Re: Why can't a non-SLR have equivalent quality?

reply@xxxxxxxxxxxxxxxxx writes:

>I'm not saying that
>either side is right or wrong, but I understand that disproving what
>Alfred said is going to be difficult because it takes more than
>individual camera lenses as counterexamples. You'd need pairs of
>similar lenses, or better, a proof based on a bit of knowledge of
>lens design and some math. With luck it wouldn't require more than
>algebra and trig. Maybe Roger (or Ilya if he keeps the math clear
>and understandable) could help solve what's actually an interesting
>puzzle, though not for everyone's taste I suppose.

It doesn't take *much* knowledge of lens design. Reducing the diameter
of the elements in a thin lens does not reduce coverage at all - it
just reduces f/number. In a more complex design with multple elements,
you can still reduce the f/number while keeping the coverage the same.

If the lens is a long one for the format, so it only covers a few
degrees of the subject, just scaling all the lenses and mechanical
mounts by the change in f/number is sufficient. With a normal or
wide-angle design, some elements are large to achieve angular coverage
instead of light throughput, and the designer would have to trace ray
paths and keep some elements larger than proportionality to avoid
vignetting - but coverage *still* isn't directly linked to element
diameter.

And, in general, lens diameters do not scale up by as much as image
size does in the real world, because designers choose smaller apertures
and different designs to keep the lenses practical. There are view
camera lenses with very wide field of view and large image coverage
while the lens diameter is still much smaller than the image circle.
You just won't find view camera lenses with large apertures or zoom.

I don't think it's necessary to present a lot of math to prove something
that is one of the most basic ideas of imaging optics. Lens diameter is
not related to coverage (except in telecentric lenses, which are a very
specialized design).

Dave
.

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