Hi Nicholas,
That's got me puzzled ... Why would a lens that is marked as f/1.4 but is actually f/1.8 have the depth of field of an f/1.4 lens? Thanks, Helen
You know, now that I think about it, that's a good question. Don't really know, TTH, but that doesn't effect my propensity to potificate on matters about which I know not (it's a Guy thing, like never asking for directions):
Some of the additonal attenuation seen in a camera's meter reading is due to greater vignetting of the corners throwing the whole meter reading off, assuming a semi-averaging or matrix meter. But there is additional attenuation in the center of the field when wide open.
So, the depth of field should also be attenuated some. Wide open there are two parts to the attenuation: the attenuation seen at any stop, and the additional attenuation of the peripheral rays for whetever reason. It is the additional peripheral rays that are responsible for the larger CoC and if half of them (WAG) aren't arriving then the CoC will have fall-off at the edges and act as a smaller CoC.
If the lens is marked 1.4 and:
f/stop t/stop
f1.4 1.7 - 1/2 stop attenuation
f2.0 2.2 - 1/4 stop
.... .... - 1/4 stop
Not knowing any better, I would split the difference of the additional 1/4 stop of attenuation and figure that at wide open the lens is behaving like a
f1.4 + 1/8 stop = f1.45 lens for depth of field purposes.
This is only for the center of the image. The corners have additional attenuation because the exit pupil is cut-off by vignetting. Whether this reduction in effective aperture causes an additional decrease in dof at the corners I really don't know.