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Old Mar 16, 2011, 7:11 PM   #1
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Default What does f number stand for?

I understand the concept of aperture, but what does the number actually stand for? I also know it's a fraction (ie f/2.8 = 1/2.8th of whatever) But what is it actually a fraction of?
And how can you have a f/9.5 lens if that means 1.95th?

Thanks!
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Old Mar 16, 2011, 8:14 PM   #2
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"f" is for "Focal Length". The F-Number is the ratio of the focal length to the diameter of the aperture, as in, 100mm / 25mm, or f/4.0. When you put the focal length in place of the 'f', as in, 100mm/4.0, you get the diameter of the aperture, or 25mm.
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Old Mar 16, 2011, 8:31 PM   #3
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The F-Number is a Dimensionless Number. A Dimensionless Number is a number that doesn’t have a physical dimension to it. The numbers used in the calculation usually have a dimension, such as millimeters or kilograms. The F-Number and also the Reynolds number are two examples of dimensionless numbers.

The method used to display the F-Number is as a division that signifies the focal length divided by the F-Number. If you have a 50mm lens set to an F-Number of 2, then the aperture is 25mm wide. If “f” represents the focal length of the lens (50mm in this case) then “f/2” is the same as “25mm”. But “f/2” is meaningful because of the F-Number...”25mm” is not as meaningful.

You can have an f/0.95 lens because there’s no rule that says the diameter of your lens can’t be larger than the focal length of said lens. So when your lens diameter is larger than your focal length, you can have F-Numbers that are less than 1. However, such lenses are an extreme optical challenge, not only to produce but also to utilize well on digital cameras...which is probably why you don’t see any new f/0.95 lenses being produced.
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Old Mar 16, 2011, 8:35 PM   #4
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You are correct that it is a ratio. The problem is that the standard designation may be a little awkward. It is the ratio of the lens focal length divided by the diameter of the aperture. If we have a 50mm lens with a maximum aperture of 25mm the f no. would be 2. A 10mm lens with 5mm maximum aperture also has a f no. of 2 not to mention that a 4" lens with an aperture 2" will be f/2. It is a dimensionless number. In the normal way f #s are written if we substitute the actual focal length for f as in 50mm/2 it will tell us that the actual aperture diameter is 25mm.

Hope this helps.

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Old Mar 16, 2011, 8:48 PM   #5
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Quote:
Originally Posted by Graystar View Post
However, such lenses are an extreme optical challenge, not only to produce but also to utilize well on digital cameras...which is probably why you don’t see any new f/0.95 lenses being produced.
Cosina under the Voigtlander name has just introduced a 25mm f/0.95 in a micro 4:3 mount. It's manual focus but initial reports are that it is absolutly outstanding optically.

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Old Mar 16, 2011, 9:35 PM   #6
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F 2.0 has a shallow depth of field, allows more light in
F 22. has a greater depth of field, allows less light in,
Depending on the surrounding environment/light, etc. this effects how fast the shutter can be.....that is the more important...not the above answers
One could say, 'the dimension of the f no. is in direct opposition of the fraction of the reverse of the value of the F no.' and it wouldn't mean a rats arse...more value is placed on the above as I have written it...IMHO
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Old Mar 16, 2011, 9:38 PM   #7
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Thanks everyone (especially tcav)! T
hat's what I was looking for!
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Old Mar 17, 2011, 4:23 AM   #8
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It maybe important to remember that the F-Number represents the effective diameter of the aperture, and may not be the actual diameter of the diaphram in the lens. In a simple lens design, with a single element, the maximum aperture is the based on the diameter of that element. The adjustable diaphragm is behind that element, where the light path has begun to constrict, so the actual diameter of the diaphram is smaller than the effective diameter. In more complex lens designs, the diaphragm is typically in the middle of multiple elements, where the light path is most constricted, so the actual diamenter of the aperture will be very much smaller than the effective diameter.
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Old Mar 17, 2011, 8:06 AM   #9
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Isn't the aperture of a 500mm f/4 lens 10 times as large across as a 50mm f/4? I.e., 500mm/4 = 125mm 50mm/4 = 12.5mm.
Your last statement is the diameter of the aperture in relation to the diameter of the lens, not the focal length, correct?
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Old Mar 17, 2011, 8:29 AM   #10
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Quote:
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Isn't the aperture of a 500mm f/4 lens 10 times as large across as a 50mm f/4? I.e., 500mm/4 = 125mm 50mm/4 = 12.5mm.
Your last statement is the diameter of the aperture in relation to the diameter of the lens, not the focal length, correct?
Your math is correct but it has no relation to how the maximum aperture is determined. The diameter of the lens represents the maximum aperture possible, as the lens itself is the biggest possible hole for the light that passes through it. So if you have a lens with a 50mm focal length but the lens itself is only 25mm in diameter, then f/2 is the largest aperture for that lens. Likewise, a 50mm lens that can be set to f/0.95 must be at least 52.63mm in diameter. This is why very fast lenses are big, heavy, and expensive. Lots of glass means lots of pounds...and it is costly to grind big glass accurately.

Personally I always try to avoid talking about the physical diameter of the diaphragm. The model of focal length and aperture that is used most often is very simplistic. The optics of real lenses are far more complex, and I have found no benefit to my photography in knowing them.
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