A question I can't get out of my head

Hi all, I have a question that I should probably know by now, but I don't so I'm going to ask it anyway. The only stupid question is one that goes unasked....(or something like that!) Problem is I'm not even sure the best way to ask it....so here goes.

F/4 is F/4 no?

So lets say I have x amount of light in a scene and I need to use F4 at 125th to get a correct exposure. ISO being a constant.

So I hook up my nifty 50 at F/4 at 125th and take a shot. Now I put on my 135 and set it to F/4 shutter at 125th and take a shot.

Given at F/4 the actual aperture size on the 135 is 33.75mm and on the nifty it is 12.5 mm, don't I have more light entering with the 135? I do just divide the focal length by the F number to calc physical size don't I?

I just know I am missing a critical, probably very simple part of this, but I need to know!!

I did read this on wiki, but I'm having a hard time understanding the last part in bold text:

Aperture values are ratios. f/4 isn't the same physical size on all lenses but it'll produce the same exposure (within manufacturing tolerances).

I know that bit, but why is it the same exposure given my point that the actual hole is larger on the 135 than the 50.

go to extremes, 10mm focal length at F/4 =2.5mm physical hole that the light enters through. 800mm F/4 =200mm physical hole for the light to enter....yet both end up with the same exposure, camera, shutter speed, and ISO being constants

It has to do with optics, physics, and other stuff beyond my brain's capabilities. Something to do with the size of the front element or something. I dunno.

oookaaay then...


anybody else maybe?

Maybe this will help?

en.wikipedia.org/wiki/​F-number wrote:
=http://en.wikipedia.or​g/wiki/F-number]T-stops

F-stops are purely geometrical, the ratio of aperture to focal length, regardless of actual light transmitted. Since all lenses absorb some portion of the light passing through them (particularly zoom lenses containing many elements), f-numbers do not accurately correlate with light transmitted. F-numbers corrected to measure light transmission rather than aperture ratio, called T-stops (for Transmission-stops), are sometimes used instead of f-stops for determining exposure. [5] A real lens set to a particular T-stop will, by definition, transmit the same amount of light as an ideal lens with 100% transmission at the corresponding f-stop.

Use of f-numbers leads to exposure inaccuracy, particularly for lenses with many elements. This is particularly problematical in cinematography, where many images are seen in rapid succession and even small changes in exposure will be noticeable. To avoid the problem, lenses used in cinematography were bench-tested individually for actual light transmission and calibrated in T-stops, allowing fixed-focal-length turret-mounted lenses to be changed without affecting the overall scene brightness due to differences in transmission for the same f-number. Many modern cinematographic lenses are factory-calibrated in T-stops. In still photography, without the need for rigorous consistency of all lenses and cameras used, slight changes in exposure are less important, and are largely masked except for the highest-absorption lenses by film and sensor exposure latitude.

Since all lenses absorb some light, the T-number of any given aperture on a lens will always be greater (less light transmission) than the f-number. The T-stop corrects exposure for absorption of light, but the depth of field is determined by geometry; consequently, the depth of field for a given T-number will be slightly less than for the corresponding f-number; the discrepancy will be different for lenses of different degrees of absorption.

ok, I see you replaced your last post, but you refered me to F-number wiki page originally.

right, so the question is now why will it "produce the same illuminace in the focal plane"...... which is basically what we all know and have already said.....but in the written word at least, the article doesn't explain the "why" if it's all hidden in the mathematics then fine, but I was hoping maybe somebody who knows may like to explain it in "simple terms"...if indeed it can be explained easily.

Not sure T-stop explains it either, but is interesting to know nonetheless

I think it really is hidden in the math. Unfortunately. I'm not a moron I just never took physics so if it weren't for the math, I'd most likely get it. To me it's like trying to understand why 1s and 0s as opposed to Xs and Ys represent data. People say "because 1 is on and 0 is off" but that doesn't really help. It's also like trying to understand how a copper cable can transmit signals. "Because electrons can flow through it easily." Okay but why can electrons flow through it easily? "They just do."

Ya know?

It's similar to the inverse square law. I don't know if you are familiar with this in relation to lighting but if you have a light illuminating an area at say 3ft, then move the light away to 6ft then the illumination over that same area will only be a quarter of what it was before So twice the distance will need 4 times as much light to get the same illumination.

Now think about a lens, the longer the lens the further it is away from the sensor so at any given aperture the physical opening needs to be larger to provide the same level of light reaching the sensor.

cdifoto,

You mentioned electricity and copper cable.

Your answer was amazing, because "electricity" is still not really understood by those who should know. All "explanations" are actually excuses, made up stories to satisfy the children.

Why do electrons flow through as cable? Do they actually flow?

The standard story of putting one more electron into one end of a cable and another falling out the other end, driven by a potential difference (voltage) exciting the existing electrons to the point that they want to drift in a certain direction is total fiction but the problem is, no one knows any better. No one at all.

Wouldn't all of this have to do with how many elements the light has to pass through? With a longer focal length, I would expect that there are element groups for light to pass through, and therefore you need a larger outer element and a larger aperture to collect that much more light up front. By time the scene has been shuffled past so many elements, it is now at the point that sensor sees about the same amount of light as that from a smaller FL lens.

Isn't this why adding a 1.4x TC reduces your light by 1 stop, by putting that added element into the path?

Just an guess though, I never took optics at Rose, one of a few classes I should have taken.

f/4 on a 135mm lens is 33.75mm

f/4 on a 50mm lens is 12.5mm.

Thus, as you say, you'd expect that the amount of light hitting the sensor with the 135 would be larger. However, remember that the 135 is also collecting light over an angle of view of 18 degrees, while the 50 is collecting light over an angle of 46 degrees.

Thus, while the physical aperture of the 135 is larger than the 50, thus having the ability to let in more light, it's gathering light over a much smaller area than the 50. As it happens, these two cancel each other out exactly.

As such, f/4 gives the same exposure, regardless of focal length.

The math seems to add up...

18 deg/ 46 deg = .39

12.5 / 33.75 = .37

TeamSpeed wrote in post #13656328
The math seems to add up...

18 deg/ 46 deg = .39

12.5 / 33.75 = .37

Math has a way of doing that.

My head hurts.