# Tag Info

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In general, what you see in Jupiter's atmosphere is simply clouds: large collections of small liquid droplets floating in the atmosphere. Liquid droplets are excellent at scattering light in any direction, and indeed one can visually see clouds on Earth with some regularity. In fact, it is indeed possible to observe 3D details on these clouds. Here is an ...

8

Sometimes we do, and the phenomenon is called a light echo. What you're looking at there is NOT moving gas. It's an "echo" exactly as you describe. The problem is that you need a pulse of light. If you have a constant stream of light, the "light echos" will be exactly like what you see in fog on earth.

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For many materials the change in refractive index over the range of visible wavelengths isn't huge, so it's not a bad approximation to take a single value. The range of visible wavelengths is from about 400nm to 700nm, so the middle wavelength is 550nm. As it happens, the sodium D lines are not far from this, at 589nm, and since they are bright and easy to ...

3

The colour you see in the sky on cloudy nights is due to the reflection of city lights off the clouds. In rural areas, a cloudy night is, as you expected, significantly darker. However, the massive amount of light given off in urban areas reflects back to Earth when there is cloud cover. And so, you see a red-orange hue, similar to the overall colour ...

3

It is not that magnetic fields repeal each other, but the bodies that are origin of these fields repel if they are oriented in such a way that their fields in the space between them are pointing in opposite directions. Electromagnetic waves in vacuum obey linear wave equation, so one does not change the other; they pass each other as if the other one did ...

3

Actually in most cases you don't see the event horizon, but instead the photon sphere. For example if you are looking from some distance, if light emitted from some star goes inside the photon sphere (where light can travel theoretically in circular orbit, though the orbit is unstable), which is located outside the event horizon, it is more or less doomed to ...

3

Yes, well, sort of. Energy can be a bit tricky to keep track of in general relativity, and it's important to be precise about what we mean by energy. In this case the issue is whether the light is red shifted. The red shift does reduce the energy of individual photons, though overall the energy is not lost - it's just diluted. You probably know that the ...

2

To add to John Rennie's answer and Noldor130884's answer: a half wavelength waveplate will reverse the handedness of circularly polarized light. It's the same principle as converting linear to circularly polarized light although here you don't have the precise alignment problems (to convert linear to circularly polarized, you need to align the input light's ...

2

In short, no. It's not possible to generate a 600THz electrical signal. The sole reason for this is that electrons could not physically oscillate (vibrate) fast enough using the same methods as when generating radio waves (100MHz.) One way to have electrons (which make up electricity) have that high of a frequency, is to accelerate them to high speeds and ...

2

Everything you see that does not emit light itself is due to light from an external light source being scattered. Most objects scatter light in all directions, so no matter where you position yourself, you will see the object. If something blocks the light from the source, less light will be reflected from a particular area. This area is the shadow. Since ...

2

From the interior, a spherical mirror can be analyzed as a continuous assemblage of concave mirrors. If you were illuminated by an invisible light source, your image would be reflected from all points on the interior surface according to the mirror equation for concave spherical mirrors: (1/object distance) + (1/image distance) = (1/focal length) The ...

2

Calculating what you would see as you fell into a black hole is straightforward but tedious. Fortunately there are lots of sites that have done this for you. Actually, if you've been to the cinema recently the film Interstellar does a pretty good job of it. Less spectacularly, have a look at this site that has videos of what the journey would look like. ...

2

Be careful about your description. One distinction which can be made is that the "sky" is not red, it's the lower faces of low-hanging clouds. You've certainly noticed the fact that, near the horizon, the sun looks red. This is due to Rayleigh scattering, which is also responsible for the sky being blue. What is happening here is the sun (out of sight ...

2

Under red light, your picture never shows white. It only shows bright red, which you are interpreting as meaning white. If you take a piece of paper and put a small hole in it, then hold the hole over any part of the screen image, you will see that it is red, not white. When doing this in real life (which I suspect is what you're talking about), three ...

2

Jupiter is cloudy. It's upper atmosphere contains clouds of ammonia, ammonium hydrosulfide and water. These clouds reflect light just as clouds made up of water droplets reflect light in Earth's atmosphere. We should note that no gas is completely transparent, even if only because of Rayleigh scattering. The sky on Earth looks blue because our atmosphere ...

2

When a wave travels through a rope, the rope goes up and down, the position of all the 'rope-particles' changes, they oscillate and this makes up the wave. With light, it is indeed the electromagnetic field oscillating, but you shouldn't think of the arrows that represent that field in your first picture of light as 'extending into the rest of the space'. ...

2

The accepted airglow answer might be technically true, but it does not answer the question! The existence of an additional and very faint source of green light in the atmosphere does not explain the absence of the green light in the sunset sky gradient. I wasn't satisfied with other answers either. The only satisfactory answer I could find is this one. ...

2

The ray theory of light is equivalent to the Eikonal Equation, which in turn is essentially a slowly varying envelope approximation to Maxwell's equations. If we write the electric and magnetic field vectors as $\mathbf{E}\left(\mathbf{r}\right) = \mathbf{e}\left(\mathbf{r}\right) e^{i\,\varphi\left(\mathbf{r}\right)}$, $\mathbf{H}\left(\mathbf{r}\right) = ... 2 I think the answer is very simple if you ask another question, you are implying that both black holes generate the same gravity and that the light passes exactly through the middle between them, so the question is If the light deviates, where will it deviate to? Since given the conditions it's not possible to give an answer to this question, it means ... 2 The answer is not simple. The sensor you linked has 4 independent photodiodes, each with its own optical filter, and each gives a different result. In a perfect world, the 3 color sensors would exactly cover the overall spectrum of the white sensor, with no gaps or overlaps. AND the responsivity of each color sensor would be exactly the same as the white ... 2 Individual photons are not considered rays. Because of the wave and particle nature of photons, they are much more complicated than what they are generally thought of: a projectile of light. In fact, they do not have an exact measurable position, but do travel in straight line trajectories. What we consider rays are lines perpendicular to the wave front of ... 2 A "ray" in geometric optics is a locus of continuous propagation of light. Think of it as mapping where the energy is going in space. In principle there are an arbitrarily large number of them, but we draw a manageable number for visualization purposes. The various [letter]-rays were so named when people didn't know what they were beyond being things that ... 2 The first 2-D image you posted is a typical simplification for teaching purposes. In it, they use the height of the sine wave to represent magnitude, and the directions of the sine waves to show how the fields point relative to each other. The light itself however is not itself at all cone-like. You have to imagine this sine wave existing at multiple points ... 2 The link which NeuroFuzzy provided appears to be time lapse photography of a pulse of light from V838 Monocerotis, the most spectacular light echo in the history of astronomy, according to the European Space Agency. It is much brighter than even a supernova, but it is not exactly an explosion. The initial pulse happened in 2002. V838 Monocerotis didn't ... 2 The highly interesting thing is, brown absorbs all other colors but reflects red and green. This may be a source of your misunderstanding. The way humans perceive color is by having three types of photoreceptors in their eyes. Each of them has a broadish response function which means, that the "green" receptors react well to pure green light (as in ... 1 First of all using visible light to transmit data is definitely possible. We can argue as much as you want on the fact that the transmissions are not optimized and that the efficiency of such techniques are limited to very small distances (because of absorption and attenuation in the air). About UV I would say that there would be a lot of interference given ... 1 As you probably already know, reflective surfaces invert the handedness of a C-pol light. So one solution would basically put a source of light in front of you, and the other one at your back, so that it would be reflected more or less from the direction from which the first source radiates. About thin films, I could possibly think of something, but it ... 1 In the first of your data sets, it does look like the white value is (close to) the average of the RGB values. In the second one, the white value is significantly greater (average of RGB ~= 388, but W ~= 639). Looking at the data sheet, the W sensor includes a significant range of infrared in its spectral response (see the graph on page 5) which might ... 1 There is no simple conversion, it depends on the wavelength or color of the light. However, for the sun there is an approximate conversion of$0.0079 \, \text{W/m}^2$per Lux. To plug in numbers as an example: if we read 75,000 Lux on a light sensor, we convert that reading to$\text{W/m}^2\$ as follows: $$75,000 \times 0.0079 = 590 \, \text{W/m}^2 \, .$$ ...

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The issue may be one of purity at the surface of reflection. Silver oxide is black. The presence of black silver oxide on the surface together with un-oxidized silver may be leading to an overall gray appearance. The way to test this hypothesis is to prepare a pure sample of silver in an inert atmosphere, and another sample in an oxygen atmosphere, both ...

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