why are f-stops 2, 2.8, 4, 5.6, 8, 11?

[sevo]

i don't think this has been mentioned. but what i find interesting is that f-stops seem to double in a way. look at 1, 1.4, 2, 2.8, 4, 5.6, 8, 11, 16, 22. there are two sequences of near doubling. the first from 1 to 2 to 4 to 8 to 16. the second from 1.4 to 2.8 to 5.6 to 11 to 22.

That is one of the neat properties of one single sequence progressing by √2.

 

[David Hughes]

Hi,

As we are talking about areas of glass you have to square the f numbers and then you'll find they go (roughly) 1 2 4 8 16 etc.

But the figures are cut back to fit on the lens so for F/2.8 we ought to see f/2.8284271 etc. It works for most of us and, given the fudge factor, in the shutter speeds and the actual shutter speeds, light passed by the lens and film exposure latitude, it's minor.

Regards, David

 

[rxmd]

But the figures are cut back to fit on the lens so for F/2.8 we ought to see f/2.8284271 etc. It works for most of us and, given the fudge factor, in the shutter speeds and the actual shutter speeds, light passed by the lens and film exposure latitude, it's minor.

Rounding to two significant digits already is more precise than most mechanical shutters and chemical development processes would be.

 

[Pherdinand]

That is one of the neat properties of one single sequence progressing by √2.

it's not so surprising, really. It is self-determied by the fact that surface has units of length times length (=basic math) and the photographic requirement that we want to know "how big must the aperture be in order to double the received light" which is completely up to us to define.
If we were talking about volume, instead of surface, e.g., calculating how wide (or tall, or long) you have to make a cube-shaped box to fit a certain nr of toilet paper rolls then the series would be cubic root of 2 and would, again, double.
Or if, for whatever reason, we would be interested in tripleing the light that passes through the lens, then we'd have to use √3, and therefore there would be 2 series of tripleing intermixed 🙂

Indeed the "next stop" in this learning process is T-stops 😀
but that loses relevance since the invention of proper anti-reflection coatings, coz the transmission of the lens elements is almost 100% therefore T-stops are only a tiiiny bit smaller than F-stops.

 

[DNG]

Quote: Originally Posted by DNG

it does not matter what Focal Lentgh lens you are using.
If you have a good exposure with your 50mm lens at f/4, you can put your 100mm lens on and use f/4,
and you will get the same exposure value (for the knit pickers, move back with the 100mm lens to cover the same framing)

f/4, or any f/stop, is one consistent you can count on when using different lenses for the same lighting. END QUOTE

wolves3012 said
This is reasonable in most cases but it's not quite true. Two lenses at the same f-stop are not guaranteed to give the same exposure, since it also depends on the lens' light transmission. T-stops are equal-illumination points (and therefore equal exposure for a given shutter speed), not f-stops. Of course, most lenses aren't marked in t-stops, so f-stops is the best you can do with them.

Yes, this is true also, although, I leave the ubber tech to the mathematician! in all practical none tech photographic applications, the small variance in light transmission, is a non issue. The way we arrive at number f/1.4, f/2.8 etc.. is beyond my math skills. BUT, That may be the OP question also.

 

[Dwig]

in all practical none tech photographic applications, the small variance in light transmission, is a non issue.

Actually, the variance is often fairly large (~1/2stop or greater) in very complex optical designs. For almost all still photography users, it is completely compensated for by the ubiquitous TTL metering found in cameras that accept such complex optical designs (read: zooms). It is only a real exposure concern to cine shooters, and cine lenses are frequently calibrated in T-stops rather than f-stops.