Are f-stops equivalent between large format and 35mm lenses?

Asked 12/15/2021

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I’m new to large format and have a Schneider-Kreuznach Angulon 90mm f/6.8. Coming from 35mm and medium format, I’m used to aperture ranges like f/1.8 to f/22, but this lens goes from f/6.8 to f/45.

What confuses me is that when I set this 90mm large format lens to f/16 and compare it with a 35mm-format lens also set to f/16, the visible opening looks much larger on the large format lens.

Is aperture operation the same across different formats? In other words, if I use the sunny 16 rule, will f/16 give the correct exposure on both a 35mm camera and a large format camera when each lens is set to f/16?

I’m asking about exposure equivalence, not depth of field.

Originally by Photography Stack Exchange contributor. Source · Licensed CC BY-SA 4.0

Photography Stack Exchange contributor

4y ago

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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.

Originally by user44949. Source · Licensed CC BY-SA 4.0

user44949

4y ago

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AI Answer

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Yes. For exposure, f/16 means the same thing regardless of film/sensor format.

An f-number is a ratio: focal length divided by the diameter of the entrance pupil. Because it’s a ratio, two lenses on different formats can both be at f/16 and transmit the same light per unit area to the film, even if the physical opening looks different.

Your 90mm lens will have a larger physical aperture at f/16 than a shorter 35mm-format lens at f/16, because a longer focal length needs a larger opening to maintain the same ratio.

So if the lighting calls for f/16 by the sunny 16 rule, f/16 is the correct exposure setting on either camera format, assuming normal shutter accuracy and no special factors like bellows extension compensation.

Also, when visually judging aperture size, the meaningful reference is the entrance pupil as seen from the front of the lens, not simply the opening as seen from the back.

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4y ago

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