F-Stops

[david.wagle]

Ok . . .I understand what f-stop represents.

 

I get that f/2.8 lets in twice as much light as f/5.6 and four times as much as f/8.

 

I understand that the apperature effects the depth of field, and that a wider

apperature (which is a smaller number) gives a shorter depth of field.

 

I also get the relationship between f-stop and exposure time -- that you can

halve the light and double the exposure time and still have a properly exposed

picture.

 

I even understand the effect of distance from the lense on depth of field at a

given apperature.

 

 

What I don't understand is what do the NUMBERS mean?

 

2.8 what?

[scottcondit]

You've missed out - f/4 lets in twice as much as f/5.6 and four times as much as f/8.

 

You've written part of the answer yourself - "f over 2.8". It's a ratio, the ratio of effective focal

length over the diameter of the aperture. So, talking about one specific focal length: since the

amount of light coming through the lens will be proportional to the area of the aperture, it

will be proportional to the reciprocal of the f-stop squared. So this area is half the size at f/

5.6 that it will be at f/4, and this is the reason the f-stops are (approximately) a geometric

progression in the square root of 2.

[david.wagle]

Ok, that makes sense.

 

And also explains why larger focal length lenses don't have the wider f-stops available to them . . .

 

What are the implications of this when using a manual settings on a digital camera with a CF? Or do the camera electronics take care of the CF and I don't have to worry about it? What about if using a hand-held meter?

[jimstrutz]

CF = Compact Flash<br>

CF = Custom Function<br>

CF = Close Focus<br>

CF = Cool Factor<br>

CF = Canned Food<P>

 

Which one do you mean? And why would this change on a digital camera?

[albert_richardson1]

What a great question! The f number is the ratio of the diameter of the aperture to the focal length of the lens. (Focal length / diameter of aperture) It is an early and lasting solution to the problem of marking all lenses in such a way as to allow a photographer to adjust any of them properly using the same unit of measure.

 

You can devise a simple series of calculations in Excel to prove to yourself that the system works effectively for all focal lengths. You will find that there are two interlaced sequences of numbers that represent all f stops. The seed for the first one represents a wide open lens like a magnifying glass with an aperture the same diameter as the focal length. This is f1. The second seed lets in 1/2 as much light as the first one, and is f1.4. All other f numbers turn out to be multiples of these two seeds as follows: f1, f1.4, f2, f2.8, f4, f5.6, f8, f11, f16, f22 ... and so on. Here is a sample of my spreadsheet:<div></div>

[ronaldo_r]

<a href="http://en.wikipedia.org/wiki/F_stop"> en.wikipedia.org/wiki/F_stop <a/>

[richard_cook2]

The area of a circular aperture equals PI * radius squared.

 

To double the area you multiply the radius by the square root of two.

 

Hence the ratio is

 

1.0 equals one

 

1.4 equals one * square root of two

 

2.0 equals one * square root of two * square root of two

 

2.8 equals one * square root of two * square root of two

 

4.0 equals one * square root of two * square root of two * square root of two

 

5.6 equals one * square root of two * square root of two * square root of two * square root of two

 

and so on.

[mikeseb]

Ansel Adams, <i>The Camera</i>.

[david.wagle]

<blockquote>

Jim Strutz - Anchorage, AKphoto.net patron prolific poster, Apr 01, 2007; 01:06 a.m.

 

CF = Compact Flash

CF = Custom Function

CF = Close Focus

CF = Cool Factor

CF = Canned Food

 

Which one do you mean? And why would this change on a digital camera?

</blockquote>

 

I mean "conversion factor." My digital camera's effective lens focal length is 1.6x greater than whatever the lens says because the CMOS is smaller than 35mm film.

 

I'm thinking this doesn't make a difference since what is at issue is the actual focal length, not what %-age of the surface is film and what %-age isn't film.

[jimstrutz]

Ahh, now I get it. And yes, it doesn't make a difference.

[alex_lofquist]

Literally, "f-stop" comes from f (which is focal length, not factor, as defined in the thin lens equation: 1/f=1/s1+1/s2). Therefore the relative aperture is shown as f/# where #=diameter of the entrance pupil. This is the circle that you see when looking at the front of a backlit lens. Since this circle gets smaller as you "stop down", the "f#" becomes greater.

[j_smith6]

square root of 2 = 1,41421

<br>

square root of 4 = 2

<br>

square root of 8 = 2,8284 --> 2x more light than with a root of 4 (f/2.0) and 4x more light than with root of 2 (f/1,41)

<br><br>

square root of 8 = 2,8284

<br>

square root of 16 = 4

<br>

square root of 32 = 5,656

<br><br>

32:8=4

<br><br>

difference between f/2.8 and f/5.6 --> 4x

[j_smith6]

It's like recounting the guide number of your flash for other ISOs.

[j_smith6]

http://www.uscoles.com/fstop.htm

[graybrick]

Damn, where's Bob Atkins when you need him? Look at his website, there you'll likely find more answers than you were looking for.

 

http://www.bobatkins.com/photography/technical/index.html

[matthew_newton]

As someone mentioned it doesn't make a difference. If you use a 100mm lens that has a maximum aperature of 50mm you have an f/2 lens. On a cropped frame sensor, like say an APS-c sized sensor from a 30D you have a conversion factor of 1.6x. This means that that 100mm lens an effective focal length of 160mm, but this doesn't influence the aperature at all since the actual focal length is 100mm.

 

It would be just like looking at medium format lens. A medium format lens that is 90mm and an aperature of 22.5mm (making it an F/4 lens) might have a conversion factor of say x.55 compared to 35mm/full frame format...so it would behave like a 50mm lens on the medium format camera, but it would still have an aperature to focal length ratio of f/4