Sure, this is a technical curiosity question. Does anyone have a simple way to explain how the "speed of a lens" is calculated? I do not mean the basic answer that a f/2.8 is faster than a f/4.o. I know that part. Here is a "for instance": I have binoculars. These are about 350mm focal length, based on the magnification. The exit pupil is about 6mm, meaning these binoculars are about f/52. This is a small aperture. (Good dof but not a lot of light compared to a camera lens.) Still, this does not tell me anything about the speed of the lens. The objective lens(the big one on the front end) is 35mm. This HAS to figure in the speed of the lens. I use the binoculars only as an example. I have a 300mm zoom prime lens for my camera. This is an f/4.0 throughout the zoom range. The objective is 82mm, or close to 3.2 inches. Apertrue is adjustable as in any camera lens. Now, how in the world is the effective "speed of the lens" figured out? I assume the larger the objective lens the faster the lens might be. I assume an f/2.8 300 mm lens would have a larger objective lens...getting really large. Can anyone help me on this question which is almost irrevelant to actually shooting the pic? This is my curiosity working. Thanks. If no replies, I do understand. TomDart.

Attached Link: How to calculate Exposure Values

Not sure , but does this help any.

"The happiness of a man in this life does not consist in the absence but in the mastery of his passions."

This is not exactly what I am looking for but gets close..thanks very much! My searched yielded little or nothing. The link is appreciated. Tom.

Well you got my interest up on that too so I go searching all over to read anything I can. Did not think it was exactly what you want, but maybe has some links avail too.

"The happiness of a man in this life does not consist in the absence but in the mastery of his passions."

After reading your post I was compelled to refer to an article about lenses in my Photographic magazine(02/04). I think the following excerpts are relevent to this post: "The aperature ring (or camera LCD panel, when aperaturs are set using the camera dial) is calibrated in f-numbers. F-numberes represent ratios: the ratio between the diameter of the aperature and the focal length of the lens.....Lenses with large maximum aperatures are termed "fast".....Faster lenses provide brigher viewfinder images for eaisier focusing, and they permit using faster shutter speeds in dimmer light...." According to the article, "Fast" sounds like photography lingo to me; the term is not calculated like the F-stop. I do wonder what is considered fast as opposed to slow. What may be considered fast for one person may be too slow for someone else. I assume the term "fast" lens term is probably established by the someone else. -Jeff

Here's my explanation, don't know if this helps: The equation to calculate the f-stop (aperture ratio) is focal length divided by the diameter of the lens. So the bigger the aperture/diameter of the lens at a given focal length the faster the lens.

Coolj and zahn pretty much hit what you're asking, I think. Fast in photography terminology is really a bit subjective, but normally for 35mm format cameras, F/2.8 and larger aperture lenses are considered fast (some would say f/2.0 and larger). The F-stop is derived by the formula mentioned above. Theoretically, this ratio lets the same amount of light through to the film plane regardless of the focal length of the lens. So, the amount of light let through at F/2.0 on a 50mm lens is the same as F/2.0 on a 200mm lens. In the real world, you have to deal with light loss through the multi-object optics so I'm sure there's some fudging going on in that ratio somewhere.