Amazing lens, incredible sharpness

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. 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. 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.

Now I totally agree that all else equal, ff according to theory will have higher s/n than a sensor with smaller pixels. I just think the scaling with pixel size is complicated.

...surely 1.8 is 1.8.

Well, 75mm f/1.8 has an aperture (entrance pupil) diameter of 75mm / 1.8 = 42mm, and 150mm f/3.6 also has an aperture diameter of 150mm / 3.6 = 42mm. So, f/1.8 on mFT is equivalent to f/3.6 on FF, where by "equivalent to", I mean results in the same diagonal angle of view, the same DOF for a given perspective, framing, and display size, and projects the same total amount of light on the sensor for a given shutter speed, resulting in the same noise for equally efficient sensors.