New answers tagged

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This is caused by chromatic aberration in the eye: short waves (blue) are refracted more (by the cornea and the crystalline lens) than long waves (red light). This isn't visible in normal conditions, but can be observed using your method. A good overview of the eye can be found in this ref., see page 49 for chromatic aberration. Because of it, a ...


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I think this question does not actually address the photoelectric effect because this would be strongly dependent on the frequency of the light source. To me this seems to be a question regarding the relationship between intensity, power and work. While this is technically not how an electron is emitted in real life, one could calculate the power incident ...


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I believe the dyes behave similar to glow in the dark materials; They fluoresce a certain wavelength over time after being radiated. In the case of these shirts you'll have different chemicals for different colours, but which are all "activated" by UV radiation. There is a pretty nice explanation on Wikipedia, but basically this: You can excite an ...


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To discriminate between the other two answers. Lifetime spectroscopy would be capable of distinguishing between photochromism (in this case, reversible ultraviolet-switchable reflectance, example: hexaarylbiimidazole where the transition time is milliseconds to seconds) and ultraviolet fluorescence (where the "transition time" is zero, although the excited ...


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There is an explanation on the Del Sol web site but it omits the technical details. This is probably because Del Sol regard it as commercially valuable confidential intellectual property, and I have to concede that they are probably correct. Anyhow, some Googling has turned up a suggestion for how it works but there is no proof for this idea so treat it as ...


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A quick look at the Del Sol How it Works page shows their explanation, HOW DOES IT WORK? The Spectrachrome® crystal reveals color upon irradiation by ultraviolet waves; i.e., sunlight. When a flower blooms, the result is the exposure of the inherent color of the flower. A Spectrachrome® crystal is similar in that an energy-shift occurs causing the ...


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Basic quantum mechanics implies that a photon is both a particle and a wave and any experiment you perform, and the result you get, depends on how you set up the experiment. Quantum field theories, or QFTs, are a combination of basic QM and Special Relativity. The main idea is that a field, (one for each particle) exists throughout the universe. A photon is ...


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If the sun were a true black body, then it would be black. The extent to which it differs from a true black body can be seen from the sun-light spectrum outside the earth's atmosphere. Overall, it is pretty close to a 5778 K black body, so the sun will likely be very dark if you remove its own thermal radiation. Upon closer inspection, you can see that it ...


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Whoa...whats that?sound waves travel faster in liquid than air; 4.3 times faster dependent on the temperature, The higher the elastic constant of the medium the faster sound travels and The tighter the molecular bond the higher the elastic constant. Light travels faster in air than water, simply because air is less denser than water.The denser the medium the ...


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For each wavelength you have been given (exactly) two values $\theta$, so $$\begin{align}n\lambda &= d\sin\theta_1\\ (n+1)\lambda &= d\sin\theta_2\end{align}$$ subtracting these two equations, we get $$\lambda = d\left(\sin\theta_2-\sin\theta_1\right)$$ You should be able to figure it out from there...


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No. If you go to a random spot in the visible universe, you will usually be far from any galaxies because the separation between galaxies is large compared to the size of the galaxies themselves. Since distant galaxies are so dim that we can't even see them, you certainly cannot see your reflection by them.


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You are just seeing differently illuminated clouds, as should be clear from this picture of an aircraft flying over a thunderstorm: Clouds are made of water droplets or ice crystal, which are transparent. When light is incident upon a cloud, it is scattered at the same wavelength (a process known as Mie scattering). This means that it remains of the same ...


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Wavelength is used as a convenience. It's much easier to imagine a photon with a 500 nm wavelength than to comprehend a photon oscillating 600 trillion times per second. But in reality that's all it is is a photon moving at the speed of light and oscillating 600 trillion times per second as it goes along. The photon completes one cycle every 500 nm. Many on ...


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Oxygen and Nitrogen do absorb 'light' but only in the ultra-violet region of the spectrum below approx 200nm, an area invisible to our eyes but easily observed by photomultipliers and similar detectors. This absorption is caused when an electron from a molecule's ground state is promoted to one of several electronically excited states. These states have such ...


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If you have a beam of light that is a millimeter wide (1/25th of an inch), then its width is approximately 2000 times the wavelength (since the wavelength of visible light is ~400-700 billionths of a meter, depending on color). Due to diffraction, that means that the beam will spread out by roughly 1 unit for every 2000 units it travels: send the beam down ...


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This is probably mainly because of two reasons: Psychological reasons. You want to see a lot of details on the sculpture, which it might be difficult to do if it is small unless you paint shadows on it. If you took a close-up photo of a miniature it would probably be easier too fool someone that it was real-sized if it didn't have shadows painted on it. ...


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What we call the "dark side" of the moon is just the side that is not visible from Earth. That side is illuminated by the sun just as often as the side that we can see. But when it is in the light, we don't see it. As for your other question: conventional flashlights work by sending a current through a light source (incandescent bulb, or these days ...


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When objects are very small, every source of illumination will appear to be "extended" - which softens the shadows and makes it harder to see the contours of the surface. By painting highlights and shadows, you reduce the impact of the extended source. See for example Why don't fluorescent lights produce shadows?


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There is a phenomenon that generates light without generating heat as by-product*. The phenomenom is called Electroluminescence, and is used in LEDs which are used in many electric equipment (the tiny lights in your computer to let you know it's on). We can stretch it a bit more with Fluorescence and other forms of creating liminous paint Also if you will, ...


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There are some materials, called Birefringent in which the index of refraction depends on the direction in which light is moving (often also dependence on the particular polarization of the light). This is due to the crystallin structure of the material which allows faster propagation in one direction or another. Another example would be a magnetized ...


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The surface of the Sun, as you can see from the following picture, is not uniform: But it is impossible to see it with your naked eye. You will need a telescope and specialized solar filters (more information here). Notice also that the picture is not showing visible light, but extreme-ultraviolet light. You will need a special telescope for that. Or you ...


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Short answer, the sun isn't on fire. Flames can flicker with wind or with pockets of fuel that might collect and burn in spurts or bits of water that can steam when the flame touches them, causing movement, or due to small bits of turbulence as the heat expands away from the flame. The visible surface of the sun is much more like a hot iron that has a ...


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The answer should be zero since each source is emitting light of same wavelength and same intensity thus same amplitude. The path difference between $S_4$ and $S_3$ is 10.5$\lambda$ thus their phase differnce is π and hence they interfere destructively giving 0 amplitude. Same is the case for source $S_2$ and $S_1$ which gives zero amplitude. Now net ...


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The first thing you need to do is make an assumption about the relative phase of the waves emitted by each source which with no other information you make zero. Then you need to add the amplitudes from all four sources taking account of the relative phase due to the distances the waves have travelled and then square the result to get the intensity.


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You haven't seen these effects because your eyes are rather insensitive to wavelength changes. A moderately resolving spectrometer can detect these changes quite easily. It's called "Stark effect" and it can be observed in atomic spectroscopy: https://en.wikipedia.org/wiki/Stark_effect. For magnetic fields the analog is called "Zeeman effect": ...


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The surface has aged more : events move faster/happen quicker on the surface than in the core. So there would be a blue-shift in the light reaching the core. Light coming out of the core would be red-shifted. But in the case of the Earth (and even the Sun) the effect is incredibly small. https://en.wikipedia.org/wiki/Blueshift#Gravitational_blueshift


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Lots of questions! I'll try to get each of them: the object become infinitely heavy and requires infinite energy to move Light (photons) have no rest mass - ten, a hundred or a million times nothing is still nothing, so the (non-existent) mass isn't relevant - as you say further down. (Light does, counter-intuitively, have momentum, but that's ...


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The graph of angle of deviation vs angle of incidence is a $U$ shape. The fact that the angle of deviation is the same for these two rays means that a ray which is incident at $40$ degrees to the normal will emerge at $60$ degrees to the normal. This allows us to find the refractive index. $$\sin i_1 = \sin 40 = n\sin r_1$$ $$n\sin i_2 = \sin r_2 = \sin ...


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The refraction angle will be different for different materials provided these materials have a different index of refraction. Reversely, by measuring the angle you can determine the index of refraction of the material. You can look up Snell's law.


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The most direct answer I found is that N2 and O2 are very simple molecules. 2 atoms, tightly bound, no angles. Source From Source: Because the nitrogen gas molecule is so simple, it cannot do very much with the light energy that it absorbs. It can spin or vibrate only a little bit by stretching and pulling. Oxygen acts pretty much the same ...


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Quantum mechanics says that only very specific wavelengths are acceptable for exciting an atom, wavelengths that are not those (most of the spectrum) passes by the atom without interacting much (just some scattering in the case of visible light). The allowed wavelengths are the ones that have energy extremely close to the energy difference between two ...


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Yes the singularity of a black hole would be incredibly bright once you get to the center because the light and energy can't escape the black so it is all compressed into and infinity dense small point and even if it is just normal matter going inside the black hole it would even make it more bright because the black hole tidal forces would rip the atoms ...


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Let us dive into the light clock thought experiment, Special relativity is based on two postulates, There is no such thing as absolute motion. Phrased another way, all laws of physics should be invariant under changes in inertial frame. The speed of light is measured to be the same value in all inertial reference frames. Let's say you are on the train ...


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In the classical electromagnetic theory the word "photon" has no meaning. Nature as we know it from our experiments is basically "quantum" , i.e. it follows the equations and postulates of quantum mechanics. The equations and laws of classical mechanics can be shown to emerge from the quantum mechanical underlying framework. Classical mechanics is very ...


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First find the wavelength of light which corresponds to the OP, by solving the diffraction formula: $mλ = d \sin(θ)$. The corresponding wavelength of the electrons impinging upon the sample is given by the de Broglie relationship: $\lambda = hp = h\gamma m_e \beta $c ; where the symbols have the usual meanings: $h$ is Planck's constant, $c$ is the speed ...


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Any static setup will produce a static output. As mentioned in the comments, a diffusing plate will mix things up reasonably well, but with annoying side effects such as beam angular dispersal. What we actually used in some setups was a rotating phase plate. First we built or bought (I forget which) a phase plate with more or less random phase patches, as ...


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You need to mix them incoherently. If the beams are from different sources, then you're pretty much there, as they won't stay in phase. You can't represent incoherent light with Jones vectors (2 components), but you can do it with density matrices (this approach has a huge importance in quantum mechanics, where incoherent mixtures are most important). So, ...


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I expect the answer is no because unpolarized is a 'random' mixture of all sorts of different polarizations mixed together. Much easier to make polarized from unpolarized with a filter. Perhaps in theory possible to make something that looks like unpolarized from polarized but requiring lots and lots of beam splitters wave plates etc. to make up the ...


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One way to look at this is to go to the definition of the index of refraction $$n = \frac{c}{v}$$ v is related to the wavelength and frequency as $$v = \lambda f$$ So the index of refraction becomes $$n = \frac{c}{\lambda f}$$ Frequency remains the same when light goes into a new medium, so the only difference between light of different colors is the ...


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The limit c comes from Maxwell's equations for electromagnetic waves in vacuum. This does not know about photons, is pure mathematics given E and B fields and permittivity and permeability of free space. $$c= \frac1{\sqrt{\mu_0\varepsilon_0}}= 2.99792458\times 10^8~\mathrm{ms^{-1}}\;.$$ If photons do have a mass within the experimental limit, Quantum ...


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Interesting question. Taking a stab at it - not absolutely sure this is correct, but let the comments begin. In the frame of reference of Earth, the light travels straight out to the reflector, and straight back. You are asking about the case where an observer is in a reference frame that is moving with respect to Earth/moon, and the picture would have to ...


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Agarose is a polysaccharide, and like all polysaccharides it is bristling with polar hydroxyl groups. This means the agarose chains interact very strongly with each other by hydrogen bonding. If the solution is heated to a high enough temperature to break the hydrogen bonds (around 90-100ºC) then the agarose molecules behave pretty much like any other ...


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This question raises several concerns. From a purely signal processing point-of-view, where the aforementioned pulse is not a photon, but rather a time-domain pulse of a certain duration (here less than 10^-100 seconds)--then NO: you cannot have pulses shorter than the period of wave in question--as the wave an be defined. Consider sound: when a 440Hz A note ...


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Forget for a while about Snell's Law, Maxwell's equations and take a look in the Figure. As Feynman said : "In Figure, our problem is again to go from A to B in the shortest time. To illustrate that the best thing to do is not just to go in a straight line, let us imagine that a beautiful girl has fallen out of a boat, and she is screaming for help in the ...


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The other answers are not strictly true, as "whiteness" is a neurological perception. We can perceive a neutral color when the red, green, and blue retinal cones have roughly equal stimulation, but that color is only "white" when it exceeds some magnitude. Otherwise it's what we call 'gray.' Further, in very dim conditions, only the retinal rods respond, ...


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White light is made of visible photons of many colors (frequencies). Our eyes mix the different frequency and interpret it as white. Photons come in all frequencies and through evolution our eyes have evolved to register visible photons from red to violet. Blue photons are higher frequency than red photons. White light is not a photon but a mix of photons


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The microscopic mechanism of emitting photon in a solid is the transformation of kinetic energy of atoms into EM energy. If an atom is in an excited state due to collisions among other atoms, then it will emit photon when it jumps into the ground state, and the energy of the photon is $$ E=\varepsilon(\text{excited state})-\varepsilon(\text{ground ...


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No, each color in the spectrum has a characteristic frequency. Every light source has a so called spectrum of frequencies. The relative intensity of these frequencies determines what color you see (or not). For example, the sun looks yellow because it's peak intensity is in the yellow wavelength. White light comes from a source consisting of a very broad ...


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To add to the answers so far, here's a outline of a pretty crude calculation you could do. Let's assume the house is in space. That way, we won't have to worry about the other parts of the Earth equilibriating with the house. Now let's assume your house is a perfect blackbody (which may or may not be a good approximation depending on how your house is ...


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Yes. All matter that interacts with light absorbs it to some degree. This is true whether you're discussing UV or infrared or visible light. For example, the earth wouldn't be nearly as warm as it is if it didn't absorb visible light.



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