Amazing lens, incredible sharpness

Great Bustard wrote:

thk0 wrote:

Great Bustard wrote:

thk0 wrote:

Great Bustard wrote:

tinternaut wrote:

In terms of DoF, yes it's a 3.6 equivalent...

I'm glad you agree.

...but in terms of light gathering (i.e. the shutter speed it enables one to achieve at a given aperture for correct exposure)...

Anders explained it nicely in his post above, but I'll take a stab at it, too. The exposure is the amount of light per area that falls on the sensor, not the amount of light gathered. So, f/1.8 on mFT results in 4x as much light per area as f/3.6 on FF, but a FF sensor has 4x the area, so the two neatly cancel each other out, and the light gathering of f/1.8 on mFT is the same as the light gathering as f/3.6 on FF.

I hear this often, but I don't think its correct.

It is. The photos here demonstrate it fairly well:

http://www.dpreview.com/forums/post/51782704

An imaging sensor needs to compute the difference in the number of photons hitting one site and the number hitting another site. The average is easily taken over the entire sensor. Having a 4 times larger sensing pixel does not give you 4 times better ability to compute that difference.

Pixel size has a secondary, and minor, effect on image noise, as a comparison between the D600 and D800 will clearly demonstrate.

Also its worth noting that a lot of image noise is not poisson photon counting; rather it is thermal noise related to electrical currents required to make the sensor work, and that does not scale in the same was as noise owing to pixel size.

If you mean read noise, yes, that absolutely has an effect (more so at higher ISOs such as those in the link above), but this all comes under the "equally efficient sensors" clause for noise equivalence.

By the way, we would all agree that a 75 / 1.8 on an EM5 is equivalent to a 75 / 1.8 on the more noisy G1, right? That is, we wouldn't say that f/1.8 on an EM5 is "equivalent to" f/1.4 on a G1, or something like that simply because the noise levels were not exactly the same, would we?

Now I totally agree that all else equal, ff according to theory will have higher s/n than a sensor with smaller pixels.

For a given exposure, but the same noise for a given DOF and shutter speed (all else equal, of course).

I just think the scaling with pixel size is complicated.

Not complicated, but largely irrelevant, again, as a comparison between the D600 and D800 will clearly demonstrate.

Ah. I see I misread your post. You are only making a point about total light gathered, not how it might translate into IQ.

Actually, I'm saying equivalent lenses will put the same total amount of light on the sensor for a given shutter speed, and that this will result in the same noise for equally efficient sensors.

I was questioning whether eg 4x more total light (ff at same f-stop) gives exactly 4x less noise.

4x as much (2 stops more) light will result in 1/2 as much (1 stop less) photon noise. The other player is read noise (the additional noise added by the sensor and supporting hardware).

The read noise becomes dominant only in the portions of the photo made with very little light -- most of the noise in the photo is dominated by photon noise.

True. But a) the portions of the photo made with very little light are those where quality shortcomings are most visible and b) the lower the overall light level, the greater the portion of the photo where read noise makes its presence known. Both consideration are of considerable importance, in my opinion.

(But I've not thought about it in great detail, so grain of salt.)