Is aperture the same on different format of lenses?

Question

I have just acquired a large format lens (a Schneider-Kreuznach Angulon 1:6,8/90) for the first time, coming from 35mm and medium format, and I am a bit puzzled by its aperture. I want to make sure I am not mistaken regarding its operation.

First its range is quite different from what I am used to with 35mm lenses; it goes from 6.8 to 45. (I am more used to seeing ranges like 1.8 to 22)

When I stop it down to F/16 for example, and I compare it with a 35mm lens also stopped down at F/16, I can see a clear difference in the resulting opening size. It's quite bigger on the large format lens.
I guess the larger format needs to let in more light to cover the bigger film at the same aperture that the smaller format. But I'm really just guessing here.

Can anyone please confirm if they are equivalent in terms of operation?
For example, if I follow the sunny 16 rule and the lighting conditions result in F/16, will it result in the correct exposure on both a 35mm format lens and a large format lens both stopped down at F/16 (both mounted on their respective format)?

Edit: the question in not about depth of field.

Answer 1

One of the challenges early photographers faced was, how can we set the lens so it delivers a predicable amount of exposing light and how we interchange this exposing intensity between different cameras?

Ratio to the rescue: The amount of light energy that traverses a lens is based in part on the surface area of the lens opening (aperture). Since most aperture opening are circular, we can fall back on the geometry of circles. If we multiply the diameter of any circle by the square root of 2, we calculate a revised circle diameter that doubles the surface area. Conversely, if we divide the diameter of any circle by the square root of 2, we compute a revised diameter that halves the surface area. In other words, this magic value is 1.4. This value is the bases of the f-number system.

OK, we can compute the f-number set using the 1.4 factor, we derive this number set: 1 – 1.4 – 2- 2.8 – 4 – 5.4 – 8 – 11 – 16 – 22 – 32 – 45 – 64. Each going right is its neighbor on the left multiplied by 1.4. This sequence causes the iris (named for the Greek goddess of the rainbow) to close down thus cutting the exposure 2X (halving) the light energy of the exposure.

Conversely, going left, each value is its neighbor on the right divided by 1.4. Going left opens up the iris, each click is a 2X (doubling) of the exposure energy. This set calculates what we call “full f-stops”. We can mathematically compute a number set that is in 1/2 or 1/3 f-stop increments.

Opening and closing down the iris to control exposure is not the only factor involved. The other factor is the focal length of the lens. Long focus lenses magnify (telephoto). With this increased image size that comes with longer focal length is a loss of light energy. You need to know, a magnified image is spread out over more area, thus each doubling of the focal length results in a 4X reduction of exposing energy.

What I am trying to tell you, we need a system that sets the exposing energy universally. One that takes into account the area of the iris and intertwines focal length. Ratio to the rescue. In math a ratio is dimensionless. Supposed a lens with a focal length of 100mm has an iris diameter of 6.25mm. We can divide 100 by 6.25, the answer is 16. We call this value the focal ratio or f-number for short. Now suppose a giant camera with a 1000mm lens is operating with an iris diameter of 62.5mm. What is its focal ratio? 1000 ÷ 62.5 = 16 (written as f/16). Both lenses, the big one and the small one operates at the same focal ratio of f/16. That means we can set both cameras to f/16 and the film or digital sensor will receive the same exposing light.

This is the magic of the focal ratio; it works regardless of focal length or iris diameter. In other words, f/11 on one camera lets in the same exposing energy as f/11 on any other camera regardless of the dimensions. The ratio method overcomes the actual dimensions of the lens.

Answer 2

You need to remember that the f-stop number is the ratio of the diameter of the entrance pupil (virtual image of the aperture opening seen from the front of the lens) to the focal length. So if the entrance pupil appears to be e.g. 5mm, on a 35mm lens that's roughly f/7, where on a 90mm lens it's roughly f/18 for the same apparent aperture size.

This is why it's difficult to get low f-numbers on very long lenses - for example, f/1.2 on a 600mm lens would require an entrance pupil of 500mm, with a correspondingly large front element.

Answer 3

You should not be comparing the visible aperture size from the back of the lens (the exit pupil); you should be comparing it from the objective side (entrance pupil).

If comparing it between lenses of the same focal length they should be about the same; allowing for some rounding errors (in both stated F# and FL).

The same F# will give the same exposure regardless of format...

Large format lenses tend towards smaller apertures because they are required in order to record equivalent depths of field (w/o using tilt).

The same lens will project/record a larger image circle on a larger format sensor, and record a wider FoV. And because large format camera's place the lens farther away from the image plane the details are larger (the projected cone is larger in diameter), and therefore the DoF is less.

E.g. different formats have a different "Normal Focal Length," but they all record an ~ 55˚ diagonal Field of View (FoV)... the FoV is measured diagonally as that correlates w/ the diameter of the required image circle.

oversimplified drawing: