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I never understood what the f really stands for in the f-stop values, like f/1.8. Wikipedia explains it in various articles, but I still find it very confusing. What I understand, though, is that it has something to do with focal length.

Should I assume, therefore, that bigger focal lengths capture more light? For example, should 85 mm f/1.8 lens capture a lot more light than for example 24 mm f/1.8?

References to back up my confusion:

The article about aperture goes on to claim:

The amount of light captured by a lens is proportional to the area of the aperture, equal to:

enter image description here

Where f is focal length and N is the f-number.

The "f-number" is mentioned here. But the article about F-number claims:

In optics, the f-number of an optical system expresses the diameter of the entrance pupil in terms of the focal length of the lens; in simpler terms, the f-number is the focal length divided by the "effective" aperture diameter.

This seems very recursive. Why does the aperture article refers to both the focal length and the f-number, when the latter article claims that f-number already carries the property of focal length?

What's going on here?

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5 Answers 5

up vote 12 down vote accepted

The f-number is in use to express how much light a lens can capture, so the 85mm f/1.8 and 24mm f/1.8 can capture the same amount. Here, f is the focal length, and f/1.8 means that maximum aperture diameter is 47.2mm in first example and 13.3mm in second.

What you have to consider here is that the 85mm lens has a much narrower field of view, therefore it has to gather the same amount of light from a much smaller area - to compensate the narrow view, aperture has to be bigger.

How much the aperture has to be bigger is linearly correlated to focal length. We could say that a 24mm lens with 13.3mm aperture can gather as much light as a 85mm lens with 47.2mm aperture, but talking about the F-number makes this much easier to notice.

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Does this mean that a constant-aperture F4 zoom actually has an increasing absolute aperture (in square mm) as you zoom in? And that a normal kit lens, say f/3.5-5.6 has the same absolute aperture (in square mm) throughout the zoom range, and the reduction in the aperture as measured as f/something is only due to the increase in f as you zoom in? –  Kartick Vaddadi Jan 5 '14 at 16:02
    
Even the kit lens has an increasing aperture (especially if you consider kit lenses with longer zoom ranges, e.g 18-135), it just increases somewhat slower than focal length. It should be mentioned though that it's the effective aperture discussed here; physically, the blades are placed on the optical path somewhere where the light rays are condensed (otherwise there would be no room for aperture blades when they are fully open). In variable aperture zooms, the width of rays at aperture depends on zoom; in fixed aperture lenses, it doesn't. –  Imre Jan 5 '14 at 21:26

The terminology is confusing, isn't it?

The f-number of an f/1.8 lens is simply 1.8. That is given by N, not f. f is the focal length.

f/1.8 literally means "focal length divided by N". So if you refer to f/1.8, that's not the f-number, that's focal length divided by f-number.

To explain the equations further:

The second equation says that the focal length f divided by the aperture diameter d is the f-number N. Or f/N = d. So f/1.8 literally means "focal length divided by 1.8 (equals diameter).

If you look at the top equation since f/N = d, then f/2N must be d/2 (and d/2 is half the diameter, so is the radius of the aperture). And we know that the area of a circle is pi r squared. So pi times f/2N squared is simply the area of the aperture opening.

So one equation tells you the diameter of the aperture opening, and the other uses basic geometry to calculate the area of the aperture opening, which is a circle.

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I think the confusion comes from two different things the explanations erroneously lump together as "amount of light".

The true amount of light a lens lets in is only a function of the aperture area. Since area goes with diameter squared, this is proportional to the square of the diameter.

However, what is more relevant to exposure is not the total amount of light a lens can gather, but the brightness of the focused image it produces. This is where focal length gets involved. Let's say you have a 100 mm lens with a 25 mm diameter aperture (or a adjustable aperture set to 25 mm). Now compare that to a 200 mm lens. If the 200 mm lens also has a 25 mm aperture, then it will let in the same amount of light. However, that same amount of light coming from the subject is now focused twice as large, thereby taking 4 times the area. That means the 200 mm lens with 25 mm aperture creates a image 1/4 as bright (2 f-stops down) compared to the 100 mm lens with the same 25 mm aperture.

Notice that the brightness of the focused image goes down with the square of the focal length, but goes up with the square of the aperture diameter. That means if we were to take the ratio of the two, we'd get a normalized measure of how bright the focused image will be for exposure purposes. That ratio is exactly what f-stops are. These are commonly written as f/n, like f/8.0 or f/11 for example. That is just a expression. The full equation is:

aperture = focallength / n

In the first example of 100 mm lens with 25 mm aperture, that is:

25mm = 100mm / 4

Since that gets cumbersome to write and say all the time and the point is to not have to care what the absolute focal length and aperture are, this is abbreviated to "f/4", with "f" referring to the focal length of the lens and "4" being the ratio of that focal length to the aperture diameter. The second example was:

25mm = 200mm / 8

or "f/8". Other than minor light losses and other subtle effects you can ignore most of the time, one lens with its aperture set to f/8 is going to make the same brightness focused image as another lens at f/8 regardless of focal length. This also explains why long lenses tend to be larger in diameter. A 50 mm lens only needs a 12.5 mm aperture to achieve f/4. A 300 mm lens on the other hand requires a 75 mm diameter aperture to make the same brightness image of the same subject. That means basic physics says a 300 mm lens needs to be at least 3 inches in diameter somehow to achieve f/4.

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Any lens at f/2.8 should deliver the same amount of light to the camera. Certain lenses have to work harder to get there though.

It's not technically accurate, but I found the best way to get my head around the f-stop number is too think of it as representing the amount of light lost. So at f/2.8 you are only loosing 2.8 x the amount of light, while at f/11 you are loosing 11 x the amount of light. *

A tele lens, by it's nature is dealing with less light than a wide angle lens. So the harder the lens has to work to avoid loosing light the more glass you need to capture as much light as possible, so you can have a 55-250 f/4-5.6 that weighs about 1 pound and is 6-8 inches, while a 70-200mm f/2.8 weights 6 pounds and is over 12 inches long.

*This is not the way it works from a math point of view, but it may help you with a practical working understanding.

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5  
Not bad intuition. Some of it we can fix; some we have to change. You can put your response on a more rigorous footing--and explain otherwise paradoxical things like f/0.95 lenses--by recognizing that an f/1 lens is actually letting in only about (1-sqrt(3)/2)/2 = 0.067 of all the light. But the thing you must change to make this reply correct is to recognize that the amount of light admitted scales with the inverse square of the f/stop, not the f-stop itself; e.g., f/11 lets in 1/121 times as much light as f/1. This is an essential thing to know when choosing exposures in photography. –  whuber Mar 14 '12 at 19:00

What f stands for in the f-stop values

f stands for the focal length of the lens. An f/1.8 lens has the entrance pupil diameter of D = f/1.8. An 85 mm lens at f/1.8 aperture will have the entrance pupil diameter of 85/1.8=47.2 mm. A 24 mm lens will have the pupil diameter of 24/1.8=13.3 mm. Since the amount of light passing through the lens is proportional to the area of the entrance pupil and the latter is proportional to the square of its diameter, the 85 mm lens will apparently collect

(47.2/13.3)^2 = (85/24)^2 = 12.5

times more light. However, this consideration is only true for the amount of light, collected from each individual point of the object, not the total amount of light arriving from the object space.

Same f-number, same exposure (independent of f or D)

One thing that I also used to find confusing is that the amount of light collected at the sensor with the same shutter speed by different lenses with the same f-number is the same. How comes if one lens is clearly larger than the other?

Here is an illustration of what happens in the camera:

enter image description here

For simplicity, the object is assumed to be at infinity, so that all rays from the same object point are coming parallel to each other. The red solid rays enter the lens parallel to its axis, and are all focussed in the centre of the frame. The blue dashed rays are parallel to each other but not parallel to the axis. They all focus at the edge of the frame. Thus, the frame size together with the focal distance of the lens determine the field of view of the lens.

(Note that since I made the object distance infinite, the field of view in the object space is an angular one.)

If we change the lens to the one with a longer focal length while keeping the frame size the same, the field of view of the lens decreases:

enter image description here

Thus while the lens still collects the same amount of light from each point in the object space, the size of this space is smaller, so the total amount of light reaching the film or detector is reduced.

This reduction is proportional to the increase of the focal length, i.e. the amount of light with the same D is reduced by a factor of (f2/f1)^2. (It is squared because we need to take into account the reduction of the field of view in both directions.)

If we now increase D by f2/f1, we will again collect the old amount of light (since it's proportional to D^2). The f-number will become: D2/f2 = [D1*(f2/f1)] / f2 = D1/f1. Thus, if we want to collect the same amount of light while changing the focal length, we need to keep the f-number constant.

Frame size matters

The last parameter of interest is the frame size. Take a compact camera with the same f-number lens as a full-frame SLR. If the size of both the lens and the sensor are scaled down proportionally to the focal length, the two cameras will have the same field of view. The compact camera will collect less light than the SLR because its lens is smaller. However, it will still give the same exposure value on the sensor because exposure is the amount of light per unit area.

If the two cameras have the same resolution, the exposure will be the same but the actual amount of light on each pixel will be greater with the bigger SLR camera, resulting in lower noise.

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