# Tag Info

74

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 ...

15

First, it must be said that the picture you provided in your question is extreme. The concept of light bending is true, but the amount that the light bends is nowhere near as large as the picture shows it. The quantification of how much light bends when transferring from one medium to another is called the "index of refraction," and air's index of ...

12

Do keep in mind that the frequency of light is reference frame dependent. So, for example, the cosmic background microwave radiation would appear as a concentrated gamma radiation source 'in front' to an observer with ultra-relativistic speed relative to the CMB. In other words, light emitted from a body of a particular frequency in that body's frame of ...

11

Yes, X-ray, UV, and even radio-waves ares made of photons. The differences is the Energy (or equivalently the wavelength). See the picture of the Electromagnetic spectra . The different nomination comes from the time of the discovery. Youre eyes can see the visible part. the radiowaves can be observes with antenna etc... The only differences is the way we ...

9

An object is white when it reflects more or less any color – any frequency – and it reflects it in a random direction: the light gets scattered. For this reason, the photons arriving from a specific point of the white object (e.g. a point on the paper) are photons that arrived there from random directions in the space and they have random colors. Because ...

8

The distance where light has a circular orbit is actually $1.5r_s$ not the event horizon. This distance is known as the photon sphere. In principle a shell observer hovering at this distance could indeed see their own back. The proper distance is indeed just $2\pi r$, however the object would look bigger than expected because the curvature of spacetime has ...

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 ...

7

You don't even need highly specialised equipment to see the colour separation of the sun at low sun angles, a decent zoom lens on a camera will see it, and it's the origin of the "green flash" effect as the sun drops below the horizon. This site offers a good image: http://www.atoptics.co.uk/atoptics/gf15.htm

7

White Light is Just an Illusion Pretty much any light source which is emitting "white" light has a different spectrum. There are a few reasons: blackbody radiation and the method of producing light. When our eyes see radiation of several different wavelengths at once, we can combine them and make them "white" or "pink" or whatever color it is. Blackbody ...

7

Shouldn't pure white light have a unique spectrum no matter what? White is not a spectral color. It's a perceived color. The human eye has three kinds of color receptors, commonly called red, green, and blue. (Source: The hyperphysics page on "The Color-Sensitive Cones") Note that there's no receptor for yellow. A spectral yellow light source will ...

6

Infinity as long as you have a detector strong enough to detect it. Light keeps on travelling in a straight line forever as long as it doesn't bounce off some object. The problem with the lights you are talking about is that their intensity is really low and you can't resolve them because of the other stronger sources of lights you have around yourself. ...

6

The electromagnetic spectrum does range between (almost) zero and (almost) infinity. It's just that your eyes are sensitive to a very small part of it (from about 380 nm to about 800 nm). At the lowest frequencies, it becomes difficult to recognize the signal from background fluctuations. From this site: "Gamma-rays are detected by observing the effects ...

5

In quantum mechanics you calculate the charge density by taking the square of the wave function. If you do this for a hydrogen atom in a superposition of the ground state the first excited state (1s and 2p) you get an oscillating charge density. If you analyze this oscillating charge using Maxwell's equations, you get all the properties of the hydrogen atom: ...

5

There are better answers than this, but I just want to contribute. As Joshua said, The quantification of how much light bends when transferring from one medium to another is called the "index of refraction," and air's index of refraction is very very close to that of a vacuum, so the bending of the light is very small, and the spreading apart of the ...

5

Theoretically, the shortest wavelengths of light would be limited by the Planck length, at some point the space 'closed' by the wavelength would be so small that gravitational effects would dominate, in the same way that black holes can bend light passing near their event horizon at very small scales the wavelength would be so small that it might be at the ...

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 ...

4

Your observation is linked to the "Optical window in biological tissue". Like you already suspected, the absorption of blue light in tissue is higher than the absorption for red light. Best read the related wikipedia article, where all relevant effects are nicely illustrated. http://en.wikipedia.org/wiki/Optical_window_in_biological_tissue

4

White light consists of many wavelengths. Hence, the interference pattern using white light appears different than that for monochromatic light. At the center point, all the waves travel the same length and hence no path difference is produced at the center point. Thus at the center point we get the maxima of all wavelengths and we obtain the maximum for ...

3

$r=1.5r_s$ for the Schwarzschild solution corresponds to the unstable maximum of the effective potential for a photon, therefore you won't be able to see much in practice, since practically every photon on this orbit will either fall in the black hole or escape to infinity.

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 ...

3

I guess you have this image in mind: This hods just for a single photon, or elementary packet of light. An unpolarized beam of light contains a bunch of photons with many different polarizations. The total electric field will randomly jump all around, still the interactions with matter typically involve a single photon at a time. This means that if you ...

3

That's right, running away from a gamma source fast enough would shift them into the visible portion of the spectrum. It goes without saying that he'd have to run quite fast: $$\frac{\lambda_{\rm obs}}{\lambda_{\rm emit}} = \sqrt{\frac{1+\frac{v}{c}}{1-\frac{v}{c}}}$$ Picking rough round numbers for gamma radiation at $\lambda=10\;\rm pm$ and visible at ...

3

Yes, to the approximate extent allowed in the real world. The condition that the material doesn't reflect visible light means that the material looks black. So consider various black coatings, for example, and ask what they do with electromagnetic waves at different frequencies than visible light. Of course that you find out that they generally reflect ...

2

This is similar to why a glass of water and a glass of milk look different. We can use geometric optics (used in 3D animation ray tracing) to explain this. When a light ray (of any color) hits a mirror, it reflects as another light ray. But for a white nonreflective surface, it is scattered in all directions. From the perception side, a eye gathers ...

2

No, it doesn't mean that. One must distinguish two things: "laws of physics that apply to an object" and "laws of physics formulated from an object's viewpoint". These are two different things. Laws of physics apply to all objects. And the behavior of the objects may be described relatively to many coordinate systems or "frames of reference". The special ...

2

They will look all over the place. If you take a particular photon, you will see it in a particular polarisation state, but it will have nothing to do with the photon next to it, or the one before. Actually, getting a perfectly unpolarised source of light is quite difficult. One of the best cheap options are the sodium discharge lamps, used extensively in ...

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 = ...

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