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Additional possibility: when light travels through a medium such as water, it slows down substantially. If light leaves water and enters air, it does in fact speed up when it enters the air. I have no idea how to calculate the time required for this "speed up" to happen, but if the time interval can be calculated or measured, you can calculate an ...


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Imagine sitting on an airplane and playing with a laser pointer. As you fire the laser beam upward perpendicular to the motion of the plane you see it go straight up to the ceiling. The same would be true on a spaceship from the point of view of its passengers. As for the people watching this happen on Earth, they will see you fire the photon at an angle ...


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The speed of light is constant in a vacuum. However, it can change direction in the presence of gravity so in a sense it does accelerate.


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Okay so I just witnessed this phenomenon and I'm not 100% convinced its entirely from diffraction, because, like Carl Witthoft said, "...an individual ray of light is not something you could see as a line, because it'd be pointing straight at your eyeball." I was looking at an orange colored light. I noticed that when my eye focused at a certain distance ...


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Black is not a color, it is a shade. In physics, we call something "black" when it does not reflect any of the incident light. However, all black bodies radiate. The frequency of that radiation, the black body spectrum, is a function of the temperature of the object and follows Planck's Law: $$B(\lambda, T) = ...


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You can find the conversion matrices for a variety of colorspaces here: http://www.cs.rit.edu/~ncs/color/t_convert.html and the format for integrating the input spectrum at RIT. Quoting from the latter: So, I'm guessing the part you're missing is the x,y,z curves in the second row there. Go to Wikipedia or some other optics-related site to get as ...


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I think you will have an easier time viewing the index of refraction from a speed-point of view. Consider the following: The energy of a given photon is determined by its frequency (color): $E = h \nu $ (h being the Planck constant) Assuming the photon does not lose energy when entering the material, its frequency must be conserved. However, as light is an ...


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You are mixing up two different things. The refractive index is usually defined in terms of the velocity of light: $$ n = \frac{c}{v} $$ where $v$ is the velocity in the medium. However the velocity is related to the frequency and wavelength by: $$ v = \lambda f $$ so: $$ n = \frac{\lambda_0 f_0}{\lambda f} $$ The frequency of the light, $f$, doesn't ...


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Using a camera that can capture "Motion at a Trillion Frames Per Second", this can be done at the laboratory scale. The technique used has been called femto-photography. (Image credit to Ramesh Raskar, Associate Professor, MIT Media Lab) Of course a camera that literally takes one trillion full frames per second is totally impossible with today's ...


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Let's say you build a ping pong ball counter. It increments the count every time a ping pong ball hits the sensor. You throw a ball, and it hits the sensor: Detected! You throw a ball across the sensor from left to right... no detection, because you didn't hit the sensor. Your eyeball is a light sensor, which creates pictures from the light that hits ...


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Mirrors flip in the dimension looking into/out of the mirror, not in the others. What changes left/right is what we present to the mirror. If you stand in front of another person, their right is on your left and vice versa, but their top is your top and their bottom is your bottom. That's not the mirror, that's convention for left/right. If you stand ...


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If some of the light is reflected off the dust at such an angle that it is diverted to reach the observer, the observer will see that light. However, those specific photons reaching the observer will not reach B (unless they are reflected there by the observer). Similarly, unless the observer is at point B (which is not the case in the question as asked), ...


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My masters project was on something like this (though with hydrogen alpha emission lines for gas clouds between galaxies rather than dust between stars) and the answer in that case (and almost certainly in this one too) is that you can't see it because it's just too dim, though using hundreds of telescope hours can get you close (maybe). Again, this is not ...


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It would be possible to see the progress of photons through space if the light pulse were exceedingly intense, and if the dust cloud from which they reflect were positioned and shaped to reflect the light toward us. Rather than shooting a beam from Point A to Point B, it would be better if the light source were between us and the dust cloud, as light ...


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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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What do light wave oscillations look like? You know what ocean waves look like. Imagine you're in a gin-clear ocean, and there is no surface. Now imagine an ocean wave passes you by. A light wave looks like that. By which I mean it doesn't look like anything. I know that light waves oscillate, but I don't know how. They don't really oscillate. ...


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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 ...


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This is due to the fact that different colors of light refract through glass (or whatever your glasses are made of) at different angles. Plain white light contains all other colors within it. When a beam of white light hits your glasses at an angle the yellow/red and blue/violet light separate from the main beam due to their different energy levels ...


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The ocean is blue because blue light is at the end of the visible spectrum. The wavelengths(lower energy) that produce red, yellow, green etc. are readily absorbed while the the photons that penetrate the depths correspond to blue and reflect that way. In shallow waters, green will be reflected from the sand or whatever color the sediment is.


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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 ...


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This is an old question, but I'm very surprised no one has mentioned this: You CAN'T see red-green. According to multiple websites, some studies showed that the human eye cannot see both red and green simultaneously, as the red cone and the green cone send out signals that cancel each other out. Similarly, blue-yellow is equally impossible to perceive. ...


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If you look at one specific point in space of your light wave, the electric and the magnetic field which is perpendicular to it are both in their turn perpendicular to the direction of propagation. So, both fields lie in a plane perpendicular to the direction of propagation. Both fields are varying with time in a sinusoidal way. Different polarizations are ...


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This is possibly not a full answer, but the Wikipedia article Polarisation of Light below seems to suggest to me that, because of the chance of partially polarised light occurring, there may not be a well defined answer to your question: So what do light wave oscillations look like? because the mathematical appearance of the light wave oscillations ...


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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 ...


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Following Einstein's comment that photons are indivisible units it is not really to explain why a photon has to be red shifted when it emerge from a gravitational potential. It does not radiate, hence it does not loose energy. More intuitive is the explanation that dependent from the gravitational potential, where a body is located, this body radiates ...


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The two effects have slightly different origins. Polarization of reflected light is a function of the angle of incidence and the refractive indices of the two media on either side of the surface - as described in this reference. Specifically, there is an angle (called the Brewster angle) where the reflected light is entirely polarized with the electrical ...


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Is it accurate that light loses energy in the absence of gravity and gains energy in its presence? No. Conservation of energy applies, even when it comes to gravitational redshift and blueshift. If you send a 511keV photon into a black hole, the black hole mass increases by 511keV/c². It doesn't increase by more than this. The descending photon doesn't ...


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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 ...


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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 ...


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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. ...


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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 ...


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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 ...


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The event horizon or 'point of no return' is the point where the escape velocity of an object falling into the black hole has to be greater than the speed of light (about 299,792,458 m/s). That concludes that light can't escape it, therefore there is a sharp, clear distinction between the event horizon itself and the background. If you moved towards the ...


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A black body is an ideal physical body that absorbs all incident radiation, regardless of frequency or angle of incidence. Thus, there is no light reflecting from it; for example, Black Holes. The definition of "Black" is different in Physics from Everyday Life. In Physics, color is a term that anglophones use to describe the various frequencies of ...


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I think your question is a matter of semantics. Our eyes are sensitive to radiation between approx. 400-700nm wavelength those frequency limits correspond to blue and red respectively. But the perception of color is actually quite a bit more nuanced than that - many things we see don't emit/reflect light at a single frequency, but our eyes/brain still ...


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Okay, so here's the deal about so-called "black bodies": we can in some ways define a thing's color by saying, "if I shine a light on it, what are the frequencies of the extra light that I get back?" In this precise sense, the Sun is black. The Sun can't help but shine the exact amount of light that it shines; if you shine a light on the surface of the Sun, ...


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Wikipedia says: The total amount of energy received at ground level from the Sun at the zenith depends on the distance to the Sun and thus on the time of year. It is about 3.3% higher than average in January and 3.3% lower in July (see below). If the extraterrestrial solar radiation is 1367 watts per square meter (the value when the Earth–Sun distance is ...


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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 ...


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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 ...


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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 ...


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To add to WhatRoughBeast's answer, this is because you are not really viewing the image under pure red light. The "red light" you used to create the image has small amounts of green and blue wavelengths, causing the white areas to appear more "white" than the red areas, even though they are both just slightly different shades of red. To quote WhatRoughBeast, ...


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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 ...


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Absolutely, though the accuracy might not be the best because clouds don't exactly have hard edges and it might be hard to differentiate between the red light from scattering and the tail end of the reddened sunlight hitting the clouds. Refraction might have an effect too, though thinking about it, I don't think this would change the calculation much. ...


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You can't see the light coming towards you. But you can see it as soon as it reaches your position. That means, that if you're in space and there is a star which is 10 times far than the speed of light in seconds. And if that star emits the light "now", then you won't be able to know that whether the light is emitted "now" or not. Instead, after 10 seconds, ...


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The simple physical model of the eye (or indeed a typical camera) is that it records just three values for each pixel. In your eye, this is because you have three different types of cone cells, called S, M, and L, peaking in blue, green, and red wavelengths. In a camera, the light is passed through blue, green, and red filters before having its intensity ...


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If two things pass through one another, it simply means they don't interact, or they interact very weakly. In many ways, this is the "natural" state of affairs: we only tend to think no two things can pass through each other because of the strong interactions between what makes up our everyday world. People, walls and chairs are made of stuff that interacts ...


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if for example a light beam coming from a single point of a galaxy, of a single wavelength, crashes with another wave coming from a different direction and angle, with a different wavelength, how is that they just pass one another and continue as if nothing has happen? In classical electrodynamics, where we treat light as electromagnetic waves that ...


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Your first question can be explained by the classical laws of electromagnetism. Classically, electric fields can be in a superposition where they either add or subtract. This means that waves which criss cross do in fact interact in the sense that they add up to a point of constructive or destructive interference in that region. For your second question ...


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Why clouds scatter red wavelenght? I think it is kind of unusual clouds. There is two processes going: absorbtion $A$ and reflection $S$ (scatter). I think that these type of clouds have density which is causing ratio of $S/A$ to be big enough for thin lower border so we could see that red light. I personally think that this type of clouds consists of ...


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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 ...



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