A great physics problem for senior year students:

In J. R. R. Tolkien’s The Lord of the Rings (volume 2, p. 32), Legolas the Elf claims to be able to accurately count horsemen and discern their hair color (yellow) 5 leagues away on a bright, sunny day.

*“Riders!” cried Aragorn, springing to his feet. “Many riders on swift steeds are coming towards us!”*

*“Yes,” said Legolas,”there are one hundred and five. Yellow is their hair, and bright are their spears. Their leader is very tall.”*

*Aragorn smiled. “Keen are the eyes of the Elves,” he said.*

*“Nay! The riders are little more than five leagues distant,” said Legolas.”*

Make appropriate estimates and argue that Legolas must have very strange-looking eyes, have some means of non-visual perception, or have made a lucky guess. (1 league ~ 3.0 mi.)

On land, the league is most commonly defined as three miles, though the length of a mile could vary from place to place and depending on the era.

At sea, a league is three nautical miles (3.452 miles; 5.556 kilometres).

Several solutions are possible, depending on the estimating assumptions

I.

By Chad Orzel is an Associate Professor in the Department of Physics and Astronomy at Union College in Schenectady, NY

The limiting factor here is the wave nature of light– light passing through any aperture will interfere with itself, and produce a pattern of bright and dark spots.

So even an infinitesimally small point source of light will appear slightly spread out, and two closely spaced point sources will begin to run into one another.

The usual standard for determining whether two nearby sources can be distinguished from one another is the Rayleigh criterion:

sine of the angular separation between two objects = 1.22 x ratio of the light wavelength to the diameter of the (circular) aperture, through which the light passes.

To get better resolution, you need either a smaller wavelength or a larger aperture.

Legolas says that the riders are “little more than five leagues distant.”

A league is something like three miles, which would be around 5000 meters, so let’s call it 25,000 meters from Legolas to the Riders.

Visible light has an average wavelength of around 500 nm, which is a little more green than the blond hair of the Riders, but close enough for our purposes.

The sine of a small angle can be approximated by the angle itself.

The angle = the size of the separation between objects divided by the distance from the objects to the viewer.

Putting it all together, Legolas’s pupils would need to be 0.015 m in diameter.

That’s a centimeter and a half, which is reasonable, provided he’s an anime character. I don’t think Tolkien’s Elves are described as having eyes the size of teacups, though.

We made some simplifying assumptions to get that answer, but relaxing them only makes things worse. Putting the Riders farther away, and using yellower light would require Legolas’s eyes to be even bigger. And the details he claims to see are almost certainly on scales smaller than one meter, which would bump things up even more.

Any mathematical objections to these assumptions? Sean Barrett writes:

“The sine of a small angle can be approximated by the angle itself, which in turn is given, for this case, by the size of the separation between objects divided by the distance from the objects to the viewer.”

Technically this is not quite right; the separation divided by the distance is not the angle itself, but rather the tangent of the angle. (SOHCAHTOA: sin = opposite/hypoteneuse; tangent = opposite/adjacent.)

Because the cos of a very small angle is very nearly 1, however, the tangent is just as nearly equal the angle as is the sine. But that doesn’t mean you can just skip that step. And there’s really not much need to even mention the angle; with such a very tiny angle, clearly the hypoteneuse and the adjacent side have essentially the same length (the distance to either separated point is also essentially 25K meters), and so you can correctly say that the sine itself is in this case approximated by the separation divided by the distance, and never mention the angle at all.

(You could break out a calculator to be on the safe side, but if you’re going to do that you need to know the actual formulation to compute the angle, not compute it as opposite/adjacent! But, yes, both angle (in radians) and the sine are also 1/25000 to about 10 sig figs.)

II. Another solution

Using the Rayleigh Criterion. In order for two things, x distance apart, to be discernible as separate, at an angular distance θ, to an instrument with a circular aperture with diameter a:

### θ > arcsin(1.22 λ/a)

5 leagues is approximately 24000 m.

Sssume that each horse is ~2 m apart from each other

So arctan (1/12000) ≅ θ.

We can use the small-angle approximation (sin(θ) ≅ tan(θ) ≅ θ when θ is small)

So we get 1/12000 ≅ 1.22 λ/a

Yellow light has wavelengths between 570 and 590 nm, so we’ll use 580.

a ≅ 1.22 * (580E-9 m)* 12000 ≅ .0085 m.

8 mm is about as far as a human pupil will dilate, so for Legolas to have pupils this big in broad daylight must be pretty odd-looking.

Edit: The book is Six Ideas that Shaped Physics: Unit Q, by Thomas Moore

III. Great discussion on the Physics StackExchange

Could Legolas actually see that far? Physics StackExchange discussion

Here, Kyle Oman writes:

For a human-like eye, which has a maximum pupil diameter of about 9 mm and choosing the shortest wavelength in the visible spectrum of about 390 nm, the angular resolution works out to about 5.3×10−5 (radians, of course).

At a distance of 24 km, this corresponds to a linear resolution (θd, where d is the distance) of about 1.2m1. So counting mounted riders seems plausible since they are probably separated by one to a few times this resolution.

Comparing their heights which are on the order of the resolution would be more difficult, but might still be possible with dithering.

Does Legolas perhaps wiggle his head around a lot while he’s counting? Dithering only helps when the image sampling (in this case, by elven photoreceptors) is worse than the resolution of the optics. Human eyes apparently have an equivalent pixel spacing of something like a few tenths of an arcminute, while the diffraction limited resolution is about a tenth of an arcminute, so dithering or some other technique would be necessary to take full advantage of the optics.

An interferometer has an angular resolution equal to a telescope with a diameter equal to the separation between the two most widely separated detectors. Legolas has two detectors (eyeballs) separated by about 10 times the diameter of his pupils, 75 mm or so at most. This would give him a linear resolution of about 15cm at a distance of 24 km, probably sufficient to compare the heights of mounted riders.

However, interferometry is a bit more complicated than that. With only two detectors and a single fixed separation, *only* features with angular separations equal to the resolution are resolved, and direction is important as well.

If Legolas’ eyes are oriented horizontally, he won’t be able to resolve structure in the vertical direction using interferometric techniques. So he’d at the very least need to tilt his head sideways, and probably also jiggle it around a lot (including some rotation) again to get decent sampling of different baseline orientations. Still, it seems like with a sufficiently sophisticated processor (elf brain?) he could achieve the reported observation.

Luboš Motl points out some other possible difficulties with interferometry in his answer, primarily that the combination of a polychromatic source and a detector spacing many times larger than the observed wavelength lead to no correlation in the phase of the light entering the two detectors. While true, Legolas may be able to get around this if his eyes (specifically the photoreceptors) are sufficiently sophisticated so as to act as a simultaneous high-resolution imaging spectrometer or integral field spectrograph and interferometer. This way he could pick out signals of a given wavelength and use them in his interferometric processing.

A couple of the other answers and comments mention the potential difficulty drawing a sight line to a point 24 km away due to the curvature of the Earth. As has been pointed out, Legolas just needs to have an advantage in elevation of about 90 meters (the radial distance from a circle 6400 km in radius to a tangent 24 km along the circumference; Middle-Earth is apparently about Earth-sized, or may be Earth in the past, though I can’t really nail this down with a canonical source after a quick search). He doesn’t need to be on a mountaintop or anything, so it seems reasonable to just assume that the geography allows a line of sight.

Finally a bit about “clean air”. In astronomy (if you haven’t guessed my field yet, now you know…) we refer to distortions caused by the atmosphere as “seeing”.

Seeing is often measured in arcseconds (3600 arcsec = 60 arcmin = 1∘3600 arcsec = 60arcmin = 1∘), referring to the limit imposed on angular resolution by atmospheric distortions.

The best seeing, achieved from mountaintops in perfect conditions, is about 1 arcsec,

or in radians 4.8×10−6 . This is about the same angular resolution as Legolas’ amazing interferometric eyes.

I’m not sure what seeing would be like horizontally across a distance of 24 km. On the one hand there is a lot more air than looking up vertically; the atmosphere is thicker than 24 km but its density drops rapidly with altitude. On the other hand the relatively uniform density and temperature at fixed altitude would cause less variation in refractive index than in the vertical direction, which might improve seeing.

If I had to guess, I’d say that for very still air at uniform temperature he might get seeing as good as 1 arcsec, but with more realistic conditions with the Sun shining, mirage-like effects probably take over limiting the resolution that Legolas can achieve.

IV. Also on StackExchange, the famous Luboš Motl writes:

Let’s first substitute the numbers to see what is the required diameter of the pupil according to the simple formula:

*narrower*because the resolution allowed by the diffraction formula would get worse. The significantly more blurrier images are no helpful as additions to the sharpest image. We know that in the real world of humans, too. If someone’s vision is much sharper than the vision of someone else, the second person is pretty much useless in refining the information about some hard-to-see objects.

The atmospheric effects are likely to worsen the resolution relatively to the simple expectation above. Even if we have the cleanest air – it’s not just about the clean air; we need the uniform air with a constant temperature, and so on, and it is never so uniform and static – it still distorts the propagation of light and implies some additional deterioration. All these considerations are of course completely academic for me who could reasonably ponder whether I see people sharply enough from 24 *meters* to count them. 😉

Even if the atmosphere worsens the resolution by a factor of 5 or so, the knights may still induce the minimal “blurry dots” at the retina, and as long as the distance between knights is greater than the distance from the (worsened) resolution, like 10 meters, one will be able to count them.

In general, the photoreceptor cells are indeed dense enough so that they don’t really worsen the estimated resolution. They’re dense enough so that the eye fully exploits the limits imposed by the diffraction formula, I think. Evolution has probably worked up to the limit because it’s not so hard for Nature to make the retinas dense and Nature would be wasting an opportunity not to give the mammals the sharpest vision they can get.

Concerning the tricks to improve the resolution or to circumvent the diffraction limit, there aren’t almost any. The long-term observations don’t help unless one could observe the location of the dots with the precision better than the distance of the photoreceptor cells. Mammals’ organs just can’t be this static. Image processing using many unavoidably blurry images at fluctuating locations just cannot produce a sharp image.

The trick from the Very Large Array doesn’t work, either. It’s because the Very Large Array only helps for radio (i.e. long) waves so that the individual elements in the array measure the phase of the wave and the information about the relative phase is used to sharpen the information about the source. The phase of the visible light – unless it’s coming from lasers, and even in that case, it is questionable – is completely uncorrelated in the two eyes because the light is not monochromatic and the distance between the two eyes is vastly greater than the average wavelength. So the two eyes only have the virtue of doubling the overall intensity; and to give us the 3D stereo vision. The latter is clearly irrelevant at the distance of 24 kilometers, too. The angle at which the two eyes are looking to see the 24 km distant object are measurably different from the parallel directions. But once the muscles adapt into this slightly non-parallel angles, what the two eyes see from the 24 km distance is indistinguishable.

V. Analyzed in “How Far Can Legolas See?” by minutephysics (Henry Reich)