- This topic is empty.
F2.8 @5.6 v F5.6 @ 5.6
-
Folks,
I’m steadily building a nice little arsenal of lenses and I’m currently waiting on delivery of a Bigma (thanks for the advice Aidan!). I’m a relative beginner at this artform but I’ve a question that has been bugging me for the last couple of weeks. I got a good deal on a Tokina AT-X 80-200mm f2.8 and couldn’t resist but I already have a Minolta 70-210 f4-5.6. What is the difference in using the Tokina at f5.6 over using the Minolta at f5.6 ignoring brand differences? Surely on face value they’re the same thing??
Sorry for the query but this to me is like the “if a tree fell in the forest and there was no one there, would it make a noise” type question to everyone else!
Most lenses perform best when stopped down from their maximum apertures, therefore by the time you have stopped down the 2.8 lens to 5.6 it is starting to reach closer to it’s sharpness ‘sweet spot’ whereas the 5.6 lens is just at its maximum and wont start sharpening up until F8 and over. There’s also the fact that the viewfinder will be brighter looking through a 2.8 lens than a 5.6 lens across the range of aperture selected since they only stop down when the shutter is pressed.
There is another very important point to consider which is often forgotten by users. Rather than me explaining what an f/number is, take a look over at Wikipedia, where it’s explained very well. The opening paragraph there states that
“In optics, the f-number (sometimes called focal ratio, f-ratio, or relative aperture[1]) of an optical system expresses the diameter of the entrance pupil in terms of the effective focal length of the lens. It is the quantitative measure of lens speed, an important concept in photography.”
The piece then goes on to describe how the f/number is derived mathematically. A lot of people will make the mistake of saying that if they were to follow all that through, then they could calculate the physical diameter of the diaphragm for any given aperture. But they would be in for a surprise if they actually tried to measure it, because f/no’s relate to the effective diameter rather than an actual diameter. If you were to remove the diaphragm (or aperture blades) from a lens altogether, the lens would still have an f/no.
A simpler, although less exact way of explaining the importance of this would be to imagine your f/5.6 lens working wide open – the full surface of each lens element is utilised. Now consider your f/2.8 lens stopped down to f/5.6 – you’re effectively blocking the light that’s travelling through the outer edges of the lenses and only the better quality inner parts of the lens elements are transmitting light through them so you end up with a better quality image straight away.
Hope I haven’t confused you more now! :(
Thorsten wrote:
A simpler, although less exact way of explaining the importance of this would be to imagine your f/5.6 lens working wide open – the full surface of each lens element is utilised. Now consider your f/2.8 lens stopped down to f/5.6 – you’re effectively blocking the light that’s travelling through the outer edges of the lenses and only the better quality inner parts of the lens elements are transmitting light through them so you end up with a better quality image straight away.
Hope I haven’t confused you more now! :(
Beautifully explained as always Thorsten :wink: . And that’s why fast glass
is so much more expensive, quality.Rob.
Ah yes! Simple enough really, isn’t it.
Thanks for that folks. Tune in next week when Thorsten will explain how jacobs get the figs into the fig rolls. :lol:
And as for the tree in the forest I think we’ll have to contact Joe_Elway & his hide to prove that one.
Really, thanks for your explanation. It’s easier to get a good photograph when you understand how your equipment works…seemingly.
You must be logged in to reply to this topic.
