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 Post subject: How much is 1 f-stop ?
PostPosted: Sun Aug 26, 2007 4:55 pm 
Sorry if this question is kind of stupid but how much (in f numbers) is a stop ?

For example if you got a 2.8 lens and you narrow the aperature 1 stop you get f4 (1.2 difference). If you stop down again you get 5.6 (1.6 difference).

So that's confusing me a bit :)

Please help me
THX

bart


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PostPosted: Sun Aug 26, 2007 5:23 pm 
Bart Michiels wrote:
For example if you got a 2.8 lens and you narrow the aperature 1 stop you get f4 (1.2 difference). If you stop down again you get 5.6 (1.6 difference).


The f-number is a notation used to know how much the diaphragm is open.

Big numbers mean the diaphragm is small (it lets little light in), while small f-numbers means the diaphragm is big (it lets much light in).

With each increase of the number, you are letting half the light in as before. So changing from f/4 to f/5,6 decreases the intake of light by half.

For some perfectly sound mathematical reason that I cannot explain (but I'm sure someone will step in and do so), f-numbers increase by powers of the square root of 2.

So, for the first step, start with the square root of 2:

step 1 = sqrt(2) ^ 1 = 1,4

Now, for the next steps, use sqrt(2) elevated to the step number:

step 2 = sqrt(2) ^ 2 = 2
step 3 = sqrt(2) ^ 3 = 2,8
step 4 = sqrt(2) ^ 4 = 4
step 5 = sqrt(2) ^ 5 = 5,6

Thus, step(n) = sqrt(2) ^ n

So that's why f-numbers have such numeric differences. The square root of 2 is to blame. :wink:


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 Post subject:
PostPosted: Sun Aug 26, 2007 5:32 pm 
thanks for the quick response , i see things much clearer now :)


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 Post subject:
PostPosted: Tue Aug 28, 2007 3:45 pm 
The explanation for this is simple enough, each f-stop increase means that the light that get through the lens decreases by half.

Though the diafragm blades make not a perfect circle, we can safely asume that the light passes through the lens like a perfect circle. In order to decrease light by half we must decrease the circle area by half as well. The area of a circle is PI*(radius)^2.

Thus, if given a circle, we want to obtain another with exactly half the area of the first one:

Area of circle 1: A1
Radius of circle 1: r1
Therefore A1 = PI * (r1)^2

Area of circle 2: A2
Radius of circle 2: r2
Therefore A2 = PI * (r2)^2

We know this: A1 = 2A2 (the area of circle 2 is half the area of circle 1), so:

PI *(r1)^2 = 2 * PI * (r2)^2
(r1)^2 = 2 * (r2)^2
r1 = square_root(2 * (r2)^2)
r1 = square_root(2) * r2

Therefore, if we want to half (or double) the area of a given circle, we must diminish (or enhance) its radius by a factor of square_root(2).

Note that f-stops are the relation between the focal lenght and the diameter of the diafragm, the diameter of a circle is twice it's radius, it is a direct conversion so it does not affect the previous calculations in any way.

Bye


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