F-stop light values equal on diff. lenses?

Hello all - I'm trying to wrap my head around the following:

Comparing two lenses, say the 50mm f/28 and the 17-40 f28L. The 50mm has a lens diameter of 2.7 in. and the 17-40 has a diameter of 3.3.

If both lenses (on the same camera body) are set to the same f-stop, say f4, will the larger diameter lens let in more light (i.e. 20% more light if diameter is 20% larger?)

Or, if my thinking is invalid, does an f-stop of f4 on specific camera body let in a set amount of light no matter the lens (this does not make sense to me.)

Can someone please enlighten me?

Thanks,

David
_______________
Canon EOS A2E
Canon EOS 10D
50mm 1.8 II
135mm 2.8 Soft Focus
28-105mm 3.5-4.6

As I understand it, f/2.8 means the same thing lens-to-lens, Canon/Nikon/..., even SLR compared to tiny digicams. It defines, along with the shutter speed an absolute exposure level. The lens diameter is just one component of aperture. Length of the lens is another. Zoom lenses always seem to be slower than primes. I don't know the answer why, maybe someone else can answer.

your understanding with regards to the physical diameter of the aperture is correct. i.e., at identical f values, a telephoto lens has a bigger apperture diameter than a wideangle lens, because the diameter = focal length / f value.

however, the amount of light coming in is the same. light coming in thru a telephoto lens needs to travel a longer distance, simply put, this sort of "cancels" out the effect of a larger aperture opening.

remember the intensity of the light is proportional to the area of the aperture opening and is inversely proportional to the square root of the distance the light has to travel. that's why two different focal length lenses at same f-stop let in the same amount of light.
-alex

ok here's some math to illustrate this:

we'll use two lenses here:

(1) 100mm lens at f/4
(2) 40mm lens at f/4

for lens (1):
aperture diameter at f/4 = 100/4 = 25mm
aperture radius = 25/2 = 12.5mm
area of aperture = 12.5*12.5*pi = 156.25pi
the intensity of light: area/distance^2 = 156.25pi/(100mm * 100mm) = 0.015625

for lens (2):
aperture diameter at f/4 = 40/4 = 10mm
aperture radiu = 10/2 = 5mm
area of aperture = 5*5*pi = 25pi
the intensity of light: area/distance^2 = 25pi/(40mm * 40mm) = 0.015625

-alex

OK
I'm officially impressed.

Wow..... I've thought about that, but.... didn't know there was a mathmatical answer to it....

Rentboi, very impressive!

I've used that math before, least the first portion of it, not because of this question, but to prove to someone that a longer prime at the same aperture is more shallow a depth of field than a wider lens, since the opening is wider on a longer lens.

A basic photography class is a great idea, as it covers helpful stuff like this. Nice maths Alex

basc photography class was never offered...! that said I understand the above...cool!

Mmm, I'm taking a basic photography class sometime soon. Free classes came with my 20D purchase. I just gotta figure out where and when the classes take place.

Either a class, or you could just pick up a book and read it yourself.

Goooooooooooood point! Books are amazing.

Goooooooooooood point! Books are amazing.