Tamron 90mm SP AF f/2.8 Di

Discussion in '35mm Cameras' started by pkokolis, Dec 25, 2004.

  1. pkokolis

    pkokolis Guest

    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
     
    pkokolis, Dec 25, 2004
    #1
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  2. pkokolis

    pkokolis Guest

    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
     
    pkokolis, Dec 25, 2004
    #2
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  3. 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.
     
    Michael Turner, Dec 26, 2004
    #3
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