Originally posted by devorama Isn't the amount of the diffraction dependent on the actual aperture size rather than the F-stop? F/4 is a pretty tiny opening on an 8.5mm lens, but is not so tiny an a 300mm lens.
Girl unclothed, so I can answer this
No.
Diffraction is caused by interference and is due to the angular difference of light rays hitting a pixel (actually, path length differences between all the rays coming from within the lens aperture). And that difference is indeed proportional to the angle
theta and aperture diameter
a. Destructive interference happens where this difference becomes half the wavelength,
lambda/2 ~ theta * a/2.
The point is that for any given focal length, you want to keep
theta * f < eps smaller than a given limit
eps, say a pixel.
Substituting
theta, this means:
lamda * f / a < eps. And
f/a is called the f-stop
N:
N < eps / lambda .
You have to care more about diffraction in a Q because its pixels are smaller (
eps is smaller). The effect of the smaller aperture diameter in mm is offset by a shorter path towards the sensor.