This how I see it:
a lens with F2.8 from FF or aps-c camera has a bigger aperture than the f2.8 lense made for the Q. (the physical opening of aperture is much bigger) and the lenses that are made for the Q are much smaller with tiny aperture (opening).
So when a FF lens is mounted on the Q (wich has much bigger F2.8 opening than the F2.8 made for the Q) should allow more light and not less.
I know most of the light will fall around the sensor, but that loss of light is to cover FF image and is not needed for the Q sensor anyway. It's a crop sensor and it should have its fill of light. The unit of light per area should be the same or even more on the Q sensor.
The only thing that would make sense to me is a loss of light because of the greater distance of the optics from the FF lens to the Q sensor.
I am saying this knowing that I could be completley wrong, but this is how I see it, sombody please tell me what I'm missing.
Hi Jacob,
I've been thinking about this since my last post (see under the "wrong" subthread of this thread). The amount of light is actually the same. There aren't different standards for different formats. f stop values have to correspond with Exposure values so all the rules of photography work. There are, of course, sample variations and calibration variations for metering systems, and sometimes these are possibly tweaked by mfgs so they might be able to market a quality -- like when a particular model's stated ISO setting is actually faster than the actual value to make the noise performance look better at a given setting. They all have to be within some certain tolerances to be credible though.
Think about this:
You have a 12MP FF (24x36mm) body and a 12MP APS-C (16x24mm) body plus one f2.8 lens that covers the 135 frame and has a mount that fits both. Both bodies expose exactly the same with the same aperture, shutter speed, and ISO setting, as you'd expect. The APS-C has a sensor that has less than half the area of the the other, and the same number of photosites since they both are 12MP. The lens and register distance is the same for both bodies or the lens wouldn't focus from the same MFD to infinity on both bodies.
The "Laws of Physics" (I'm being sarcastic. . .this is an overused and specious term as usually applied on photo fora -- it's actually the Rules of Math) dictate that the APS-C sensor would need to have photosites that are at most proportional (less than half) the area of those on the FF sensor each (assuming that the position, shape and size of the photosites are proportional), so if the amount of light was cut roughly in half by the smaller photo sites, why do the cameras expose the same scene correctly with the same shutter, aperture, and ISO? Let's say you also have a 12 MP P&S camera with an f2.8 of the same FL equivalent lens and a 1/1.23" sensor, and when you meter the same scene and take the shot, it also exposes correctly at the same shutter speed, aperture, and ISO with only that tine sensor. Pretty weird right? -- not really, because the photography industry adheres (within tolerances) to the same standards of Ev and exposure.
The FF lens has a larger physical aperture than the P&S sensor lens only because it must cover a larger image circle. It's also farther away from its respective sensor (due to the longer registration distance used by the lens' and body's design) than the P&S Cam's lens. Neither of these make a difference, because the intensity of the light that
gets to the sensor with equivalent Ev and f-stop values
is the same between the respective cameras, or they would be out of tolerance for the industry standard and the rules of photography.
Any differences in the number of photons of light that reach each photosite for a given Ev are adjusted for in the design of the camera. Max Aperture value is dependent on the lens's FL and Aperture (diameter of the front element of the lens or entrance pupil) with no other obstruction of the light path. Smaller Avs for the lens are calibrated from the Max Av, with each stop reducing the area of the opening in respect to the previous by factor of 2 (if you have an f2.8 lens, then at f4, the aperture blades close down to the point that the area of the hole left is 1/2 of the area wide open, and f5.6 is 1/2 the area of f4, and so on. . .). Shutter speed is measurable against the international standard. ISO is calibrated against a measured standard light source using the calculated Av and the measured shutter speed and the values are set in firmware.
The designers of the camera decide the ISO range available in the camera given the sensor and their standards for IQ after test images are evaluated after processing by the camera. This is why the Pentax K-5 and the Nikon D7000, both with essentially the same sensor, have different available ISO ranges.
Sorry for the long winded reply, but wanted to try to explain this thoroughly given my understanding.
Scott