# Why are full fstops the numbers they are?

Discussion in 'Beginner Questions' started by http://www.photo.net/martinw, Mar 27, 2018.

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1. ### Andrew Garrard

Well, I did remember "two little ducks" and didn't remember "the Lord is my shepherd", but that might just mean my interests are more ornithological than ecclesiastical.

23 is the smallest prime number that's not two away from another prime number. It's the number of chromosomes we typically have. 23 is the number of people you need to have before it's likely that at least two share a birthday. It's the lane number that the protagonists used in the Big Lebowski.

22 is relatively boring - it's just twice a prime. The best way I've got to remember it is Catch 22.

Does that help?

Down with 22, up with 23.

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2. ### glen_h

I do like 22 better than 23. Once you get used to it, it is hard to change.

Though, not so obvious, as well as I know two stops down from f/22 is f/45. Why not f/44?

It may have been somewhat later that f/45 lenses came out. I think the ones I know that
have it are Nikon lenses designed to use with extension tubes. Maybe MF and LF also have f/45.

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3. ### paul_noble

Technically, f/22 is really f/22.623 (sqrt(2)**9). Two stops is sqrt(2)**11, which is 45.2548.

4. ### BeBu Lamar

We know but we are discussing if you round f/22.623 to 2 significant digit should it be 22 or 23?

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5. ### Stock-Photos

I could be wrong but I thought an F stop number indicates the ratio of light going into the front of the lens versus how much comes out the back. A 1.0 lens would be zero light loss. At F8, the light intensity is 8 time less intense as it exits as it was entering.

6. ### Andrew Garrard

Sorry, Stock-Photos, that's not how it works.

The f-stop describes the ratio of the entrance aperture (the area at the front of the lens which gathers light) to the focal length of the lens. I think of the effect on illuminating the image as standing at the bottom of a well - you get the same amount of light reaching you if the well is twice as deep but also twice as wide (the hole you can see at the top of the well looks the same size).

T-stop is similar, but allows for the amount of light loss passing through the lens due to coatings and internal reflections - so a lens's f-stop of 2.8 might be a T-stop of 3, for example.

Actual results aren't quite that simple - all lenses have a bit of illumination fall-off towards the edge of the image circle, for example.

What you're describing seems to be the ratio between the f-stop and T-stop, assuming your definition of "light going into the front of the lens" already allows for the f-stop aperture.

I hope that helps.

7. ### glen_h

I think it isn't quite right, but does have some ideas that are often not considered.

First, I think there should be a cos(theta) for the angle that the light hits the film at,
which will be more and more important at larger apertures.

But also, how much light hits the film at a give f/ value compared to how much with no lens at all.

Consider a hemispherical uniform light source, which through a lens should uniformly illuminate
the film. Next illuminate film with a hemispherical source and no lens. At what f/ value are
they equal?

At small apertures, the solid angle of light collected is about pi(d/2)^2/f^2

The largest solid angle that you can collect is 2pi, but pi(d/2)^2/f^2 can, for large enough d,
be greater than 2pi.

8. ### Norman 202i am the light

it seems to me the relationship between the input and output of a lens is simple, i.e. the amount of light exiting a lens at a given aperture is the same no matter what the focal length. is that correct?

9. ### Andrew Garrard

Ignoring issues such as the sensor having reduced sensitivity to light arriving at an angle, there's no f-stop that corresponds to capturing all the light for a hemisphere. Or rather, there is, but it's f/0 with a focal length of 0, so the maths says 0/0 - something described in computers as "not a number". Of course you can capture more than 360 degrees of incoming light with a fish-eye lens, but that leads to a much more complicated (and non-standard) definition of "focal length" compared with standard rectilinear projections.

If we take field of view out of the equation, we could look at a point light source, and ask how we can capture a hemisphere of the light being emitted by it and funnel the result to a point on the focal plane. That concept retains a projection that could be considered to have a focal length, but it does lead to the front of the "lens" being bowl-shaped, with the light point at the centre. (If you wanted to capture all of the light even emitted by the back of the source, you could put an elliptical mirror around the subject and focal plane, put the light at one elliptical focus, and the focal plane at the other - but it would only work perfectly for one point.) I'm trying to imagine the optics to get this effect (for half the light) through refraction without wrapping mirrors around the back of the plane, but that's complicated geometry for what, for me, is early in the morning.

10. ### Andrew Garrard

To a good approximation, that's correct - which is why we use f-stops in photography. In reality, a combination of light fall-off across the frame (which tends to be affected by aperture but also the rear nodal point) and optical inefficiency in the lens means it's probably only an approximation. But it's a good enough one to be getting on with.

11. ### Norman 202i am the light

Thanks Andrew, Briefly, as I know itโs early, is the relationship between input and output

linear
nonlinear (hard)
nonlinear (easy via taylor series expansion)