# Tag Info

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I bet that the wire isn't a constant diameter, varying slightly along its length. This means that the reflection angle will also vary, so that any one point on your circle of light may be receiving reflections from multiple locations on the wire. Bingo: diffraction.

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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 ...

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There are a couple of ways of knowing the wavelength of laser pointer. 1) Using Snell'law (law of refraction): A light passing the border between two media whose refractive indexes vary. The incident light PO of wavelegth, $\lambda_{1}$ travelling in a media of refractive index, $n_{1}$ is refracted in to another media of refractive index, $n_{2}$, with a ...

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From your description, I think you are talking about the diffraction pattern from a pinhole, and that is an Airy Disk. https://en.wikipedia.org/wiki/Airy_disk It shows a central maximum with surrounding rings which decrease in intensity as you get further from the center of the pattern.

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Parallel rays reflecting on a concave mirror do intersect at one point, the focus, if the mirror is a parabola (in 2d plane geometry) or paraboloid (in 3d space geometry).

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TVOC 1.pdf Snell's Law.pdf Optical WedgeA.pdf http://www.raydextech.com/led-collimation.html

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Intensity is an objectively measurable attribute of light. It is the rate at which energy is delivered to a surface. Intensity is energy delivered per unit time per unit area. The intensity of light is a measurement of photon irradiance, which is the number of photons delivered per square meter per second. You can measure intensity with a photoelement, ...

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When we say a red ball absorbs other colors, it doesn't mean it absorbs them perfectly. It can reflect non-red light, just less strongly than it reflects red. Beyond that, color vision is extremely complex. Your brain does amazing things trying to compensate for the color of light in a scene.

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I believe the point Mew is trying to make is that 'red' is a range of wavelength. Although your laser has a very narrow range of red wavelength, I suspect the range of wavelength reflection in the green pigment of the ball is much broader and may even have a range of red in it. An interesting, although perhaps unrealizable experiment would be to find a ...

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One half of the problem is light. Light is an oscillating electromagnetic field. The frequency of oscillation determines the color. Higher frequencies have shorter wavelengths and are blue. Lower frequencies have longer wavelengths are red. Light is a mix of a range of wavelengths. Sunlight contains a wide range of wavelengths, including some that are too ...

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Both operations are equivalent, up to a local phase in the second mode. In particular, if you shift the second basis vector's phase by $i$, then you will turn $H$ into $A$. In a beam splitter this is perfectly natural, because the phases of the output modes are not particularly well defined, and you can always model the difference between the two operations ...

2

I think it might be one of those things where people do something because everybody does. I agree with you, a figure of merit that includes noise would make more sense. But, as the circuit designer that I am, I could also say that that wouldn't be the end of it. For example, in the classic trans-impedance amplifier used for these kind of detectors the ...

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The "blacker" things are the less they reflect. That doesn't necessarily mean they are less visible - just less reflective. Matt coatings are probably your best bet. There is "the blackest surface there is" - nanotubes that trap the incident light and don't let it out again. Essentially they make the photons interact with the surface multiple times - if you ...

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I have a 405nm laser pointer, which emits light right on the edge of the visible spectrum. It is easy to see that this wavelength of light appears out of focus through the eye likely due to chromatic aberration. I would imagine this is because our eyes are not made to see that kind of light. I've also noticed that mercury vapor streetlights have a faint ...

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What emission lines? Unless you are looking at the chromosphere or corona the Sun does not have emission lines. Of course you can estimate photospheric abundances from a photospheric absorption line spectrum. That is how the solar abundances are estimated for most elements. However, you need a good spectrum to perform a detailed analysis. It should have ...

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There are photons traveling in all directions, not just the dozen or so you show. The further from the source the telescope is, the smaller the amount of solid angle it covers and the fewer photons it will gather. A $1 m^2$ telescope pointed at the sun will receive about $1.4 kW$. Taking a typical photon energy of $2 eV$ that is about $4.2E21$ ...

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Using the basic equipment at your disposal, this will not be possible. A naive approach would be to take the measured line amplitudes, divide by the line-integrated absorption coefficients at the wavelength of each line for some nominal set of parameters characteristic of the solar atmosphere, then divide out the emission coefficients and compare the ...

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This question is too broad. It involves ALL the objects in the universe which have a surface, i.e., everything. I'm going to avoid giving a lecture here. In some liquids and most gases the electronic structure of each individual atom or molecule is enough to describe their spectra. The "property" you are looking for in the case of solids is the band ...

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Several answers here already talk in great detail about how electron orbitals affect if a photon will be absorbed or not, but this is not the whole story. The color from reflected radiation is indeed the only factor if the surface is completely flat and perfectly reflective, excluding the black-body radiation, but most surfaces are not. Take for example all ...

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Your iPhone is a pretty good grating. I just did a simple experiment with an iPhone, a green laser pointer and a sheet of graph paper. This was the result: The display of the iPhone 6 has a resolution of 326 ppi - meaning we have a "grating spacing" of 25.4/326=0.0779 mm. Different models have different resolutions - make sure you find out what your ...

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Generally and fundamentally, the answer to your question is no. Photography / 2D imaging is the prototypical example of the mathematical notion of projection, which, by its definition, destroys information by extracting "class" information from sets of objects: forgetting about fine differences and reporting only the class. Here, for example, if $z$ is the ...

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Laser modes are the eigen-modes of a laser resonator: only specific distributions of electro-magnetic field can "resonate" in each particular resonator. Due to the 3D nature of our space, each mode is described by 3 numbers, or indices, $m$, $n$, $q$. The latter is the longitudinal mode number, and is easy to understand: to form a standing wave, the ...

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If you've got a grating with known distance between the slits, you can use diffraction: let the light fall in perpendicular to the grating and place a screen a few meters further away. You'll find the maximum intensities (the light dots if you've got a grating with 100 or more slits per mm) under an angle of: $$\sin\theta_m = n\frac{\lambda}{d}$$ with $n$ ...

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The atoms in the lattice can be thought of as coherent re-radiators of the incident photons. This is not unlike the scenario we have in a double slit experiment, where a Huygens construction of the wave front considers each point in the slit as a radiation source. So it might be "opinion" but I think that diffraction is an appropriate word to use.

1

As ACuriousOne commented, if the smallest gradation on a ruler or other measurement device that you can observe is $1 \mu$ that's one definition. In my world of optics and imaging, there are similar definitions. For example, if you have two imaged spots of light separated just enough to satisfy the Rayleigh Criterion , their spatial separation is ...

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An endoscope uses a bundle of multimode fibers. The number of fibers determines the resolution of the image transported through the fiber bundle. The angle of light isn't preserved in a single fiber. Usually - depending on the length and bending of a such a multimode fiber - angles/modes will be mixed in some way; therefore you need a bunch of fibers to ...

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Reflecting energy back into the photosphere using mirrors is equivalent to asking what happens if we artificially increase the opacity of the photosphere - akin to covering the star with large starspots - because by reflecting energy back, you are limiting how much (net) flux can actually escape from the photosphere The phenomenon could be treated in a ...

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Nonlinear optical elements are called nonlinear precisely because of the behaviour you note: because the optical response of the material does not depend linearly on the driving fields. The response may then have a quadratic or higher dependence on the driver, which is usually written in the form $$\mathbf P =\varepsilon_0 \chi^{(1)} \mathbf E + ... 2 Non-Linearity means that the dispersion relation becomes non-linear. Linearity is an assumption which only holds true for low intensities. Almost every material has some non-linear effects if the light source is only powerful enough. The polarization vector for example becomes: P = P_0 + \varepsilon_0 \chi^{(1)} E + \varepsilon_0 \chi^{(2)} E^2 + ... 0 If the two sides of the prism are parallel, then the ray will exit the prism at the same angle it entered, just offset. If you need the offset, or the faces are not parallel, then you need to calculate the exact path the ray takes through the material. That requires knowing the index of refraction for the material. If you have that, then you can use ... 1 Your answer appears to be correct, bar the lack of any electric charge in your formula. If the applied force is solely due to a cosinusoidally varying electric field at a given position (you can ignore the magnetic component of the Lorentz force only if the charge moves non-relativistically), then so is the acceleration. Integrating this with respect to ... 2 Actually, people have answered the question you seem to have meant, rather than the question you've asked. When a change in index of refraction causes a change in the direction of a ray of light, this is refraction. It is not dispersion, and has nothing (immediately) to do with dispersive mediums. Dispersion (rays of different wavelengths being refracted ... 0 q is a parameter which describes the distribution of a Gaussian beam with respect to the optical axis. You can think of the q parameter as a bundle (a 'pencil' in Born & Wolf parlance) of optical rays, each described by its own position r_i and slope \theta_i. So, using the transformation on the q parameter that you describe is all you need ... 0 I think this originated with Hadamard and his Method of Descent. See Lectures on Cauchys Problem in Linear Partial differential Equations--starting on page 7. His results were that waves in two dimensions did not propagate sharply, but had a wake (a tail, ..). Eg. a circular wave propagating in two dimensional space vs. a spherical wave propagating in three ... 0 Think of it this way: You are way up in the sky and the distance you see as a centimeter could be meters long since you see objects getting smaller as you go farther away. Assuming that an ordinary water wave travels with a velocity of 3 or 4 m/s at a maximum, it is not hard to imagine that you are seeing them as if they are standing on the ocean. ... 1 I imagine this effect has to do with the fact that velocity is relative. When you're on the shore, you gauge the velocity of the waves with respect to the shore. When you're in a plane, you're likely gauging the velocity with respect to the other wave crests, which are moving at the same velocity and so there is no apparent movement. 1 There are essentially two ways of generating coherent radiation from solid-state devices: Classical electronic oscillators, in which charge is made to oscillate back and forth within a device... the frequency of radiation corresponds to the frequency of charge oscillation. Solid-state lasers, in which charge-carriers undergo a transition between two ... 0 The light rays coming from one point on the object are indeed parallel. However, the sets of parallel rays from different points come from DIFFERENT DIRECTIONS. By the way, the object "at infinity" is a generalization. Of course, all real objects we observe are positioned at a finite distance from our eye. It is just that the divergence of the rays is ... 0 The rays from an infinite object are parallel for every point on the object but are not parallel for two different points on the object. Hence different points on the object form an image at different points on retina 1 The focal length of the eye's lens changes based on what is being looked at. Ciliary muscles add or remove tensions from the lens to changes its shape, and thus its focal length. The closer the object, the shorter the focal length. https://en.wikipedia.org/wiki/Ciliary_muscle https://en.wikipedia.org/wiki/Accommodation_%28eye%29 0 It is not possible. If you could build such an optical system, you would have a perpetual motion machine of the second kind. Suppose you would have a passive (no batteries allowed) optical system capable of creating images brighter than the source at which you aim it. You aim this device at the sun and thereby create an image brighter and hotter than the ... 0 In the 2 glasses there are 4 surfaces, i.e interface air/glass, and 8 surface orientations (a..h) and plenty room for interference between reflections and the main beam. LASER (air) a1b (glass) c2d (air) e3f (glass) g4h At each interface the is reflection that will be reflected forward again (self-interference) look for iridiscence in ... 2 Imaginary wavevectors are possible and, as ptomato's answer correctly points out betoken evanescence. I'd like to add a few words to his answer that might help clear up your confusion. Imaginary wavenumbers always betoken Evanescence. Sometimes the vague term "nearfield" is used to connote something not propagating. Evanescence is NOT dissipative; this is ... 2 Yes, indeed the out-of-plane component of the wave vector of a surface plasmon is imaginary. A purely imaginary wave vector means the wave does not radiate in that direction, but instead is evanescent. (That's what you get if you plug in a purely imaginary k_z = -i\alpha into the formula$$ E(z, t) = e^{i (k_z z - \omega t)} = e^{-\alpha z} e^{-i\omega ...

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If energy is lost, it must go somewhere. So, there won't be any "interference" where the beams overlap which saps the power. Unless, and this does happen, the lasers are powerful enough to ionize the air, in which case the light is blocked at that point and converted to heat, which will never make it to the surface. See this article on laser-induced ...

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When the mean number of photons is huge, the Heisenberg uncertainty becomes negligible and "disappears" (formally it looks like $\hbar\to 0$). Thus, such a coherent state becomes quite classical one.

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In this particular example the photoluminescence is of quantum wells. The reason for the asymmetry is because the density of states is not symmetric. At the low energy side the density of states has a excition Lorentzian line shape to the absorptivity. At higher energies the density of states becomes step like. The photoluminescence intensity,  I(E) ...

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A Gaussian beam has a width that changes with distance because of diffraction, which is an effect that takes place in any wave phenomenon. It has a pretty similar description to the Heisenberg uncertainty principle in QM if you're familiar with that. Namely, as the position in the $x$ and $y$ directions (with the optical axis pointing in the $z$ direction) ...

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The problem is that you'd need your flat mirrors to be the same sort of size, if not larger, than your liquid mirror. But the whole point of liquid mirrors is that you can make very large mirrors without the problems of the mirrors deforming under their own weight. So your suggestion just exchanges the problem of making a very large parabolic mirror for the ...

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At a guess, the effect rises from the fact that your interferometer is not properly aligned. The presence of linear, rather than circular, fringes suggests that there is an angular misalignment. Then moving the wedge causes a lateral shift in the intersection point of the beam and the angled slide, which results in a shift in the apparent position of the ...

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