared to the full-size body, and receive only a quarter of the photons. So, with less photons available, the same-resolution m43 camera needs a higher amplification to achieve the same ISO value, typically resulting in a higher noise level.


No one thinks about the "total" amount of light. What counts is the amount of light enough to produce an image on the projection plane.

The lens has no idea on the sensor size, if it is a film, if it is electronic, if it has a piece of cardboard blocking some areas cropping it.

The aperture is a relation between the aperture diameter and the focal length. It does not take into account the sensor size at all.

Do I use the crop factor in calculating aperture size and area?

Here are some similar questions, with different premises:

Macro amplification of the lens: What does "magnification" mean?

The focal length of the sensor. Effective focal length with crop sensor, and detail of image

  • \$\begingroup\$ The total amount of light matters, but the reason isn't obvious. It's because increasing the amount of light decreases the S/N ratio due to Photon Shot Noise. It turns out shot noise is the dominant source of noise in modern cameras. There's no magic sensor technology that will ever eliminate it, because it's a property of the light itself. \$\endgroup\$ Commented Aug 29, 2021 at 23:17
  • \$\begingroup\$ Increasing the amount of light increases the S/N ratio because the amount of shot noise only increases as the square root of the increase in total luminance. If you multiply the amount of light (signal) by a factor of 9, the shot noise only increases by a factor of 3. Thus your S/N ratio is higher the more light you add (until you reach full well capacity). \$\endgroup\$
    – Michael C
    Commented Aug 31, 2021 at 6:14
  • \$\begingroup\$ @MichaelC you're right, I said decreases where I meant increases. The point is that more light means less relative noise. And if you're not considering total light, perhaps you should start. \$\endgroup\$ Commented Sep 2, 2021 at 4:07

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