If I set a lens to, say, f/2 but my focus is set to its closest point, my effective f-number won't be f/2 any more but it will be higher by a factor that depends on focal length, image magnification and ratio of entrance and exit pupil. Does that mean that if I move my lens from my closest focus to infinity the brightness of my image (assuming I'm looking at the same object that in both cases is in focus) increases by a factor that depends on the difference f-number at infinity vs effective f-number?

  • \$\begingroup\$ It's not clear from this question what rule you're referring to or what you actually want to know about it. Rewording it may help you get the answer you're looking for. \$\endgroup\$ Oct 23, 2013 at 9:42
  • \$\begingroup\$ I think he's on about this: dpreview.com/forums/post/50081957 \$\endgroup\$ Oct 23, 2013 at 10:18
  • \$\begingroup\$ @JamesSnell Hope it's more clear now \$\endgroup\$
    – maupertius
    Oct 23, 2013 at 10:26
  • \$\begingroup\$ Lens focal length is a function of the lens - not of focusing or distance to target. A zoom lens will vary the focal length - f-number achieved will depend on how the manufacturer has designed the lens. eg you can but a 70-200 lens that is f/2.8 across the range. Or an eg 17-250mm that has f that varies from f/3.5 to f/6.3 as the focal length increases. In both cases the f number depends on the design. \$\endgroup\$ Oct 23, 2013 at 10:48
  • 1
    \$\begingroup\$ @RussellMcMahon The f-number you are referring to is still the infinity f-number. \$\endgroup\$
    – maupertius
    Oct 23, 2013 at 11:03

2 Answers 2


The common f-number we use in photography embodies the assumption that the magnification is much less than 1. When you're taking a picture of a mountain 3 miles wide and that will be imaged 30 mm wide on the sensor, the magnification is so small that the common f-number approximation is quite valid.

However, this breaks down as magnification approaches 1, and fails dramatically above 1. The "real" amount of light you get is described by the f-number times 1 / (1 + M)2. M is the magnification, which is the linear dimension of the actual object divided by the linear dimension of its image on the image plane. As you can see, this correction factor is close enough to 1 as long as the magnification stays well below 1. At exactly 1:1 magnification, note that this factor is 1/4, which means you are two f-stops down.

So how close to macro is close enough so that you have to consider this effect? We can work the equation backwards and see what magnification results in the minimum factor we care about. Let's say anything less than a 1/4 f-stop is slop so that we don't care. 1/4 f-stop means the equation must result in 2-1/4 = .841, which results in M of .095, or a reduction factor of 11. So as long as you aren't closer than 11:1, you can pretty much ignore this effect. For a "35mm" frame, which is 36x24 mm in size, this means you are fine as long as the subject is 400x270mm, or about 16x11 inches.

Of course any thru the lens metering will see the light actually delivered thru the lens and automatically compensate for this effect. Since just about every camera has that nowadays, your pictures will still be properly exposed, but you might notice the increasingly worse tradeoff between f-stop and shutter speed as you get more into macro range.

  • \$\begingroup\$ Thank you, that is very clear! That also explains why I did the test myself and couldn't see much difference between infinity and closer focus (I am not using a macro lens). The numerical example was very useful too! \$\endgroup\$
    – maupertius
    Oct 23, 2013 at 13:04
  • \$\begingroup\$ I find this statement somewhat confusing: “The ‘real’ amount of light you get is described by the f-number times 1 / (1 + M)².” You should probably clarify that the correction factor applies to the amount of light, not to the f-number. The “effective” f-number is the f-number multiplied by (1 + M). \$\endgroup\$ Nov 24, 2014 at 8:51

Edit 2: You learn something new every day! As stated by Olin Lathrop the old rule that the light hitting the sensor is just determind by the apature and shutter-speed is not accurate when using macro.

The brightness of an image (that is the amount of light hitting the sensor) is defined by three things: The aperture (f-number), shutter-speed and ISO - nothing else. That means that an "effective" f-number, focal length or focus-point is not affecting the brightness of the image.

Edit: To make the above 100% correct:

  • The brightness of an image is defined by three things: The aperture (f-number), shutter-speed and ISO.
  • The amount of light hitting the sensor is defined by two things: The aperture (f-number) and shutter-speed.

Exactly what an "effective f-number" is, I do not know. There are some discussion over at DPS, but I do not think our photos will be any different even if we knew :-)

  • \$\begingroup\$ Thanks. In other words you are saying that no matter at what distance you are working at, a lens will collect the same amount of light when set to a certain f-number. I then wonder why the concept of effective f-number has been introduced. \$\endgroup\$
    – maupertius
    Oct 23, 2013 at 10:43
  • \$\begingroup\$ Sure, mine is more a theoretical curiosity rather than a practical consideration \$\endgroup\$
    – maupertius
    Oct 23, 2013 at 11:04
  • \$\begingroup\$ @maupertius it is not the case that a lens will collect the same amount of light at all focus distances when set to a certain f-number, you do have to consider the effective f-stop when shooting close up. \$\endgroup\$
    – Matt Grum
    Oct 23, 2013 at 11:31
  • \$\begingroup\$ @MattGrum That is my point (I just call it 'effective f-number' instead of 'effective f-stop'). \$\endgroup\$
    – maupertius
    Oct 23, 2013 at 12:20
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    \$\begingroup\$ This answer is just plain wrong in that it ignores the very effect that OP is asking about. \$\endgroup\$ Oct 23, 2013 at 12:45

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