Is there such a thing as a maximum aperture that a lens can be open to? What about a minimum aperture that it can be closed to? Do these concepts even make any sense? Is there a lens with the narrowest aperture in the world? Is there one with the widest?

  • \$\begingroup\$ Absolute aperture can be arbitrarily big (long wavelength, long focal length, like Hubble and FAST) or arbitrarily small (short wavelength, short focal length, like electron microscopes). I would say the small relative aperture f number is easier because it's easier to make long focal length systems, e.g. Hubble has a f=24. But anything smaller than f=1 would be difficult. X ray fluoroscopy lenses are f=0.5 - f =1 but resolution is awful. \$\endgroup\$ Aug 11, 2017 at 16:59

6 Answers 6


An aperture could be closed which is effectively an infinitely large f-stop number since no light gets through. The fastest possible (smallest f number) is a bit harder. The speed of a lens is limited by the ratio of the entrance pupil to the focal length of the lens. The longer the focal length, the bigger the entrance pupil must be. In theory you could make one very very large, but eventually the amount of glass is going to make it so you physically would lose more light than you were gaining.

There "record" for fastest lens is arguably the f/.33 Super-Q-Gigantar 40mm, but it was really just a marketing gimmick and only one was ever made. It isn't actually functional. There is a functional f/.7 lens of which 10 were made. Six were purchased by Nasa, Carl Zeiss kept one for himself and 3 of them were purchased by Stanley Kubrick and used in the film Barry Lyndon.

In theory, it should be possible to design lenses faster than this, but the cost and benefit are simply not worth it. The lenses become too costly and complex and don't offer any significant benefit for the effort since the difficulty goes up faster than exponentially. (Since each f/stop requires a doubling of the size and physical issues make it more that twice as complicated for each additional f-stop.)


Physics plays a role in answering your question and that information is out there. The basics from that linked discussion are that the index of refraction of the lens material will affect the maximum aperture you can achieve, so for pure glass that has an index of refraction of 1.5, the maximum aperture would be f/0.5 or thereabouts. Better substances, such as diamonds, with an index of refraction of 2.417 can give you an aperture of f/0.235 with a corresponding insane cost of ownership (consider just how much a lens of pure diamond might cost). The lensmaker's equation is the basis for the numbers.

As for minimum aperture, you could basically get down to what amounts to an atomic size hole, large enough for one photon to pass through, but that's useless for, well, anything. For a lot of lenses, getting to some place around f/11 or higher results in loss of sharpness as a function of diffraction, so f/32 is about the top out point for 35mm lenses though they can get smaller for larger formats and do so. Pinhole lenses are often in the smaller range, as much as f/177 (Lensbaby has one like this). Still, even if the optics were perfectly able to handle something like this, consider what the ISO and shutter speeds would need to be to get an image, so at some point, the value of this is pretty much zero unless you're into blurry abstracts.

  • \$\begingroup\$ f/0.5 is the theoretical limit that was taught to me for photographic lenses. On the other end of the continuum was f/64 that was common on view camera lens boards. I have seen f/128 but not smaller. Group f/64 was a club of focus fanatics started by Van Dyke that included Adams, Cunningham, Weston, Stieglitz and other South Western US photographers. \$\endgroup\$
    – Stan
    Aug 11, 2017 at 17:40
  • \$\begingroup\$ @Stan The photographers in Group f/64 were using 8x10 LF cameras. f/64 on an 8x10 camera is basically equivalent to about f/8 in the 135 format ("35mm" or "FF"). \$\endgroup\$
    – Michael C
    Apr 18, 2019 at 18:31
  • \$\begingroup\$ @Stan "theoretical limit" entirely depends on optical density of the glass \$\endgroup\$ Nov 14, 2023 at 10:48

There are many terms related to aperture, but let's get the most interesting to us: after wikipedia: "angular aperture N of a lens is expressed by the f-number, written f/, which is the ratio of the focal length f to the diameter of the entrance pupil D:"

N = f/D

So, minimum aperture this is simple: you just close the hole and have aperture of zero (f/∞).

But you can quite easily get below magical f/1 by the clever design. No need for diamond lenses, as Joanne C greatly explains. You can just grab a lot of light with front element as big as you want (D) to and squeeze it to the considered image (which relates to focal length).

In today's world you can meet this effect when using for example Metabones T Speed Booster 0.64 or 0.71 converter. It multiplies an focal length your lens by the number specified. So, if you get the beautiful Leica Noctilux f=50mm lens f/0.9 after using the Metabones 0.64 converter you get the effective f = 50mm * 0.64 = 32mm. Entrance pupil (as well as f) is proportional to sensor size d at given angle of view. So we move our lens+converter to a camera with d=35mm0.64 which gives ~23mm (sensor longer edge) - this appears to be micro fourd thirds system!. On this system our f gets back to 50mm, but D also gets multiplied by 0.64, so we have = f/(0.90.64) = f/0.576.

So were's the catch, you ask? Of course converter is not magic wand. It squeezes available light on smaller image circle, so you can use your Leica only on micro four thirds cameras. And added lens set affects image quality, but this is another story :)

This effect is also explained in cambridgeincolor lenses tutorial

  • \$\begingroup\$ Metabone Boosters in principle could give f/0.576; but in practice they don't, due to vignetting etc. In the spec they list f/ 0.8 as max cone getting through \$\endgroup\$
    – patta
    Nov 13, 2023 at 20:53

Maximum aperture f/number of conventional lens is 0.35 = SQRT(2)/4, one step faster than 0,5.

Not zero, not 0.5, not 0.33. Hope bloggers, commentators and wikipedia will learn that once and for all.

This 0.35 is when the sensor surface gets illuminated from all directions, covering an half-sphere. The half-sphere is then the max surface that an entrance pupil can have. I've done this estimate by Lagrange invariant - the max amount of light that can be concentrated over a finite area. By f-number here is meant diameter of flat entrance pupil divided by focal length.

This geometric limit can be cheated in several ways; for example getting illumination also from backside, like a lamp light collector. In this case (full-sphere) the max should be 0,25 = 1/4, one more step faster than above.

f number is well defined for small apertures (up till f/2) but beyond that it gets messy and counterintuitive.

  • \$\begingroup\$ Hope bloggers, commentators and wikipedia will learn that once and for all. That's a bit dismissive, perhaps even arrogant. I'm going to throw up [citation needed] over your claims. re: Lagrange invariant, William Smith's textbook (that I cited in my answer) specifically invokes and states that the optical invariant supports his derivation of f/0.5 being the fastest aperture possible. \$\endgroup\$
    – scottbb
    Nov 16, 2023 at 18:46
  • \$\begingroup\$ oh yes I was pissed because couldn't find any reasonable explanation nor calculation online. At least I've got you on the task! Thanks for the book hint and post. Now check the math... \$\endgroup\$
    – patta
    Nov 17, 2023 at 19:47
  • \$\begingroup\$ Are you asking me to check the math? I quoted the math directly. It's a published source, verifiable what it said (right or wrong). But you haven't posted your math. \$\endgroup\$
    – scottbb
    Nov 20, 2023 at 18:02

Since this is 2 questions (narrowest aperture, and widest aperture, I'm answering this in 2 answers

Narrowest Aperture

For imaging systems in non-scientific applications (i.e., cameras taking photographs), there isn't a practical limit to the smallest aperture. Ultimately, the limit on the smallest aperture is determined by physics, namely the longest wavelength of light being recorded. Cameras with lenses that are limited in this way are called diffraction limited systems.

In photography, the smallest relative apertures are usually just found in pinhole-aperture cameras. From How to calculate the optimal pinhole size?, the aperture diameter \$d\$ is:

$$ d = c\sqrt{f\lambda} $$

where \$c\$ is a constant, usually between ~1.4–2, \$f\$ is the focal length (from the aperture to the film/sensor plane), and \$\lambda\$ is the wavelength of light that the pinhole is optimized for.

Notice that here, the absolute aperture diameter is what we care about, not relative f-number \$N = f/d\$. But for comparison or interest purposes, we can simply calculate

$$ \begin{align} N &= {f\over d} = \frac{f}{c\sqrt{f\lambda}} \\ &= {1\over c}\sqrt{f\over\lambda} \end{align} $$

Using \$\lambda = 550\,\mathrm{nm}\$, which is green-yellow, roughly the middle of the visible light spectrum, and a focal length of, say, \$f = 60\,\mathrm{mm}\$ (the diameter of a can of soda; hobbyists often make soda-can pinhole cameras), the largest f-number N, and hence the narrowest aperture, is around \$f/165\$\$f/234\$.

For any lens, including refractive lenses with glass, and hence different index of refraction than air, we can use the diffraction of light to determine when further stopping down (that is, choosing smaller apertures, larger f-numbers) will cause unavoidable blurring of light, depending on the color (wavelength or frequency) of that light.

The waves from a point source of light traveling through a circular aperture of diameter \$d\$ will interact with themselves, producing a larger-than-point sized image of concentric rings of diminishing brightness called an Airy disk.

Airy disk, from Wikimedia Commons Computer-generated image of an Airy pattern. By English Wikipedia user Sakurambo, from Wikimedia Commons. Public domain.

If two point sources of light, such as two stars in the sky, are too close together, then the center main areas of their Airy disks on the image plane will overlap, and the images cannot be resolved as separate objects; this means light coming near the same area will begin to blur or "smear" together. This results in reduced resolution that we call diffraction limited.

The radius of the central maximum of the Airy disk (half the diameter \$d\$) is approximately

$$ {d\over2} = 1.22 \lambda N $$

where 1.22 is the value of the first zero of the \$J_1\$ Bessel function (modulo \$\pi\$). By the way, the pinhole aperture equation is a reworked version of this equation (just square both sides of the pinhole equation, substitute \$N=f/d\$, and rearrange).

Anyhow, then two Airy disks are said to be just resolved when their center-to-center separation is at least the sum of their radii. Therefore the Airy-disk determined diffraction-limited aperture occurs when the smallest resolution of the film or image sensor is the size of the Airy disk:

$$ d = 2.88 \lambda N $$

But this doesn't mean apertures smaller than the DLA are impractical, or even not useful. Lenses are sharper now than they ever have been, and in the SLR film era, f-numbers larger than f/11 or f/16 were commonly used when the exposure or depth of field requirements dictated. And large format film photographers regularly shot at f/64 and smaller.

But since digital cameras have overtaken film, a lot has been written about DLAs, and how high-resolution digital cameras "over resolve" their lenses, using the Airy diffraction limit in their reasoning.

Let's look at the same \$550\,\mathrm{nm}\$ green wavelength of light as sort of the nominal "center" color we're taking pictures of. Using the DLA calculation, for an Airy diameter of \$d = 4.3\,\mathrm{\mu m}\$ (the effective sensor pixel size of the Nikon Z9 camera), the diffraction limit is apparently \$N = 2.7\$. For smaller apertures (many consumer lens are no faster than f/2.8), the camera supposedly over-resolves the lenses, and it's wasted resolution.

There's more to photography than pixel-peeping and chasing sharpness ratings though.

  • \$\begingroup\$ I thought that the aperture can be smaller than that. Approximated, the diffraction spot on the sensor is as big, in microns, as the aperture value. So f/8 means you have a 8 micron blur (that's why closing down the aperture makes things blurrier). So if you have a sensor which is APS-C 24x16mm = 24000x16000 microns and you want around 4 units of information (otherwise it's not even a camera), then 16000/2=8000. The maximum "practical" aperture is f/8000 or a bit less than that. \$\endgroup\$
    – FarO
    Nov 17, 2023 at 8:33
  • \$\begingroup\$ @FarO I was referring only to pinholes in my original edit. But to your comment, it depends on the film/sensor size, right? The Polaroid 20×24 camera (that's inches, not mm) takes 1:1 macro head and shoulders shots, starting at f/90. That's absolutely necessary to have useful DoF of a person's face at that magnification. But that's an Airy diameter of 0.14 mm! Still, on a 500x600 mm print, I don't think the diffraction limit is concerning. \$\endgroup\$
    – scottbb
    Nov 17, 2023 at 19:35

Since this is 2 questions (narrowest aperture, and widest aperture, I'm answering this in 2 answers

Widest Aperture (fastest f-number)

TL;DR: For a photographic camera and lens in air, the fastest aperture is \$f/0.5\$, based on the numerical aperture (NA).

The reason people usually ask what the fastest (widest) aperture possible is because we're taught that the f-number is defined as \$N = {f\over D}\$. The thinking is, "well, if I just keep making \$D\$ bigger, then I get a faster aperture, right?" But here's the kicker: the f-number assumes a system that:

  1. is described by the paraxial approximation; and
  2. is designed to be corrected for spherical and coma aberrations.

Photo-SE user Edgar Bonet wrote an excellent explanation and derivation of why \$f/0.5\$ is the fastest lens achievable (when in air) due to conservation of étendue.

There's not much to add to Edgar's explanation from an étendue standpoint. Instead, I'm just going to cite and quote from a textbook. In Modern Optical Engineering: The Design of Optical Systems, 2nd ed. (1990), pp. 259–261, Warren Smith states:

Fig. 9.16. A simple radiometer with an objective lens which forms an image of the radiation source directly on the detector cell. A simple radiometer with an objective lens which forms an image of the radiation source directly on the detector cell. From Warren J. Smith (1990). Modern Optical Engineering, p. 259.

A simple radiometer is sketched in Fig. 9.16. The detector, with a diameter \$D\$, is located at the focus of an objective with a focal distance \$F\$ and a diameter \$A\$. The half-field of view of the system is \$\alpha\$, and since the detector is at the focus of the system, it is apparent that the half-field of view is given by

$$ \alpha = {D\over 2F} \tag{9.19} $$

And then proceeds with this explanation:

The relationship between \$A\$ and \$F\$ is, as we have previously noted, a limited one. If the optical system is to be aplanatic* (that is, free of spherical aberration and coma, the second principal surface (or "plane") must be spherical; for this reason, the effective diameter \$A\$ cannot exceed twice the focal distance \$F\$, and the slope of the marginal ray at the image cannot exceed 90°. This limits the numerical aperture of the system to \$\mathrm{NA} = N'\sin 90° = N'\$; for systems in air with distant sources the limiting relative aperture becomes \$f/0.5\$.

(Emphasis mine) From a simple geometric argument, granting that the secondary (rear) principal surface is spherical and not planar and that the distance from the center of the detector to the principal surface, the diameter of the aperture, \$A\$ (in Smith's notation), must be less than, or equal to in the extreme marginal ray angle of 90º, \$2F\$.

There are other limits imposed on the speed of the objective lens; the design of the system may be incapable of whatever resolution is required at large aperture ratios, or physical limitations (or predetermined relationships) may limit the acceptable speed of the objective.

We can introduce the effective \$f/\#\$ of the objective by multiplying both sides of Eq. 9.19 by \$A\$; setting \$(f/\#) = F/A\$ and rearranging, to get, for systems in air,

$$ (f/\#) = {D\over 2A\alpha} \tag{9.20}$$

or for systems with the final image in a medium of index \$N'\$

$$ \mathrm{NA} = N'\sin u' = {A\alpha\over D} \tag{9.21} $$

Equation 9.21 can also be demonstrated by setting the optical invariant (Eq. 2.54) at the objective (\$I = A\alpha/2\$) equal to the invariant at the image (\$I = ½DN'u'\$) and substituting \$\sin u'\$ for \$u'\$ (in accordance with our requirement for aplanatism).

The optical invariant relates any pair of rays through (and exiting) an optical system; any pair of rays satisfy the invariant. For more on Smith's use of the optical invariant, see Smith 1990, p. 47.

By the way, the ray-tracing optical invariant Smith describes is exactly analogous to the object-size and image-side étendue and illuminance that Edgar Bonet's answer derives. The thermodynamic conservation of étendue and the ray-tracing optical invariant lead to the same conclusion regarding fastest lens speed.

Since the \$(f/\#)\$ cannot be less than 0.5 and \$\sin u\$ cannot exceed 1.0, it is apparent that the objective aperture \$A\$, half-field angle \$\alpha\$, and detector size \$D\$, are related by

$$ \left|\frac{A\alpha}{N'D}\right| \le 1.0 \tag{9.22} $$

It should be noted that Eq. 9.22, since it can be derived by way of the optical invariant with no assumptions as to the system between object and detector, is valid for all types of optical systems, including reflecting and retracting objectives with or without field lenses, immersion lenses, light pipes, etc. It is thus quite futile to attempt a design with the left member of Eq. 9.22 larger than unity; in fact, it is sometimes difficult to exceed (efficiently) a value of 0.5 when good imagery is required.


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