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1

Legolas probably only needs one eye if he has enough time and can make sufficiently accurate spectral measurements. First, note that Legolas was watching on a sunny day; we'll assume that between incident intensity and albedo that object were reflecting on the order of $100 \mathrm{W}/\mathrm{m}^2$ light, which is about $10^{22}$ photons per second. At 24 ...

7

Take the following idealized situation: the person of interest is standing perfectly still, and is of a fixed homogeneous color the background (grass) is of a fixed homogeneous color (significantly different from the person). Legolas knows the proprotions of people, and the colors of the person of interest and the background Legolas knows the PSF of his ...

4

One thing that you failed to take into account. The curve of the planet (Middle Earth is similar in size and curvature to Earth). You can only see 3 miles to the horizon of the ocean at 6 feet tall. To see 24 km, you would need to be almost 100m above the objects being viewed. So unless Legolas was atop a very (very) tall hill or mountain, he would not have ...

4

Deconvolution can work but it only works well in case of point sources as e.g. pointed out here. The principle is simple; the blurring due to the finite aperture is a known mathematical mapping that maps a hypothetically infinite resolution image to one with finite resolution. Given the blurred image, you can then attempt to invert this mapping. The blurred ...

71

Fun question! As you pointed out, $$\theta \approx 1.22\frac{\lambda}{D}$$ For a human-like eye, which has a maximum pupil diameter of about $9\rm mm$ and choosing the shortest wavelength in the visible spectrum of about $390\rm nm$, the angular resolution works out to about $5.3\times10^{-5}$ (radians, of course). At a distance of $24\rm km$, this ...

15

Let's first substitute the numbers to see what is the required diameter of the pupil according to the simple formula: $$\theta = 1.22 \frac{0.4\,\mu{\rm m}}{D} = \frac{2\,{\rm m}}{24\,{\rm km}}$$ I've substituted the minimal (violet...) wavelength because that color allowed me a better resolution i.e. smaller $\theta$. The height of the knights is two ...

2

In the spirit of your question, having two eyes and assuming you can use them as an array (which requires measuring the phase of the light-something eyes don't do) allows you to use the distance between them for $D$ in the resolution equation. I don't know the spacing of an elf's eyes, so will use $6 cm$ for convenience. With violet light of \$\lambda = ...

3

I have a strong reason to believe I have found the correct answer to my own question, you may correct me if I'm wrong. But this image seems to explain everything about my question in one single hit: These are results from Bowmaker & Dartnall (1980). Relevant reference: Bowmaker, J.K., & Dartnall, H.J.A. Visual pigments of rods and cones in a human ...

3

So the actual problem here is: Purple is the color at the very shortest wavelength we can see. Purple is an additive mix between what we see as red light and blue light. That just doesn't make any sense. I don't see how our brain can possibly perceive this as being the same color. Shouldn't both purple colors actually be different colors ...

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