My language is German, so forgive my lack in English words, especially with regard to photography.

I tried to understand at a very basic level the relation between aperture number and exposure and find me in some trouble:

f/# number N is the ratio of focal length f to aperture diameter D:


For simplicity I assume a thin lenses, where the physical pupil is at the lens plain. In the pictures I have two lenses, where the focal lengths are in ratio 1:2

Edit: sorry, there was a construction error in my image for the lens; I did it now with an animated tool and changed the text accordingly:

Here a lens with focal length 2: The object size is 3, object distance=10, image size is 0.75. enter image description here

Here a lens with a focal length 4 and the other parameters same as before: Image size is 2 now. enter image description here

Assuming that the object is a homogenously illuminated wall, all light passing from that wall within circle of radius equal to the object arrow finds its way to a corresponding image circle on the film, whose radius is equal to the image arrow. From the diagram it is clear, that the amounts of light fetched to the image are the same. However the ratio of the areas of these circles is (2/0,75)² ≈ 7.1, which gives a ratio of exposure 1:7.1. I would have expected 1:4 instead.

Is this law only an approximation for object distances much bigger than focal length (which is of course the "regular practical usage"...)? Because for growing object distances the ratio is more and more as expected.

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    \$\begingroup\$ I think you meant to say f-number is the ratio of focal length f to aperture diameter D. N=f/D. D is actually the diameter of the entrance pupil, too. \$\endgroup\$
    – Michael C
    Commented Oct 31, 2021 at 18:21
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    \$\begingroup\$ I'm not seeing in your ray diagram where the aperture of the green lens is twice the diameter of the red lens. I'm also not understanding why the rays for the red lens do not converge at FL1 nor do the rays for the green lens converge at FL2? \$\endgroup\$
    – Michael C
    Commented Oct 31, 2021 at 18:23
  • \$\begingroup\$ Your ray diagram only follows the light from a single point on the blue arrow. You need to consider that rays from all points on the blue arrow are also passing through the lens. And since the subject being considered is a uniformly lit wall, you also need to consider that your two dimensional diagram can not show the third dimension of the total light from across the z-axis of the diagram. \$\endgroup\$
    – Michael C
    Commented Oct 31, 2021 at 18:31
  • \$\begingroup\$ @Michael: focal length of green length is twice of red lens. Why should lines converge at F? This is only the case when object distance is at infinity. \$\endgroup\$
    – MichaelW
    Commented Oct 31, 2021 at 19:15
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    \$\begingroup\$ You may know what f-number is, but you apparently do not know what the English words "radius" and "diameter" mean. \$\endgroup\$
    – Michael C
    Commented Oct 31, 2021 at 20:57

5 Answers 5


Is this law only an approximation for object distances much bigger than focal length (which is of course the "regular practical usage"...)?

Yes. The simple formula assumes that magnification approaches zero as object distance approaches infinity. As object distances grow smaller, the Magnification factor increases and light is lost due to the increased magnification.

If your (original) diagram uses the same linear scale on both the x and y axes, then your thin lens has an entrance pupil over twice the focal length of the shorter focal length lens and approximately the same as the focal length of the longer lens! So you're talking about f/0.5 and f/1 lenses being used at more or less macro distances!

Even with your modified drawings, you're still using very wide aperture lenses at macro distances. At your specified object distance, the magnification for lens FL1 is 0.25X and the magnification for lens FL2 is 0.67X.

The simple formulae for calculating exposure all break down at macro distances. This is due to the fact that the point of convergence at such short object distances is significantly different from the lens' focal length. These differences are not considered in simple formulas such as N=F/D, which also assume more or less infinity focus, just as the focal length of the lens does.

For more typical object distances, the difference is practically negligible. But at very short object distances the differences are significant. At full 1:1 macro reproduction ratios, the object distance is four times the focal length, and the distance from the center of a single thin lens to the point of convergence is twice the focal length of the lens (measured at infinity). Thus, at twice the distance from the lens to the film or sensor, the light is spread out over an area four times greater than when the object distance is infinity and the distance from the lens to the point of convergence is the focal length.

The actual amount of light you'll record for any image can be calculated by multiplying the amount of light you'd get at infinity focus times 1 / (1 + M)^2 where M is the magnification ratio. Another way of stating it is that the "effective" f-number is equal to N(1 + M) where N is the f-number and M is the magnification factor.

When you're taking a photo of a very large object at very great distances, the reproduction ratio is very low. You're fitting a mountain one thousand meters tall onto a film or sensor 24mm in height. That's a reproduction ratio of 41,667:1 or 0.000024X. So 1/(1+M)^2 is 1/(1.000024)^2, or 0.999952. You're only losing 0.0048 percent of the amount of light you'd have at infinity.

At 1:1, this is obviously much more significant. 1/(1+1)^2 = 1/4 = 0.25. Compared to infinity, we're losing 3/4 of the field intensity of light when the reproduction ratio is 1:1 or magnification is 1.0X.

It works out the same in terms of "effective aperture": N(1+1)=2N. When you double an f-number, you're two stops darker, or at 1/4 the field intensity.

Your f/1.0 lens has a magnification factor of 0.67 at an object distance of 10: 1(1 + 0.67) = f/1.67.

Your f/0.5 lens has a magnification factor of 0.25 at an object distance of 10: 0.05(1 + 0.25) = f/0.625

The ratio of the two "effective apertures" is 2.672:1. The square of the ratio is 7.14, which is what you approximately measured as 1:7.1.

In the real world, if we set a goal of less than 1/4 stop of light loss, we need to hold the reproduction ratio to below around 11:1.

To get the exposure difference that you seem to expect, at such short object distances you need to use the same reproduction ratio for both lenses, not the same object distance. In other words, if FL2 is twice the value of FL1, then the distance of your arrow for the longer lens should be twice the object distance used for your shorter lens.

  • \$\begingroup\$ There was a mistake in drawing the image in the original picture. I removed this and inserted better pictures. Unfortunately this happened...However the problem is still the same. It was only a mistake in drawing. But your comment answers my question. So the formula is a good approximation and not valid anymore, when object gets nearer to the lens. \$\endgroup\$
    – MichaelW
    Commented Oct 31, 2021 at 20:54
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    \$\begingroup\$ Sorry for confusing with a wrong image construction. But you understood my problem, regardless of the wrong image. \$\endgroup\$
    – MichaelW
    Commented Oct 31, 2021 at 21:01

In your image, you are using an aperture radius that is fixed. The two lenses do not have the same aperture number. Since N=f/R, the longer lens has twice the aperture number. Thus, its exposure is one fourth of the exposure of the shorter lens.

  • \$\begingroup\$ You must mean that, when f is constant, a shorter lens has a larger aperture number or a longer lens has the smaller the aperture number??? Also, f is not constant in the images given, but perhaps N is??? \$\endgroup\$ Commented Oct 31, 2021 at 18:58
  • \$\begingroup\$ I think nobody can understand my question...it might be because of poor English. In my example there are two lenses, one with focal length F1, the other with double focal length F2. Aperture the same for both lenses. According to the common definition, the f-number of the lens with double focal length is half of the f-number of the other lens. In this case, the light expose of the lens with F2 should be one fourth of the exposure of the other (F1) . By comparing the sizes of the image radii this is obviously not the case. I hope it became clear now... \$\endgroup\$
    – MichaelW
    Commented Oct 31, 2021 at 19:47
  • \$\begingroup\$ @MichaelW Rather than calling the focal lengths "F1" and "F2", which are easily confused with f-numbers/aperture numbers, perhaps you should label the focal lengths FL1 and FL2? \$\endgroup\$
    – Michael C
    Commented Oct 31, 2021 at 19:49
  • \$\begingroup\$ @MichaelW Aperture diameter does not equal aperture number. One is a linear measurement, the other is a ratio between two linear measurements. If the aperture diameter remains constant and the focal length changes, the ratio between the two linear measurements also changes. \$\endgroup\$
    – Michael C
    Commented Oct 31, 2021 at 19:52
  • \$\begingroup\$ I think the question was changed quite significantly after I answered. I'll still retain my answer; if someone wants to see the original question, the edit history is there. \$\endgroup\$
    – juhist
    Commented Nov 1, 2021 at 16:58

If you keep f fixed and vary R, you vary the aperture number N by your equation. You also vary the area of the lens. For a circular lens, A = π R^2. Assume that you allow the same light flux to the lens. Flux is a measure of photons arriving per area per time. Your film at the other end becomes exposed (developed) according to the number of photons that arrive per unit time. When you cut the area of the lens by 1/N (e.g. by cutting the diameter of the lens by 1/sqrt{N}), you must increase the exposure time by a factor of N.

So, when you decrease R by a factor of 2 (e.g. cut it in half), you decrease your aperture number N by 2, you decrease the lens area A by a factor of 4, and you must increase the exposure time at the focal point by a factor of 4 to obtain the same number of photons in any given time t.

Suppose now that you keep R fixed. Your lens has a focal point f that depends on wavelength (color). Choose for example red r and green g. You send in the same incoming flux of photons for the two different wavelengths Fir = Fig. What happens?

Since the incident flux of red or green is the same and since the area of the lens is the same, under ideal conditions, the same number of red and green photons pass through the lens. The red image focuses at a different location than the green image on the image side of the lens. But the number of photons that reach those focal plans are exactly the same in a given length of time.

Suppose that your focal length for the green wavelength is a factor M further away than the focal length for the red wavelength. Consider only a 1-D line as the object. Your green image will be a factor M larger in its focal plane. In 2-D, the green image will have a factor M^2 larger area. So, the flux at the image plane will be a factor 1/M^2 smaller.

Suppose now that you have the same number density of pixels at both red and green image planes. Suppose your image falls within the full image plane at the further focal point and that you use the same size sensor for the closer image plane. Your image fills M^2 more pixels in the furtherest plane.

Suppose that each pixel requires the same cumulative number of photons to register an event (for the film to register as being "exposed"). Clearly, since the flux at the furtherest image plane is lower by a factor of 1/M^2, the number of photons per pixel arriving in a given time is lower by the same factor. You will therefore need to increase the exposure time for the green wavelength (the furtherest focal point) by M^2 to get the same registration as being counted at the red plane.

An alternative view is to consider that the lower flux density by 1/M^2 at the image plane for the green wavelength will make the pixels in the image plane less luminous by the same factor. So your red image will appear M^2 brighter for the same exposure time.

Finally, as a visualization, consider this experiment. Create two 10x10 matrices of 1 Euro coins lying with the same side up on the table. These are your image pixels. In one matrix, make a straight line image by flipping 5 coins in a row. This is an event in time t that activated 5 pixels by 5 photons. Now pretend that your factor M for the second matrix is sqrt{2}. Make a straight line of 10 coins but only flip 5 of the coins in that line. This is the image plane further away receiving the same 5 photons in an image that is twice as big.

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    \$\begingroup\$ As you've may have noticed by now, MathML isn't enabled on this site - you may want to reformat things to avoid the somewhat ugly $ signs. \$\endgroup\$
    – Philip Kendall
    Commented Oct 31, 2021 at 19:17
  • \$\begingroup\$ @Jeffrey J Weimer: All you wrote is completely clear. But, by considering the size of the red and green radii on the film, this expected behavior is not fulfilled. It is valid only in the limit of large object distances. It was the question, why we have this deviation from the theory. \$\endgroup\$
    – MichaelW
    Commented Oct 31, 2021 at 19:51
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    \$\begingroup\$ I have updated my description to consider the two different focal planes. \$\endgroup\$ Commented Oct 31, 2021 at 20:29
  • \$\begingroup\$ @Michael C: thanks, I understand now. Sorry for my wrong image construction in the original picture. It caused a lot of confusion. \$\endgroup\$
    – MichaelW
    Commented Oct 31, 2021 at 20:58

These ratios hold for exposure:
A1/A2 = (f2/f1)^2 = SS2/SS1

An = area of the aperture, commonly a circle = πr^2
fn = f-number (f-stop) = focal_length/aperture_diameter
SSn = shutter speed in seconds

  • \$\begingroup\$ but exactly this is not the case in my drawing. This was the content of the question. \$\endgroup\$
    – MichaelW
    Commented Oct 31, 2021 at 19:16

I'll publish another answer since the question was updated significantly with numbers and after updating, it doesn't seem to be about the different exposure but rather why the exposure difference is larger than expected.

You have to remember that the focal length of a lens changes as it is focused.

Normally a lens that is twice as long gives an image twice the size. However, that's only when the lens is focused far away, practically to infinity.

If a lens is focused nearby, the effective focal length changes. The effective focal length means at what distance the image is from the lens. The longer the distance, the bigger the image. So your lens that is nominally twice as long is no longer twice as long when focused nearby.

Your f=2mm lens is 2.5mm at that focus distance.

Your f=4mm lens is 6.67mm at that focus distance.

The ratio of the image sizes (and thus, effective focal lengths) is 2.668.

The ratio of exposures is 7.12.

If the aperture diameter is 6mm, the lenses are f/0.42 and f/1.11 so their exposure ratio should be about 7 (you'll find it's 7.12 if the aperture numbers are expressed with more decimals).

I wouldn't surprised to find that this is actually a different way to explain the same effect that Michael C explained in his answer.

  • \$\begingroup\$ "the focal length of a lens changes as it is focused." I think your terminology is not the same that is usually used in photography. Focal length, as normally talked about in photography, is an unchanging property of a lens. It's simply the multiplicative inverse of the lens's power (usually in dopters). For any given single element that's treated as an ideal thin lens (such as in the question), it's constant in the thin-lens equation, 1/f = 1/di + 1/do. The focal length is determined by the lens's shape (radii of curvature, thickness (real lens), and the refractive index of the lens glass \$\endgroup\$
    – scottbb
    Commented Nov 2, 2021 at 14:27

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