What is the farthest a camera can see?

At what distance can a face no longer be identified using a camera? At what distance can a figure of a person no longer be captured?

• I think this question is too broad, since it depends on a camera resolution, lens you use, lighting conditions and so on... Commented Aug 13, 2019 at 13:25
• @Tomáš Zato well then consider ideal conditions with the best lens available but I'm actually looking at the physics with lens size not being a limitation.
– Muze
Commented Aug 13, 2019 at 14:06
• OK, lens size isn't a factor. Then how far can the Webb space telescope see? Commented Aug 13, 2019 at 16:54
• @Bob Jarvis: At the moment, probably as far as the wall of whatever building it's stored in - assuming, of course, that it's even been assembled yet :-( Commented Aug 13, 2019 at 17:42
• Does the Hubble telescope count as a camera? Commented Aug 13, 2019 at 21:55

One answer to this question is not what existing lenses & sensors can do in practice, but what an optical system can do in theory. Here 'in theory' means 'in perfect seeing conditions, with no atmospheric disturbance at all'. I suspect (but am not sure) that for relatively small optical systems like camera lenses, and relatively good atmospheric conditions the atmosphere is not limiting. It is limiting for large optical systems like telescopes although there are some deeply amazing techniques which go by the name 'adaptive optics' and involve, of course, lasers strapped to the telescope which can deal with this. Also, you can just be in space.

So, the answer to this is that the limit on the angular resolution of an optical system with a front-element diameter d, working at a wavelength of λ is given by

Δθ = 1.22 λ/d

The numerical fudge factor of 1.22 can be adjusted slightly depending on what you mean by the resolution, but not by very much. This limit is called the diffraction limit for an optical system.

If Δθ is small (which it is if you have any kind of reasonable lens) then at a distance then the length you can resolve is

Δl = 1.22 rλ/d

Rearranging this we get

r = Δl d / (1.22 λ)

This is the range at which an optical device with a front element of diameter d can resolve Δl at a wavelength of λ.

The wavelength of green light is about 500nm, and let's assume you need Δl = 1cm to be able to see any detail at all on a face (I don't know if you could identify a person at this resolution, but you could know it's a face).

Plugging in these numbers we get r = 16393 d where both r and d are in cm. If d is 5cm then r is a little under 1km. What this means is that however great the magnification, if your front element is 5cm in diameter, this is the limit of the resolution at that distance: if you magnify the image more you are just magnifying blur.

In another answer someone mentioned a Sigma 150-600mm zoom: this seems to have a front element size of 105mm. This gives r = 1.7km, so this lens is probably close to or actually diffraction-limited: it is close to being able resolve as well as it is physically possible to do so.

Also mentioned is this perhaps-mythical Canon 5200mm lens. It's hard to find specs for this, but I found somewhere which claimed overall dimensions of 500mm by 600mm by 1890mm: if those are correct then the front element is no more than 500mm in diameter so we get r = 8km approx for this lens. So, in particular, what it won't let you do is see faces tens of miles away, which the hype sort of implies it can.

You can use this formula for any purpose of course: for instance it tells you why you can't see the Apollo landing sites on the Moon from Earth with any plausible telescope: if you want to resolve 3m on the moon, which is about 250,000 miles away, in green light, you need a device with a diameter of about 80m. There are telescopes under construction which will have mirrors of more than 30m, but this is not particularly near 80m.

There is another, mostly-unrelated notion of 'how far you can see' which is 'how far can you see something on Earth?'. Again there's an oversimplified answer to this question. If you assume that

• the Earth is a perfect sphere;
• there is no refraction due to the atmosphere;
• the atmosphere is in fact either absent or perfectly transparent;

then there is a simple answer to this question.

If you are at a height h1 above the surface (which, remember, is a perfectly smooth sphere), and you want to see something at a height h2 above the surface, then the distance you can see it at is given by

d = sqrt(h1^2 + 2*R*h1) + sqrt(h2^2 + 2*R*h2)

where R is the radius of the Earth, 'sqrt' means square root and all distances should be in the same units (metres say). If R is large compared to h1 or h2 (which it usually is!) then this is well-approximated by

d = sqrt(2*R*h1) + sqrt(2*R*h2)

This distance is the length of a light ray which just grazes the horizon, so this formula also tells you the distance to the horizon: if you're at a height h above the surface then the distance to the horizon is

sqrt(h^2 + 2*R*h)

or if h is small compared to R (again, usually true unless you are in space)

sqrt(2*R*h)

In real life atmospheric refraction does matter (I think it makes the horizon further away generally), the atmosphere is not perfectly transparent, and while the Earth is a pretty good approximation to a sphere on large scales there are hills and so on.

However yesterday I spent an hour watching islands gradually disappear below the horizon as I sailed away from them, so I thought I would add this, having worked this out for my own amusement on the ship.

• "why you can't see the Apollo landing sites on the Moon from Earth with any plausible telescope!" +1, your answer also explains why even Hubble cannot resolve the Apollo sites.
– scottbb
Commented Aug 13, 2019 at 11:25
• so this lens is probably close to or actually diffraction-limited - There are metamaterials that allow the projection of details that are smaller than half the wavelength (superlenses, hyperlenses). I doubt these will ever find application in photography, but I thought it was interesting enough to mention.
– Ian
Commented Aug 14, 2019 at 6:35
• I am having trouble understanding "why you can't see the Apollo landing sites on the Moon from Earth with any plausible telescope!" Also, when you say plausible do you mean that as in a mater of money or physically engineering a lens? Please explain this to me in layman's terms so I can understand.
– Muze
Commented Aug 14, 2019 at 22:43
• If you wanted to justify the ommission of atmosphere you could refer to the metrological definitions for visibility: en.wikipedia.org/wiki/Visibility clear is 80 km with mountains and very low is less than 100 m. The local weather is a bigger issue than general atmospherics. Worse still is the Horizon which is only ~5km away en.wikipedia.org/wiki/Horizon for the more fancy lenses.
– TafT
Commented Aug 16, 2019 at 7:36
• 1.22 is not a "fudge factor". It is the numerical approximation of the first zero of the ordinary Bessel function of the first kind, divided by π. It corresponds to the diameter of the Airy Disk - the pattern made by the focal spot of a perfect lens with a circular aperture. Commented Aug 16, 2019 at 19:50

If you simply want visual examples with commonly available lenses and resolutions the webpage: "Guide to Identifying or Recognizing a Face: Resolution, Focal length, and Megapixels" has a number of examples.

Axis Communications has what they call a Pixel Density Model:

Examples of maximum distances for identification (500 px/m or 80 pixels/face). The Axis definition of the requirements for detection, recognition and identification.

There are many factors to calculate: front and rear lighting even angle, fog or smoke, color, distance, which portion of the lens the face appears in (center or corner), lens quality, sensor quality, camera angle, motion of the person (or camera shaking), image compression, etc.; that is why security camera manufacturers create charts with guaranteed recognition performance.

Under perfect conditions you should expect to see further. Also if there is a list of known people to compare the image to one can often say that it is one person rather than another. Modern software can analyze multiple images, even taken at different angles, and provide a final image with enhanced resolution. All those factors make exact mathematical calculations less helpful.

Also see the Luminous Landscape article: "Do Sensors Out Resolve Lenses?" and the Edmond's Optics Imaging Resource Guide section 4.3 which explains:

"The conclusion that the imaging system could not reliably image an object feature that is 12.4µm in size is in direct opposition to what the equations in our application note Resolution show, as mathematically the objects fall within the capabilities of the system. This contradiction highlights that first order calculations and approximations are not enough to determine whether or not an imaging system can achieve a particular resolution. Additionally, a Nyquist frequency calculation is not a solid metric on which to lay the foundation of the resolution capabilities of a system, and should only be used as a guideline of the limitations that a system will have.".

Despite making all the calculations it doesn't exactly reflect real world results.

One of the furthest away (enormous) objects ever seen with a telescope is 13.4 billion light years away (the age of the Earth is 4.54 ± 0.05 billion years old), but an object the size of a human face can not be seen clearly from very far.

Here 8000 images were combined to make an enormous zoomable image using a Canon 7D and a 400mm f/5.6 lens measuring 600,000 pixels wide, it would measure 50 meters by 100 meters if printed at photographic resolution:

It's much like having an enormous zoom lens and enhancing the image to improve the resolution. You can barely see the furthest buildings, which are obscured by the atmosphere.

The largest lens ever sold (only 3 were made) is shown in the video: "5200mm Canon Lens World's MOST powerful Super telephoto EF FD (updated upload)", described in this Petapixel article: "Ginormous 5200mm Canon Lens on eBay" as having a minimum focusing distance of 393ft/120m and weighing 220lb (100kg) without its stand. It's capable of taking photographs of objects 18 to 32 miles away (30km to 52kms away), of course that depends on the size of the object.

Here are screenshots from the video:

In the first photo the top of the building is approximately the same size as the lady's hand in the last closeup photo.

• On the other hand, reconnaissance satellites orbiting 500km (+300 miles) or higher can resolve faces. Commercial satellites can resolve 10cm today, so no telling what real 'spy' satellites can achieve. thescienceexplorer.com/technology/… Commented Aug 15, 2019 at 17:13
• @cmason outside of conspiracy theory nonsense it's not likely that spy satellites can do much better: the size of the lens (probably mirror) you need is too large. In fact the best figures are significantly worse than 10cm: WorldView-4 claims about 30cm which needs a mirror something over 1m in diameter. If Hubble was in the same orbit it could resolve about 15cm. I don't know what the biggest mirror you can fly (or could, in the shuttle era), but chances are not much bigger than Hubble's.
– user82065
Commented Aug 16, 2019 at 14:58

It depends on the lens you're using.

I have a sigma 150-600mm lens on a Nikon D850 and I can safely identify people over a distance of 1.2km

There is a CANON 5200mm lens, with a much longer reach:

The 5200mm Prime, which was manufactured in Japan, has insane zoom distances. It is designed to focus on objects 18 to 32 miles away. Basically, if the 5200mm Prime was much more powerful, the curvature of the Earth would start to affect the results

check the Video in the link for a short demonstation.

• "It is designed to focus on objects 18 to 32 miles away." That sentence makes no sense whatsoever. Commented Aug 13, 2019 at 22:13
• @DavidRicherby He's probably talking about minimum focusing distance
– GPS
Commented Aug 14, 2019 at 10:47
• @GPS Nope: the article says that the minimum focusing distance is 120m. (Metres, not miles. 🙂) Commented Aug 14, 2019 at 11:45
• @DavidRicherby I understand that for shorter distances, a smaller lens would be cheaper and more manageable. And with 32 miles of atmosphere, the lens cannot resolve anything useful anymore. Commented Aug 27, 2019 at 9:17

I took this hand-held (or maybe having support from a flat platform but not a tripod) with Nikon D750 and Tamron 150 - 600 mm at 600 mm, f/11, 1/2000 s and ISO 1600. I didn't think of the settings too much since I was just demoing the camera to a friend. ISO seems to be on a higher end for these conditions but other scenes were more in the shadows :)

The original distance was about 430 meters so I scaled this crop down to 43% of the original size to simulate what it would look like from 1 km. Arguably this result is more blurred than it should be due to such an odd scale factor.

It looks pretty recognizeable to me if you knew the person and maybe she wasn't wearing glasses. But the face's skin area is mere 14 pixels wide or so since D750 has "only" 24 Mpixels. With a D810 and the same lens you could easily recognize a friend's face from 1.5 km away, maybe even from 2 km. I hope someone does the test :)

To continue with the demonstrations... The Nikon P900 has 16MP sensor and an 83x zoom. They did a few tests, not exactly to your requirements but quite close. See the video: https://www.youtube.com/watch?v=mRp13pRzzWQ

In short, they could read large letters on a piece of paper at about 1KM. Beyond that things did go a bit wrong, and the level of zoom doesn't look like you'd be able to pick out a face terribly easily. They also have some obligatory shots of the moon, but sadly didn't mount the camera very well.

• Note that "83x zoom" is a relative measure and just means that the longest focal length is 83 times the shortest. What you actually need to know here is the longest possible focal length: an 83x lens could be, say, 10mm-830mm, or 15mm-1245mm, or whatever else, giving quite different answers. Commented Aug 15, 2019 at 22:04

A camera lens is a kind of telescope. Hence it has the known resolution limit that is equal to λ/D, where λ is the wavelength of the observed light, and D is the diameter of the objective. The obtained value is in angular units, not centimeters.

For a yellow light with a wavelength of 580 nm, a camera with 12 cm diameter objective should have about 1 arc second resolution.

Assuming you need at least 50 pixels over the face for the reasonable photo art and the face is about 24 cm (0.24 m) in diameter, this resolves to about 1000 meters with Wolfram.

Difficult to say but somewhere in high mountains the air may be transparent enough to approach this limit.

• The resolution limit is usually quoted as 1.22$\lambda$/D - the diameter of the Airy Disk. Commented Aug 16, 2019 at 19:46