I.e. a 25mm aperture diameter (entrance pupil) has an area of 490mm; and 1.4 x 25mm = 35mm with an area of 962mm... approximately double the area/light/exposure.

Also, having unbalanced ratios of distances would pin the number to choice of units. Any values of such an area-based f-number would be explicitly dependent upon choice of units used for focal length. So aperture settings on lenses with fractional-inch based focal lengths would have completely different values than for millimeter-value focal length lenses (and also for centimeter-valued focal length lenses).

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It is in relation to FL because a longer FL has a narrower field of view (FOV); it collects less light and spreads it over the same area (image circle). A longer FL must have a larger aperture area in order to have the same f# and transmit the same amount of light, achieving the same exposure.

For a given intensity \$I\$ on the subject, we set our camera's exposure settings to correctly expose the subject. Because we're talking about flash photography, let's assume ISO and shutter speed aren't really free variables available to us for exposure control (e.g., let's leave ISO fixed at 100, and shutter speed at, say, 1/200). That leaves aperture available for adjustment for correct exposure of the object.

If the distance were changed by a factor of \$k\$, then the light intensity falls by \$k^2\$. In order to keep the photometric exposure the same, we need to compensate by increasing the aperture area by \$k^2\$, or the aperture diameter by a factor of \$k\$. Thus, for constant exposure, the ratio of flash-subject distance to aperture diameter needs to remain constant.

Interestingly, the guide number concept is derived from an area relation (which at first glance would seem to support the premise of your question, but as we'll see, there's no need to use square factors). The amount of light incident on an object is inversely proportional to the square of the distance between the light source and the object (the inverse-square law): \$I \propto 1/s^2\$.

Classic ƒ numbers are entrance pupil diameter as a fraction of focal length. This seems like a slightly strange choice as exposure is proportional to area rather than diameter. Naively I would think it easier work with these as exposure stops from reference ƒ/1:

The f-number system is unique in that it is universal. In other words, any lens, on any camera, regardless of focal length or image size, when set to the same f-number, will return an identical exposure. Well, not exactly, but close enough for most every need. In cinematography, a T-stop is favored. This is an f-top that has been calibrated to take into account light loss induced by the color of the glass, influence of lens coating and inaccuracies of the aperture diameter etc. The T-stop is deemed necessary in this usage because it gives improved uniformity, scene-to-scene and lens change-to- lens change. Pictorial still photography is contented with the f-stop.

The effective pupil diameter is important as a measure of unsharpness: the entrance pupil forms the base of "unsharpness cones" that have their respective tip (indicating full sharpness) in the focus plane and widen again from there. Double the f number, and you halve the diameter of any bokeh circle visible in the image.

Math. It's because in many equations regarding simple optics, the ratio \$N = f/D\$ (where \$N\$ is the f-number, and $D$ is the lens (or more often precisely, entrance pupil) diameter) pops up a lot, or the use of the ratio simplifies the expression or understanding of the expression.

The significance then of the f/stop system is that the exposure of any one stop number, like f/4, is still f/4 on any lens of any size. Two photographers with different cameras standing side by side could use the same aperture number then. In practical terms, it made the later concept of light meters possible (for any camera lens). :) The f/stop number provided significance to exposure in any camera, more about the exposure than just about the lens.

The formula for hyperfocal distance is merely a special case of the computation of far depth-of-field when the far focus distance is infinity. The geometry that describes the depth of field equations is completely described by similar right-triangles in the cross-sectional plane through the optical axis of the lens, and the thin-lens equation relating the focal length (strength) of the lens and its object-side and image-side focus distances.

A similar aperture numbering system called the U.S. system (Uniform System) was used by the first Kodak cameras (until around 1920s). That system originated in England (1880s). Not 1, 2, 3, 4, but those stops were numbered 1, 2, 4, 8, etc, starting from todays f/4 equivalence. It was more useful than 1, 2, 3, 4 because it represented exposure increase inversely (doubling U.S. number is one stop less, doubling f/stop Number is two stops less). And exposure was considered important to photographers.

The guide number encapsulates this dependency. Because f-number \$N\$ is inversely proportional to aperture diameter, the constant exposure relation is now a product rather than a ratio: \$N\cdot s\$. And importantly, the dependence on squares of distances is not necessary. We can just use linear flash-to-subject distance and linear aperture diameter.

Aperture f-numbers are near approximations. And the 1.4x increment is approximately equal to doubling of the aperture area.

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The f-stop is actually a ratio. This is important because a ratio is dimensionless (look up ratio if you doubt). Actually f-stop is the accepted jargon for focal ratio. This value is derived by dividing the focal length of the lens by the working diameter of the entrance pupil (aperture). Thus a 100mm with a working diameter of 12.5mm = focal ratio 100 ÷ 12.5 = f/8 (written with a slash). By the way, an 8000mm lens with a working diameter of 1000mm is also an f/8. Both produce the same exposing energy if set to f/8 and pointed at the same vista.

Attempting to clarify my question, I am not confused by the 2x/2 series or doubting the benefit making it relative to focal length. My question is only in regard to the names we apply to this series. In my example scheme it doesn't matter if we call it "ƒ/1.4" or "Av1"; the use is interchangeable. So I wonder what convenience is imparted by using the fractional diameters at all?

But the f/stop system rapidly gained favor starting just before 1900, because it also factored in the lens focal length (f-number = focal length / working diameter). The working diameter is as seen as magnified through the front lens element (entrance pupil). It is named f/stop in reference to the division of focal length f by diameter.

Why the odd crazy number set? 1 – 1.4 – 2 – 2.8 – 4 – 5.6 – 8 – 11 – 16 – 22 – 32 – 45 – 64 Each number going right is its neighbor on the left multiplied by 1.4 (square root of 2). Each number going left is its neighbor on the right divided by 1.4. Why? If suppose you have a circular lens with a diameter of 2 inches. This is its working diameter. Its area, the dimension that captures light is 3.14 square inches. Now suppose you wish to double its light gathering power. To do so, you must increases area 2X. What will be the revised diameter? Answer 2 multiplied by 1.4 = 2.8 inches.

The key to understating the f-stop system, which the industry accepts as the fundamental increment of exposure, is a 2X increment. This is a doubling of halving of the light energy of exposure. In modern times, it is sometimes necessary to make finer adjustments. When needed, we can refine the f-stop, and make 1/2 or 1/3 or even 1/6 increments. Let me add, except in a laboratory situation, it is impossible to control a photo process and keep it at 1/3 f-stop tolerances.

Perhaps a better analogy for argument is the debate over what is the better circle constant, \$\tau\approx 6.28\$ vs. \$\pi\approx 3.14\$ (Tau Manifesto). The debate is really a non-debate; as long as the correct factor of 2 is used in the right places, it doesn't matter. One notation might lead to a better understanding of the geometry or physics being described by the equations, but in the end, the math doesn't change. Just the notation and more or fewer factors of 2. Just like aperture diameter vs. area.

Now while the presence of \$f^2\$ in the first hyperfocal distance equation (that includes \$N\$ in the denominator) might appear to be a result of some dependence on area, it's really just an artificial creation because of the simple algebraic substitution \$N = f/D\$. In other words, as long as the aperture diameter \$D\$ is much larger than the circle of confusion diameter \$c\$, the hyperfocal distance is linearly proportional to both \$f\$ and \$D\$, and inversely proportional to \$c\$. The equation has nothing to do with the area of the aperture that would be generated by rotating the cross-section of the thin lens of diameter \$D\$ through \$\pi\$ radians.

Normalization with respect to fundamental "figures of merit" happens all the time. The first thing to pop to my mind is in relativistic physics. We talk all the time about velocities as some fraction of the speed of light, \$c\$, which is approximately 3 x 108 m/s, or about 186,282 mi/s. We don't talk in absolute values of meters per second or miles per second. But in terms of fractions of \$c\$, it's much more useful.

If the lens is 50mm in focal length (FL), the 25mm aperture would be f/2, and the 35mm aperture would be f/1.4 (actually 35.7mm).

(if the source is not larger than the FOV's like a wall, but it is a point source instead like a street light; then the increase/decrease in size and light follows the inverse square law)

Example 1: The hyperfocal distance \$H\$ is the focal distance that theoretically maximizes total depth of field. For a lens of focal length \$f\$ set to a f-number \$N\$, then given a circle of confusion limit $c$, the hyperfocal distance is defined as

The effect on exposure is something easily folded into the metering and also accesible via counting stops, but assessing the effect on image geometry depends on proportions and geometry and if you want to do rough estimates, not having to calculate square roots helps.

It might have been more important when large and medium format cameras were more prevalent than they are now and depth of focus considerations were much more dominating imaging decisions than they are now.

I.e. a FL 2x longer has a FOV 1/2 the size, gathering 1/2 the light (infinite source), and transmitting it through an aperture 2x the size. Which results in 2x the light gathered being recorded at the image plane for the same light density/exposure (2X.5=1). I.e. 100mm vs 50mm at same f#.

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Regarding units and dimension: Note that \$N\$ is a unitless quantity, defined as the ratio of two distance-measures (i.e., millimeters divided by millimeters) that are implicitly understood to be arranged at right angles to each other. If \$N\$ were instead a ratio of focal length to entrance pupil area, the units of \$N\$ would be in [length-1], such as "per meter" or "per millimeter". Net exponents of distance in the denominator is a particularly unwieldy thing for humans to think about and get their head around, in physical models.

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$$\begin{align} H &= \frac{f^2}{Nc} + f \\ &= f\left({f\over Nc} + 1\right) \\ &= f\left({D\over c} + 1\right) \\ &\approx {fD\over c} \qquad(\text{because }D \gg c) \end{align}$$

The uniqueness of the focal ratio is that it intertwines two optical factors. The longer the focal length, the more light that is lost. Double the focal length and the light loss is 4X. The other factor is the diameter of the entrance pupil doubles its diameter, and the lens will gather 4X more light. The f-number system balances both phenomena.

I found the Wikipedia page on the APEX system and see—unsurprisingly—that the naming shown above was proposed at least by 1960, as aperture value. What that page doesn't seem to provide is a robust explanation of why the proposal never took root.