Tamron 90mm SP AF f/2.8 Di

I have recently bought this excellent lens but there is something that
I do not understand about it's behaviour. I have the lens mounted on a
Nikon F80.

The lens is f/2.8 only when it is focused near infinity. If it is
focused at the minimum distance outside the macro range it gets f/3.8
as the maximum apperture.

At the macro range we have f/5.6 at 1:1 and f/3.8 at 1:2.8

I would extremely appreciate any explanation on this behaviour.
Thanks in advance

Please ignore my posting as I just found the following thread which
explains my question very well.

Help: Tamron SP 90mm F2.8 behaves like a variable aperture zoom
Thanks again.

Merry Christmas

wrote:
I have recently bought this excellent lens but there is something that
I do not understand about it's behaviour. I have the lens mounted on a
Nikon F80.

The lens is f/2.8 only when it is focused near infinity. If it is
focused at the minimum distance outside the macro range it gets f/3.8
as the maximum apperture.

At the macro range we have f/5.6 at 1:1 and f/3.8 at 1:2.8

I would extremely appreciate any explanation on this behaviour.
Thanks in advance

This is the proper functionality. What the body is showing you is the
effective aperture of the lens. The effective aperture of the lens is
defined as:
Ae = Aa * (1 + M/p)
where
Ae = Aperture effective
Aa = Aperture actual
M = magnification ratio
p = pupil ratio

Most people assume p=1 but this is not necessarily the case, though is
likely close enough in most situations (longer macro lenses tend to have
a pupil ratio of less than 1).

Thus, taking a lens that is f/2.8 normally and taking it to a 1:1
reproduction ratio and assuming a pupil ratio of 1 you get:
Ae = f/2.8 * (1 + 1/1)
Ae = f/2.8 * 2
Ae = f/5.6

It is likely easier to think of using extension tubes. If you were to
take a 50mm prime and put another 50mm of extension behind it you would
get only 1/4 of the light hitting the media when you take into
consideration the inverse square law.

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