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The formula you want is called Planck's Law. Copying Wikipedia: The spectral radiance of a body, $B_{\nu}$, describes the amount of energy it gives off as radiation of different frequencies. It is measured in terms of the power emitted per unit area of the body, per unit solid angle that the radiation is measured over, per unit frequency. $$... 11 The wavelengths of light emitted can be calculated using planks law and the temperature of the object. For your average 100W incandescent light bulb, the filament is 2823 kelvin according to google. The spectral radiance, B, is equal to$$\frac{1.2\cdot10^{52}}{\mathrm{wavelength}^{5}\cdot ...

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An object at 1000K glows red because that is basically dominating the radiation that it is emitting that we can see. Most of the radiation it emits emerges in the infrared part of the spectrum to which our eyes are not sensitive. Your edited question asks about the appearance of hot metals. The chart below is an example of a "colour-temperature" chart for ...

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This is an answer by an experimentalist who had been fitting data with mathematical models since 1968. When fitting data one goes to the simplest mathematical models. When the data display variations in time and space the Fourier expansion is extremely useful because it gives the frequencies and amplitudes that will fit a periodic data set. One gets as ...

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Photons undergo angular acceleration in very strong gravitational fields, gravitational lensing.. An acceleration can be defined in its change of direction, angular acceleration in radians/second^2, so the answer is positive, yes, light can be accelerated, but its speed will still be c, only the direction relative to the gravitational source changes.

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A model for the interaction of light with atoms and molecules treats the charge distribution as an electric dipole, because the particles consist af separate positively and negatively charged particles that can be polarised to have a non-zero electric dipole moment. Neutral particles where no (internal) charge separation is possible should not be affected by ...

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They produce heat because the surfaces on small scales are rough like canyons rather than flat like the ocean. As these rough surfaces come into contact with each other they repel. When two atoms are brought very close together they store potential energy. When they move apart that energy becomes kinetic. However, this kinetic energy generally isn't enough ...

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The Poynting vector is useful not because we say so, but because of Poynting's theorem, which in essence states that the Poynting vector can usefully model how electromagnetic energy is moved around a system of changing electromagnetic fields. More precisely, you can define a quantity $$u=\frac12\left(\varepsilon_0\mathbf E^2+\frac1{\mu_0}\mathbf ... 2 In classical electrodynamics, the process of how much light refracts, passing through the glass, and how much light reflects, is determined by the Huygens-Fresnel principle. This principle, named after Christiaan Huygens and Augustin-Jean Fresnel, is a method of analyzing the wave propagation patterns of light, especially in diffraction and refraction. It ... 2 The connection of heat to entropy in thermodynamics is through: where S is entropy Q is heat T is temperature, and it is through differential changes. This in no way means that heat is entropy . The easiest way to acquire an intuition of entropy is to read up on the statistical definition which can be proven to be the same as the thermodynamic ... 2 I will give a closed form for the integral in Chris Cundy's answer. Doing the substitution u=nx, we get$$ \sum_{n=1}^{\infty} \int_{x_{min} \cdot n}^{\infty} \frac{1}{n}\left(\frac{u}{n}\right)^3e^{-u}\mathrm{d}u \sum_{n=1}^{\infty} \frac{1}{n^4} \Gamma(4,x_{min}\cdot n)$$where \Gamma is the upper complete gamma function. We write a=x_{min} ... 2 To produce radio waves one use an AC generator and an antenna rod. The generator accelerates electrons in the rod periodically and this the acceleration is provided by photon emission from each involved electron. An antenna emit periodically photons. To send information one can overlay the generators oscillation with other modulations. What happens on the ... 2 In terms of a dielectric, it means there is a linear constitutive relation between the vectors.$${\boldsymbol D} = \epsilon_0 {\boldsymbol E} + {\boldsymbol P}$$Or$${\boldsymbol D} = \epsilon_r \epsilon_0 {\boldsymbol E}$$where \epsilon_r \epsilon_0 is a scalar relative and vacuum permitivitty. This way, there is now a linear relation between the ... 2 Because of diffraction. When the photographic plate is exposed, it blackens and change its refractive index in a spatially varying manner. When illuminated again by the reference beam it can be considered as an amplitude transmittance. In 2D you would define it as the complex function:$$t(x,y)= T(x,y)e^{i\theta(x,y)} Let's say your reference beam can ...

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I'd like to add another take on Cape Code's answer. Holography works because, given reasonable physical assumptions, solutions to the Helmholtz wave equation are uniquely defined by the values of the solutions on one plane. So if we can light a phase / amplitude mask encoding a particular wave equation solution on a plane with a plane wave from a laser, the ...

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The occurrence of pair production within the detector depends on gamma energy and detector material. The occurrence of escape peaks in addition to a given full-energy peak depends on detector geometry and sample geometry. If the gamma energy is large enough to make pair production relevant, the photon may disappear and be replaced by an electron and a ...

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Picture yourself looking into a large mirror on the wall. Now picture the mirror is made up of smaller, tiled mirrors. You will still see your reflection. If you begin to remove the tiles, so that there are only a few left, you can still use them to reconstruct the image of your face that was given by the original mirror. This is what is happening with ...

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If you had a laser you wouldn't see it unless it was aimed at your eye (ouch). Or if there is dust or such around for it to scatter off of. And scattering is the key. If you want to see something then it either has to get to your eye or it needs to deflect something towards your eye. If you have a beam of electrons you could try to get something to ...

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Photons and electrons are elementary particles and their behavior is predicted by quantum mechanical equations. Quantum mechanics predicts probabilities for a reaction to happen. Photon photon interactions have very very small probabilities of happening. Thus photons pass through each other for all measurable purposes at low energies ( light and below ...

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Greatly rewritten based on feedback in comments In order to understand this issue, it is worth considering what a telescope (or any optical / radio imaging system) really does. Taking a simple parabolic mirror, the shape is chosen such that the total path length for all rays "from infinity" to the focal point is the same. By making the path lengths the ...

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RIC involves magnetic fields oscillating at a high frequency. The system won't pick up any other frequency. A constant-in-time (DC) magnetic field magnetic field has no direct effect because 0 Hz is the wrong frequency, and it has no indirect effect by the superposition principle. I guess it's possible that there may be a magnetic material in the RIC ...

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Two different wave concepts are being confused here and called an EM wave. There exists electromagnetic radiation from light to radio waves to gamma rays , and for that see this answer of mine which explains how the classical wave of the Maxwell equations is an emergent phenomenon: it is built up by photons. And there exist the alternating currents in ...

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But I want to know, if an incoming photon is unable to excite an electron, then why not all the photons pass through glass? i.e. photons should not reflect off glass, all the photons should pass through glass. While I don't know the exact details of reflection off glass (it is related to solid state physics from what I recall, with plasmon and such ...

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Don't confuse theory with reality. Electromagnetic theory (i.e., the theory of Maxwell) talks about waves of electric and magnetic field strength. The standard model of particle physics talks about photons---discrete particles/packets of energy---whose appearance are governed by probability densities that obey wave-like laws. As far as we know, gamma rays ...

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Considering light travels at relativistic speeds much faster than you can in your car, the exposure is the same. The same logic does not apply as for rain as rain doesn't fall at relativistic speeds.

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The real underlying basis of an electromagnetic wave is a synergy of zillions of photons. In this sense it is only macroscopically that the classical theory applies. The way the build up happens, photons into an electromagnetic wave, is not simple but an example can be seen here. Hand waving: the photon as an elementary particle is a quantum mechanical ...

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I'm guessing this is related to your earlier question, Will neutral particles be affected by EM waves?, and you're puzzled that Rayleigh scattering by air and reflection/refraction by solids and liquids seem so very different. The answer is that in Rayleigh scattering each scattering object behaves as an independant scattering centre, so the scattering is ...

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If the aperture width is above cutoff width, then as you say, there are propagating modes. This means that the aperture acts as a waveguide, with a propagation constant $\beta$ with a real part much larger than its imaginary part. (The imaginary part of $\beta$ quantifies damping.) A good reference for waveguides is Snyder & Love, Optical Waveguide ...

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Many of the things you write sound OK. But I wouldn't say that the other directions in the graph are mere strengths, they indicate the actual value (strength/magnitude and direction as well) of the electric and magnetic fields. First note that technically the electric field is a vector and it is a field so it should have a vector (possibly zero at every ...

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Well the imaginative experiment you have put forth is a fairly satisfactory method I would say. The wavelength of a guitar string is simply the distance between two consecutive peaks or troughs of the oscillating string (for simplicity let the frequency be constant). The wavelength of the electromagnetic field, or light, is just the corresponding value for ...

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