>>3650571>but it would need an mtf 100 (complete black to white contrast) at double the lines/mm of the sensor in question (therefore at or above the nyquist limit).This is the kind of intuition someone with not much experience in optics might get, but it's not correct.
A couple points:
1. MTF1 means an ideal lens which cannot possibly exist
2. MTF1 at any value of lp/mm means "infinite" resolution for the lens. {Well, not exactly infinite, but the resolution limit as dictated by the Rayleigh criterion and lens diffraction, a quick formula for incoherent light is (line pairs per millimetre)=(1,000,000)/(f stop * λ), and of course it depends on wavelength}.
Another way to look at it is, MTF values when plotted against lp/mm, are smooth and continuous functions, you can't have a huge (non continuous, non smooth) drop to 0. Hence a lens with MTF 1 will take you to the optical limit, as described above.
What I'm trying to say is, MTF 1 many times is sneaked in as an "innocent" assumption, but it implies many more things, since it's a boundary value, and things behave differently at boundary values.
But let's assume you got that magical lens. So what? Now the system resolution reached the sensor resolution. How is that outresolving the sensor?
When people say "x lens outresolves y sensor" they mean that there's no benefit using a higher resolving lens than x. In practice (and theory) there's always a benefit using a higher resolving lens. A lens with MTF 1 means, there's no higher resolving lens. So then any claim about "outresolving" in that sense in moot.
Also if you want to be pedantic, the AA filter has its own MTF, the microlens array its own MTF, the bayer filter its own MTF (even the atmosphere). So I hope your calculations for the Nyquist limit take into account those too, and not just the pixel count and spacing.
The whole expression "x lens outresolves y sensor" is ill posed, like saying "the set of all sets is Z" or "the wheels outperform the engine".
