In many answers to questions about different aspects of really large aperture lenses it's pointed out that the lens mount sets a hard limit on the maximum possible aperture of the lenses to that camera (for example here and here). This may very well be true, but I can't really visualise the reason to it.

As I see it the limitation has to do with the opening physically blocking the light. I've made a drawing to demonstrate this:

enter image description here

The bottom ray strikes the lens mount and can't get to the sensor. The maximum aperture is in this case limited by the size of the lens mount.

Introducing a diverging lens

This shouldn't be a problem though since complex optics (that camera lenses are) can allow the system to converge the light rays in a plane in front of the image plane and then use a diverging (negative) lens to move the plane of focus back to the sensor/film plane without having the light interfering with the walls of the lens mount.

The following drawing uses this diverging lens and by doing so increases the maximum aperture possible despite the fact that the lens mount stays the same:

enter image description here

This is possible as long as you're not close to the physical hard limit set by refractive index. Very short focal length lenses deal with this problem all the time and I can't believe this is the reason that the lens mount acts as a hard limit of the maximum aperture.

It could also be the fact that the corrective elements required when the aperture gets too large degrades the quality too much or gets too expensive. This does not set a hard limit though, but rather a soft limit due to compromises.

Is there something I've missed? Is there really a hard limit set by the mount regarding the maximum possible aperture of a lens-camera system? If there is a limit, what is causing it?


2 Answers 2


There are two hard limits on how fast a lens can be:

The first is a thermodynamic limit. If you could make a lens arbitrarily fast, then you could point it to the sun and use it to heat your sensor (not a good idea). If you then get your sensor hotter than the surface of the Sun, you are violating the second law of thermodynamics.

This sets a hard limit at f/0.5, which can be derived from the conservation of etendue. Well, technically it's more like T/0.5. You can make lenses with f-numbers smaller than 0.5, but they will not be as fast as their f-numbers suggest: either they will work only at macro distances (with “effective” f-numbers larger than 0.5), or they will be so aberrated as to be useless for photography (like some lenses used to focus laser beams, which can only reliably focus a point at infinity on axis).

The second limit is the mount. This limits the angle of the light cone hitting the sensor. Your trick of using a diverging element dos not work. You certainly get a wider entrance pupil, but then you have a lens combination which has a longer focal length than the initial lens. Actually, your trick is very popular: it’s called a “telephoto” design. Bigger lens, same f-number.

If the lens mount allows for a maximum angle α for the light cone, then the fastest lens you can get will have an f-number equal to

$$ \begin{align} N &= \frac{1}{2\sin\left({\alpha\over 2}\right)} \\ &= {1\over 2\,\mathrm{NA}} \end{align} $$

where NA is the numerical aperture. This formula also shows the hard limit at 0.5: sin(α/2) cannot be larger than 1. Oh, BTW, if you try to derive this formula using small-angle approximations, you will get a tangent instead of a sine. Small-angle approximations are not good for very fast lenses: you should use the Abbe sine condition instead.

The same caveat about f-numbers v.s. T-numbers applies to this second limit. You can get a lens with an f-number smaller than 1/(2×sin(α/2)), but it will work as macro-only, and the bellows-corrected f-number will still be larger than the limit.


This section is intended for the mathematically inclined. Feel free to ignore it, as the relevant results are already stated above.

Here I assume that we use a lossless lens (i.e. it conserves luminance) to focus the light of an object of uniform luminance \$L\$ into an image plane. The lens is surrounded by air (index 1), and we look at the light falling on an infinitesimal area \$\mathrm{d}S\$ about, and perpendicular to, the optical axis. This light lies inside a cone of opening \$\alpha\$. We want to compute the illuminance delivered by the lens on \$\mathrm{d}S\$.

In the figure below, the marginal rays, in green, define the light cone with opening α, while the chief rays, in red, define the target area \$\mathrm{d}S\$.

diagram of lens
(source: edgar-bonet.org)

The etendue of the light beam illuminating \$\mathrm{d}S\$ is

$$ \mathrm{d}G = \mathrm{d}S\int\cos\theta\,\mathrm{d}\omega $$

where \$\mathrm{d}\omega\$ is an infinitesimal solid angle, and the integral is over \$\theta\in[0, \alpha/2]\$. The integral can be computed as

$$\begin{align} \mathrm{d}G &= \mathrm{d}S \int 2\pi \cos\theta\sin\theta\,\mathrm{d}\theta \\ &= \mathrm{d}S \int\pi\,\mathrm{d}\!\left(\sin^2\theta\right) \\ &= \mathrm{d}S \,\pi \sin^2\left({\alpha\over 2}\right) \end{align}$$

The illuminance at the image plane is then

$$ I = L\,\frac{\mathrm{d}G}{\mathrm{d}S} = L\,\pi\sin^2\left({\alpha\over 2}\right) $$

We may now define the “speed” of the lens as its ability to provide image-plane illuminance for a given object luminance, i.e.

$$ \text{speed} = {I\over L} = \frac{\mathrm{d}G}{\mathrm{d}S} = \pi\sin^2\left({\alpha\over 2}\right) $$

It is worth noting that this result is quite general, as it does not rely on any assumptions about the imaging qualities of the lens, whether it is focused, aberrated, its optical formula, focal length, f-number, subject distance, etc.

Now I add some extra assumptions that are useful for having a meaningful notion of f-number: I assume that this is a good imaging lens of focal length \$f\$, f-number \$N\$ and entrance pupil diameter \$p = f/N\$. The object is at infinity and the image plane is the focal plane. Then, the infinitesimal area \$\mathrm{d}S\$ on the image plane is conjugated with an infinitesimal portion of the object having a solid-angular size \$\mathrm{d}\Omega = \mathrm{d}S/f^2\$.

Given that the area of the entrance pupil is \$\pi\left({p\over 2}\right)^2 = {\pi\over 4}p^2\$, the etendue can be computed on the object side as

$$\begin{align} \mathrm{d}G &= \mathrm{d}\Omega\,\frac{\pi p^2}{4} \\ &= \mathrm{d}S\,\frac{\pi p^2}{4f^2} \\ &= \mathrm{d}S\,\frac{\pi}{4N^2} \end{align}$$

And thus, the speed of the lens is

$$ \text{speed} = {\mathrm{d}G\over \mathrm{d}S} = \frac{\pi}{4N^2} $$

Equating this with the speed computed on the image side yields

$$\begin{align} \require{cancel} \frac{\cancel{\pi}}{4N^2} &= \cancel{\pi} \sin^2\left({\alpha\over 2}\right) \\ N^2 &= \frac{1}{4\sin^2\left({\alpha\over 2}\right)} \\ N &= \frac{1}{2\sin\left({\alpha\over 2}\right)} \end{align}$$

I should insist here on the fact that the last assumptions I made (the lens is a proper imaging lens focused at infinity) are only needed for relating the speed to the f-number. They are not needed for relating the speed to sin(α/2). Thus, there is always a hard limit on how fast a lens can be, whereas the f-number is only limited insofar as it is a meaningful way of measuring the lens’s speed.

  • 1
    \$\begingroup\$ Great answer, two questions: 1) Do you have a reference for that formula (N = 1/(2 sin(\alpha/2)))? 2) What are typical values of \alpha on common camera mounts? \$\endgroup\$
    – Unapiedra
    Commented Nov 24, 2014 at 19:03
  • 1
    \$\begingroup\$ @Unapiedra: 1) I added a link to a Wikipedia section discussing “numerical aperture versus f-number”, but beware of their formula which has a bogus arctangent, only valid for the thin-lens approximation. Their formula is followed, however, by a useful paragraph explaining why the arctangent should not be there. On the other hand, it is not too difficult to derive the right formula directly from the conservation of etendue. \$\endgroup\$ Commented Nov 25, 2014 at 9:16
  • \$\begingroup\$ @Unapiedra: 2) I do not know. However, if you do an image search for the fastest Nikon (50/1.2) and Canon (50/1.0) lenses, you will see that their rear elements practically fill all the available room. Thus I assume those lenses reach the limits of their respective mounts. \$\endgroup\$ Commented Nov 25, 2014 at 9:17
  • \$\begingroup\$ So what happens when you use a camera mount eyepiece on a telescope? In astronomy it's all about "brightness", not magnification, and something like the Keck is a huge funnel for light. \$\endgroup\$
    – JDługosz
    Commented Nov 25, 2014 at 13:24
  • 2
    \$\begingroup\$ @jdlugosz: The straight d in dS, dG, dΩ, dω and dθ is for differentials. The slanted d in π d ²/4 is for the pupil diameter. OK, maybe this is not a very good choice... I will replace it with a “p”, like “pupil”. \$\endgroup\$ Commented Nov 27, 2014 at 8:36

I think you pretty much answered your own question, there's no hard limit as such.

If you really wanted to, you could have a huge aperture and use corrective lenses to bring everything towards the sensors, but you run into two issues:

  • price generally goes up to the square of the size of the glass, having this much would cost a lot
  • image quality would suffer.

So theoretically, there's no hard limit, it just becomes very difficult/impractical to create a lens that would actually be purchasable.

  • 1
    \$\begingroup\$ So all of the people claiming that there is a hard limit having anything to do with the lens mount in particular are simply wrong (maybe someone started the rumour and others followed)? Also just to be on the safe side, do you have any sources that can back this up? If this is the case (I need to be sure) there are many answers here on photo.SE that are wrong and unfortunately deserves to be voted down since they are misleading or just wrong. \$\endgroup\$
    – Hugo
    Commented Sep 3, 2014 at 20:21
  • \$\begingroup\$ No sources as such, but you just need to look at e.g. the canon 50mm f1.2 vs the 50mm f1.8, the 1.2 has a much larger physical aperture (larger than he lens mount) but also costs a bomb and is apparently marginally less sharp than the 1.8. Another example are lenses such as the 600mm f4 which has a (for its size) huge aperture but cost £4k+ \$\endgroup\$
    – Lenny151
    Commented Sep 4, 2014 at 6:26
  • \$\begingroup\$ In relation to the above lenses mentioned, it's worth noting that the Canon f/1 aperture is actually large enough to be obscured by the the lens mount when shooting it wide-open on a 5D (or 6D). The 1D has a larger (circular) lens mount to accomodate for the aperture. \$\endgroup\$ Commented Sep 4, 2014 at 10:49
  • \$\begingroup\$ @Lenny151 I'm a bit doubtful to this. Look at the first deagram I draw. The lens element has a larger diameter than the mount even without the diverging lens. Hence both the 50mm f1.2 and the 600mm f4 don't necessarily have to use the negative lens, given that the focal length grants a narrow enought angle of the bent light. Also you cant really draw the conclusion that the 50mm f1.2 is less sharp due to the negative lens, since it could be a result of the large elements and need for corrective elements in general. \$\endgroup\$
    – Hugo
    Commented Sep 4, 2014 at 12:42
  • 2
    \$\begingroup\$ @Lenny151 That lens is not a good example either. The Carl Zeiss Super-Q-Gigantar 40mm f/0.33 was not a working lens and the focal length and maximum aperture where arbitrarily made up. See this article for more info: petapixel.com/2013/08/06/… \$\endgroup\$
    – Hugo
    Commented Sep 4, 2014 at 22:40

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