I was trying to derive the relation between the aperture and time of exposure. I'm taking two apertures f/2 and f/4. From the formula

  A α 1/√t

I got,

  f/2 α 1/√t1    -> A


  f/4 α 1/√t2    -> B

Doing A/B, I got the result

  t2 = 4t1 (I am taking the approximation out)



Am I doing this right? If so,for the first lens with twice the aperture of the second lens, will the same amount of light will enter the first lens in just a quarter of the time it takes for the second?

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    I think you are right yes, as f4 allows 1/4 of the light through of f2, so the exposure would be 4x as long. – Digital Lightcraft Sep 3 '13 at 9:50
  • @DarkcatStudios:Thank You. Is there a constant involved in the approximation? – sjsam Sep 3 '13 at 11:02
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    Im not fully sure what you are after, but each FULL STOP (f/1, f/1.4, f/2, f/2.8, f/4, f/5.6, f/8, f/11, f/16) allows half the light through than its neighbor as the numbers rise. Therefore from f2 to f 4 are 2 stops so 1/2/2 = 1/4. – Digital Lightcraft Sep 3 '13 at 11:23
  • @sjsam See my concise and explicit mathematical answer to your explicit mathematical question. There are no definitions to confuse the issue of the interrelationship between the two variables of duration and intensity components of a photographic exposure. – Stan Sep 9 '13 at 17:10
  • @ Michaelnielsen and jrista This question asks about the relationship between the two variables. The question does not ask for a definition. I do not believe this question is a duplicate of what a "stop" is. sjsam appears to know what each variable is. – Stan Sep 9 '13 at 17:24

It's hard to tell what your equations are saying with all the undefined variables. However, the relationship is simple. Each f-stop change in aperture is a factor of 2 in light, so you need to adjust the exposure time by a factor of 2 to compensate. For example, f/2 and 1/100 second is the same exposure in terms of light level as f/4 and 1/25 second.

Note that f/4 is two f-stops smaller aperture than f/2 since f-stop numbers go in sqrt(2) sequence: f/2, f/2.8, f/4, f/5.6, f/8, f/11, etc.

Note that there are limits to this relationship. Film will generally have a non-linear response at very long exposure times. This is known as reciprocity law failure. Digital sensors tend to be linear with total light received, but pick up noise proportional to the exposure length. Therefore the signal to noise ratio goes down with exposure length. This is why many digital cameras will limit exposure length to around 30 seconds usually. If you want a longer exposure with one of those, you take a bunch of 30 second exposures and combine them in software.

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  • I'd drop the RLF stuff with film since it's so arcane and move right to the noise issue with digital sensors. Where would film RLF be now relevant? Are many shooting film, now? (I'm asking.) – Stan Sep 9 '13 at 7:11

I think what you are missing is that each stop of f equates to a doubling of light, however the amount of change for each stop varies because the f number is the ratio between the size of the aperture opening and the focal length.

The f/stops thus end up increasing when the size of the opening is cut in half relative to overall opening, which results in the f-stop progression that is defined by the square root of 2 to the nth power (the number of times it is cut in half).

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Aperture and Exposure Time Interrelationship

There is an inverse (reciprocal) relationship between the aperture and exposure time. Despite the values, the relationship is linear and needs no factoring or constant. Both are represented in the same units (stops) to allow easy exchange of combinations for different pictorial effect.

The specific pictorial effects are documented elsewhere in the knowledgebase. See tags

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