# What is the quantitative relation between f/# of the lens (aperture), exposure, and ISO?

Let's say we shoot with a DSLR camera and we have two interchangeable lenses.

Lens #1 has aperture f/X, X between 1 and 2 (see below). Lens #2 is f/4.5. Same focal length and other characteristics as previous lens.

First, we shoot using f/X lens at ISO=100, all manual controls. Second shot, we shoot using f/4.5 lens, at the same shutter speed, same light, same settings, but higher ISO.

Which ISO will I need to set on the second shot to keep the same exposure as in the first shot?

I need a formula to calculate the dependency between the ISO number and lens f-number, all other factors factor being equal, for f-numbers=1.0,1.2,1.4,1.6,1.8. Thanks.

• If you had waited another second before you deleted this from physics.SE a moderator would have migrated it for you, but this works fine as well. Mar 16, 2011 at 17:39
• See the similar question at photo.stackexchange.com/q/4157/1356 ("Manual photography cheat sheet...") Mar 16, 2011 at 22:19

For a fixed focal length, the size of the aperture (and therefore the quantity of light gathered) scales like the inverse square of F/#. So, if you double the F/#, you would need to quadruple the sensitivity of the detector. In the specific case you describe, assuming that the scene lighting doesn't change, and the shutter speed is fixed, you would be able to calculate the correct ISO as follows:

ISO=100*(4.5/X)2

Where ISO is the new sensitivity setting, and X is the starting F/#.

EDIT: photography.SE doesn't interpret LaTeX like physics.SE. My mistake!

• Unfortunately, you have your formula a bit wrong. It says (for example) that starting from f/1.4 and going to f/4.5, you'd reduce the ISO to ~10 to compensate. I believe you wanted 4.5/X instead of X/4.5. Mar 16, 2011 at 17:57
• Wait. Are ISO linear or logarithimic or exponential wrt light sensitivity ? Mar 16, 2011 at 17:58
• @Jerry Coffin is correct, you need to invert your fraction. It should be DestinationAperture/StartingAperture. As is, your formula results in a final ISO of 10 with a slower aperture (about 1/3rd of the original.) It should result in a higher ISO of 1033 (about 3 times that of the original), which would correlate with the slower aperture. Mar 16, 2011 at 18:04
• @Andrei: Technically, both. The ISO standard actually includes both what was previously the ASA number and what was previously the DIN number -- but nearly everybody ignores the DIN-like number and just uses the ASA-like number, which you have to multiply/divide to work in stops. Mar 16, 2011 at 18:05
• @Jerry: Maybe I've misunderstood something, but if you get an ISO of 100 at f/4.5, then go to f/1.4, your aperture has gotten bigger, so the ISO should go down. My equation would predict an ISO of 9.68, which is appropriately smaller than the 100 ISO you used at f/4.5. Mar 16, 2011 at 18:20

Apertures work in steps of sqrt(2) -- i.e., each time you increase the area of the aperture, you double the amount of light that can go through the lens in a given period of time.

For simplicity, let's start with, say, f/2 and f/5.6. The full stops in this case are f/2, f/2.8, f/4 and f/5.6. That means changing f/2 to f/5.6 decreases the exposure by three stops. To compensate for that, you need a change of 3 stops in the ISO or shutter speed (or a combination of the two).

ISO numbers and shutter speeds both work exponentially -- i.e., multiplying or dividing by a fixed factor changes the amount of light translated by a fixed number of stops. To use your example, starting from ISO 100, going to ISO 200 is a one-stop increase, to ISO 400 is a two-stop increase, and ISO 800 is a three-stop increase.

That leaves us with one minor detail to deal with: fractional f/stops, which are kind of a pain. f/4.5 is about 1/3rd of a stop slower than f/4. If you started from f/1.4 and went to f/4.5, that would be ~3 1/3rd stops, so you'd need to increase the ISO by ~3 1/3rd stops to compensate. Starting from ISO 100, that works out to ISO 1000 (200, 400, 800, 1000[3 1/3]). Technically, ISO 1000 isn't quite right, but it's close enough for any practical purpose and it's what your camera will (probably) provide.

• There is noting wrong with this answer, but I do want to point out a very important distinction that you've left somewhat foggy. There is a diference between the f-number, and an f-stop. f-stops are indeed spaced by sqrt(2), such that the exposure changes by a factor of 2 between stops. However, this is simply a convention by camera manufacturers that makes calculation simple for photographers. Really, a lens can be built to have any f-number, not just those sqrt(2) spaced values we are accustomed to. Mar 16, 2011 at 18:25
• Even more importantly, the ISO or f-number scale is not itself exponential. It is actually the spacing between the settings on your camera which are exponentially SPACED. As an example o what I mean, ISO 156 is exactly 1.56 times more sensitive than ISO 100, but a change of one "stop" up from ISO 100 would be ISO 200. The ISO value itself is linear in sensitivity. It is the convention we call an exposure "stop" which is in fact exponential. Mar 16, 2011 at 18:28
• @Colin K: Technically true, but talking about f/stops that aren't in powers of two is a bit like reminding people to bring enough food and water for their horses if they go on a long trip. I very carefully didn't say the ISO number itself was exponential -- only that they work exponentially, and explained exactly what I meant by that (multiplying or dividing by a fixed factor gives a fixed change in terms of stops). Mar 16, 2011 at 18:32
• I suppose so, and like I said your answer is not wrong. I'm an optical engineer with a side interest in photography, so it made me die a little on the inside to read that "ISO numbers and shutter speeds both work exponentially" when is is only "stops" which are exponential. But this isn't an engineering stackexchange so, its not really a big deal. Mar 16, 2011 at 18:35