Look at it this way: The effective aperture (more properly called the entrance pupil) is the diameter, not area, of the aperture as viewed through the front of the lens. This means:
- If you double the diameter of the aperture, you increase the amount of light allowed to pass four-fold. But all of that light is still falling on the same size image circle. This means that each point on the image circle is receiving four times the illuminance.
- As the focal length of a lens increases, the minimum diameter of the front element must increase to maintain the same f-number. An f/2 aperture is 50mm wide for a 100mm lens, so the front element of a 100mm f/2 lens must be at least 50mm wide. Otherwise you could not measure the aperture as 50mm wide as viewed through the front of the lens. A 200mm f/2 lens must have a front element at least 100mm wide.
- If we referred to apertures by total area, rather than the f-number, we would need different combinations of Tv/Av/ISO for the same amount of light at every different focal length! By using the f-ratio, it allows the correct exposure values for a subject of a specific luminosity to remain the same, regardless of the focal length.
For more on why exposure is determined by the amount of light per unit area, and not the total amount of light gathered, see Lens f-number and speed on adapted lenses
I opened that link and will read it, but I wanted to quickly say that I understand and agree with the reasons why we measure aperture in f-numbers rather than the total area. However, in this case, I find that it obscures rather than helps me understand what's going on. Namely that the FF sensor is performing better only because it's receiving more light, not because it's inherently more sensitive.
It is only receiving more light because the sensor is larger. The amount of light per cm² from a 50mm f/2 lens is exactly the same as the amount of light per cm² from a 100mm f/2 lens (assuming both are viewing the same scene).
Doesn't noise depend on the total light coming in to the sensor, and not the light per unit area? If the latter were the case, you'd find that an FF sensor with an f/2.8 lens performs no better than an APS-C sensor with an f/2.8 lens, right?
NO. Noise in the image depends on the signal to noise ratio. Since the noise is fairly constant at each pixel, the stronger the signal is at each pixel, the lower percentage the noise floor will be of the total readout of that pixel. That is why larger pixels are inherently less noisy: each pixel can collect more light/photons/signal while generating no more read noise than a smaller pixel would.
Larger sensors allow either: larger pixels for the same resolution/number of pixels or higher resolution/number of pixels for the same pixel size or a combination of both (moderately larger pixels and moderately more of them).
If the pixels are the same size (and identical in terms of other technological questions) on both the FF and APS-C sensors then you are correct at the pixel-peeping 100% large display level that the noise level in the FF and APS-C cameras would be the same. BUT: if you are then displaying the images from the different sized sensors at the same display size (i.e 8x10, or 16x20, even 36x24, or even larger), the higher magnification needed to display the 10MP APS-C sensor compared to the lower magnification needed to display the 22MP FF image would also make the perceivable noise higher.
Remember that if both sensors have the same pixel size, then at 100% viewing size on your monitor the APS-C image would be less than half the total area of the FF image.
On the other hand, if both the APS-C and FF sensors have the same number of pixels, then each pixel on the FF sensor is 2.25 times the surface area of each pixel on the APS-C sensor. That means for the same scene through the same aperture lens the FF camera collects 2.25 times more light/photons/signal per pixel than the APS-C camera does, which means the SNR at each pixel is over twice (one stop) that of the APS-C camera.
I agree fully with your last five comments, Michael. I don't think we actually have a disagreement here. To be clear, I'm viewing the photos on a fixed size computer monitor, as I wrote in my updated question, and not 100%. For this setup, since the noise depends on the total light falling on the sensor (not per unit area), I don't think it's inaccurate to say that it's not the sensor per se, but the larger entrance pupil of the lens that's responsible for the better image quality. Correct?
If you are viewing the output from two differently sized sensors on the same size monitor the the difference in magnification (between the size of each sensor and the size of your screen) will make a difference relative to noise.
"For this setup, since the noise depends on the total light falling on the sensor (not per unit area)...
Just because you keep repeating this does not make it any more correct than it was the first time you said it.
You're also ignoring the elephant in the room since you clarified that this is not a comparison of a FF and APS-C that utilize the exact same technology but rather an APS-C made by Sony and a µ4/3 made by Olympus. The difference in the way each manufacturer chose to design the sensor and how each chose to process the output from that sensor may likely have more to do with the relative performance of each.
No matter what sensor size I have, a lens with an entrance pupil of 20 sq mm is going to give me photos with lower noise under low light than a lens with entrance pupil 10 sq mm. Even if these two lenses are mounted on different sized sensors. As long as the FF-equivalent focal length of these two lenses is the same, and we don't have a mismatched system (like mounting an FF lens on APS-C without the Speed Booster, thereby wasting light, or an APS-C lens on FF body).
You're still ignoring the effect of the different magnification factors on noise. Depending on the difference between the two sensors it can be quite significant when comparing photos taken in low light.
All other things being equal, a full frame sensor will outperform a 1.5X APS-C sensor in terms of SNR by a factor of 2.25 (roughly 1.15 stops). A 50mm lens on the APS-C camera provides the same FoV as a 75mm lens on a FF camera. The 26.8mm wide e.p. on the 75mm f/2.8 translates to an f/1.9 aperture on the 50mm lens. That is roughly 1.15 stops. But here's where the "all other things being equal ends." The smaller sensor needs just as much circuitry per pixel on the sensor die as the FF sensor does, which means almost certainly the smaller sensor is using less of its total surface area to actually collect light.
Let's also assume the same sensel size. I'm clarifying since that issue came up. For that matter, one can have two APS-C sensors with a different megapixel count and hence different sensel sizes. Let's not bring that other factor in.
That "other factor" is the real world vs. the purely theoretical.