Which formula for f-numbers is correct: f/D or 1/(2NA)?

Question

I've seen two different definitions for the f-number of a general-purpose lens. Are they consistent ?

I'm assuming that subjects are distant (not macro, magnification is small), and we have a lens corrected for coma and spherical aberration.

I mean f-number in the sense that it indicates the brightness of a lens which has no internal losses.

The 1/(2 * Numerical_Aperture) definition leads to f/0.5 as a limit; the f/D definition does not.

I was looking at: Image Sensors and Signal Processing for Digital Still Cameras by Junichi Nakamura

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Addendum - Attempt to clarify:

My background: I have maths degrees, with an applied/physics bias, so I understand trig identities & approximations, though my work doesn't use so much of my maths education.

I haven't formally studied optics since high school thin spherical lens stuff.

As a photographer, I understand the everyday use of f-numbers when photographing non-macro subjects, and that T-numbers are sometimes more relevant. I'm aware of changes in effective f-number in macro cases, but I don't really do macro.

Confusion, and question:

The question concerns photographic lenses at least somewhat corrected for coma and spherical aberration, focussed near infinity, with negligible magnification, negligible internal losses, in a medium of refractive index close to 1, at points on the sensor close to the axis of the lens.

The most common formula given for f-number is: N = f/D

Formulae for f-number involving Numerical Aperture ("NA"), combined with formulae for Numerical aperture sometimes appear to give results for f-number ("N") which differ from N = f/D when f/D is small (say f-number < 2).

How should these conflicting results be reconciled ?

The NA approach makes it clear that there's a lower bound on f-number, at 0.5, because the cone angle of the light hitting the centre of the sensor cannot exceed 180 degrees. That lower bound is not immediately clear from the N = f/D formula.

My confusion is for small-ish f-numbers above this f/0.5 limit.

As I said, I don't know much optics. I wonder if the inconsistencies are related to the assumed shape of the "Second principal plane" of the lens.

If the half-cone-angle is 𝜃', I seem to get different values for 𝜃', depending on the assumed shape of the Second Principal Plane:

• If the second principal plane is assumed flat, I get tan 𝜃' = D/2f
• If the second principal plane is assumed spherical, with radius f, I get sin 𝜃' = D/2f

Perhaps, as hinted at in the comments, neither shape is a very accurate representation of a real lens, and an accurate answer can only be predicted by ray-tracing.

Is any case, is sin 𝜃' = D/2f likely to be a better approximation than tan 𝜃' = D/2f for a general-purpose photographic lens ?

[For slow lenses, 𝜃' ~= sin 𝜃' ~= tan 𝜃' ~= D/2f, where 𝜃' is in radians]

I don't really understand this, but I read that a (near) spherical second principal plane is desirable to correct spherical aberrations.

If NAi = n sin 𝜃', and f-number = 1/(2*NAi):

• If sin 𝜃' = D/2f, we get f-number = (1/n)(f/D), even for fast lenses
• If tan 𝜃' = D/2f, we get f-number = (1/n)(f/D)sqrt(1+(D/2f)^2)

Answer 1

Your formula f-number = 1/(2*NA*), where NA is the numerical aperture is not an accurate reflection of the formula presented in the book you reference: f-number = 1/(2sinΘ'), unless one assumes the index of refraction is equal to exactly 1. The refractive index of a vacuum is 1. The refractive index of air at standard temperature and pressure is 1.000277. Although a thin lens will satisfy the n=1.000277 requirement, no compound lens is perfectly corrected for aberrations such as coma and spherical aberration. Thus f-number ≈ 1/(2*NA*) is the actual formula.

f-number = f/D where f is the focal length and D is the diameter of the entrance pupil is equivalent to f-number ≈ 1/(2*NA*) within the limitations of the maximum angle at which light entering the lens is allowed to pass through the lens. I think where some get off track is that they assume if the front of the objective is enlarged that the entrance pupil will necessarily also enlarge to the increased size of the objective.

Answer 2

It's not clear exactly what you are asking, but the brightness of the image projected by a lens relative to the scene is a function of the focal length divided by the aperture diameter. This formula essentially normalizes away differences in focal length between different lenses.

For example, ignoring light absorption by the lens material and assuming small (<< 1) magnifications, a 100 mm lens with 25 mm diameter aperture will make the same brightness projection as a 200 mm lens with 50 mm aperture. In the second case, the projected image will be twice the size in linear dimension, therefore taking 4x the area. However, the 50 mm aperture has 4x the area of the 25 mm aperture, so lets in the right amount of extra light to compensate for it being spread out over the larger area.

Since saying all the above is cumbersome, we use a short notation in photography. That notation is the "f-number". The name comes from the expression f/xx used to express apertures, like f/2.0, f/2.8, f/4.0, f/5.6, f/8.0, etc. In these expressions, "f" refers to the focal length of the lens, and the overall expression indicates the aperture diameter. For a 120 mm lens, f/4 literally means the aperture is (120 mm)/4 = 30 mm in diameter.

In photography, it is convenient, and has become customary, to think of light level changes in factors of 2. The light allowed thru a lens is proportional to its aperture area. Each doubling of the area means the diameter is increased by the square root of 2. This is where the common f-numbers come from. We start with f/1 and increase the denominator by sqrt(2) = 1.414 for each 2x less light. The common sequence is therefore f/1, f/1.4, f/2, f/2.8, f/4, f/5.6, f/8, f/11, f/16, etc.

Answer 3

The camera lens acts just like a projector lens in that it projects an image of the outside world onto the surface of film or digital imaging chip. The focal length of the lens reveals its power to magnify. We are talking about the size of images of objects. A 100mm lens projects images that are twice as large as a 50mm.

Magnifications come at a price. Each doubling of the focal length forces the image forming rays to play over four times more surface area. Stated differently, double the focal length, and image brightness reduces fourfold. Conversely, if the focal length is halved, image brightness quadruples. This image brightness change occurs as described provided the working diameter of the aperture remains unchanged.

The image brightness change with a focal length change is a big deal. It makes exposure determination challenging. This is especially true, because there is a hodgepodge of different camera lenses -- all with different focal lengths and working aperture diameters. How can we cut thru the confusion?

Ratio to the rescue: A ratio is a dimensionless value. We fall back on the focal ratio of our lenses to take the chaos away. Image brightness is intertwined with focal length and the working diameter of the aperture. We divide the focal length by the working aperture diameter and calculate the “focal ratio” (f/number for short). Thus a 100mm lens with a working aperture of 25mm, functions with a focal ratio 100 ÷ 25 = 4 (written as f/4). The beauty of this method is: any lens functioning at f/4, passes the same light. This is true even if it is a giant telescope 100 meters in focal length with a working diameter of 25 meters, it projects the same image brightness as a 100mm lens with a 25mm working aperture diameter.

The focal ratio or f/number set of numbers is based on the geometry of circles. If you multiply the diameter of any circle by the square root of 2 = 1.1416 (OK to round to 1.4), you have calculated a revised circle with twice the surface area. Thus if a lens has an aperture diameter of 25mm, and you desire this lens to pass 2X more light, then 25 X 1.4 = 35. In other words a working aperture of diameter of 35mm passes 2X more light than a 25mm working aperture diameter. The result is 2X more in image brilliance.

Using the 1.4 factor, a number set emerges. 1 – 1.4 – 2 – 2.8 – 4 – 5.6 – 8 – 11 – 16 – 22 -32 Note each number going right is its neighbor on the left multiplied by 1.4. Each number going left is its neighbor on the right divided by 1.4. This is the basis of the venerable f/number set. This ratio helps us control image brightness, and the increment is a doubling or halving of light transversing the lens.

Now changing the subject (maybe): In my classroom of yesteryear, I often told this story. You are the captain of Cavalry “A” Troop. One hundred men with horses marching through the American South Western Desert on patrol. Water is a problem but you expect rain. You order the troops to bivouac for the night. You order the men to dig a circular pit 8 feet in diameter and line it with their canvas tent fabric. It rains as expected and the pit begins to collect rainwater. Due to your West Point training, you know an 8 foot diameter pit is adequate to collect rain water for your needs. Unexpectedly a lookout spots “B” Troop approaching -- another 100 men with horses. You order your men to expand the diameter of the circular pit to accumulate 200 men and horses.

How big must the revised pit be to double the amount of collected rain water?

Answer: You multiply the pit diameter (8 feet) by 1.4142. This value is the square root of 2. The answer is 11.3 (rounded it’s 11 feet). You order the pit expanded to 11 feet diameter. Surprise, this new value causes the pit to accumulate twice as much water as before. Why? The surface area (catch basin) now has double the surface area; thus it can capture twice the amount of rain.