I am confused, Classical wave theory says that Intensity of the light(wave) is the proportional to square of the amplitude. How intensity is proportional to the square of the amplitude?
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1All wave power (intensity) equations are proportional to amplitude squared; not just electromagnetic. – Carl Witthoft Jul 06 '15 at 13:03
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Actually @uhoh, it should be the other way around: your linked question should be closed as a dupe of this one. – Kyle Kanos Aug 04 '17 at 10:21
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1@KyleKanos I disagree. The answer there is both much better, and accepted. There is nothing about closing as duplicates that require it to point backwards in time. It should point to the highest quality answer. That's sort-of the fundamental goal here wouldn't you say? Take another look at this excellent answer and then look here. It's not "who answered first" - this isn't a competition. Let's point future readers to the best possible answer. – uhoh Aug 04 '17 at 10:27
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@uhoh I'd argue the answers are better here, but had one noticed initially, I bet it would have been closed as dupe of this – Kyle Kanos Aug 04 '17 at 10:27
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@KyleKanos the best answer to "Exact relationship between electric field and intensity; I often see $I(t)\propto |E(t)|^2$. What is the exact form ? Which constants are missing to make this an equality and why are they so often omitted ? is certainly there, and certainly not here. Look again. – uhoh Aug 04 '17 at 10:29
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@uhoh again, your opinion. I think the answers here suffice (and I'm leaving it at that) – Kyle Kanos Aug 04 '17 at 10:32
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@KyleKanos I don't see the kernel of the answer here anywhere; $\sqrt{\frac{\epsilon_o}{\mu_o}}$. It really is not an opinion, no. It is not here, and it is there. There is no answer to that question here. – uhoh Aug 04 '17 at 10:35
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If we take a mechanical wave, a particle on a wave oscillates with simple harmonic motion, its maximum velocity is given by $v = 2\pi\times\hbox{frequency}\times\hbox{amplitude}$. If the amplitude is doubled the maximum velocity is doubled. The kinetic energy = $\frac{1}{2}\times \hbox{mass}\times\hbox{velocity}^2$, so if you double the velocity you quadruple the kinetic energy, thereby quadrupling the intensity of the wave.
ZeroTheHero
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didas
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1Does the kinetic energy equation even apply here provided we are talking about a wave? Do you calculate the kinetic energy of each individual particle at a certain instant to its direction of movement and then integrate all particles' energy to figure out the wave's energy at that instant? But the particles move perpendicular to the wave, so the wave's net energy will be zero. And you have to presume that it is a mechanical wave. Because, as light doesn't have mass, energy will always be zero. Moreover its velocity is constant in a medium. I'm failing to make any sense of this explanation. – NurShomik Jan 06 '22 at 19:38
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It is effectively defined that way because it's the simplest form that satisfies the relevant conservation of energy equation.
As discussed by Feynman, "There are, in fact, an infinite number of different possibilities for [energy density] and [flux], and so far no one has thought of an experimental way to tell which one is right!"
lemon
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1See Eugene hecht (optics),page 48 ,intensity is average energy per unit area per unit time.so you can take average value of magnitude of Pointing vector to measure intensity and you get intensity proportional to square of the amplitude. – Paul Jul 06 '15 at 08:53
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@Paul But the definition of the Poynting vector is arbitrary. It could be that the time-average of all possible Poynting vectors necessarily give in an amplitude-square result, but I can't find any evidence for this. I invite others to correct me. – lemon Jul 06 '15 at 08:55
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@lemon: Isn't this just a gauge degree of freedom? In that case any possible definition will have the same physical result and we don't have to care, so we might as well go with the standard one. – CuriousOne Jul 06 '15 at 09:20
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Light has three properties: Wavelength, speed, and amplitude.
The wavelength determines the type of light (color, etc.). Speed is determined by whether light passes through a vacuum or some material. That leaves only amplitude as the variable available for intensity independent of the type of light and/or the medium that the light is passing through.
The more photons emitted per unit time, the greater the intensity of the light. A single photon has wavelength and speed. The photon's energy is the product of Planck's constant and the photon's frequency [E = h * f, or E = (h * c) / lambda]. Therefore the energy of a single photon is not measured by its amplitude.
However, the amplitude of a light wave depends on the number of photons per second being emitted. The greater the amplitude of a certain type of light, the greater the number of photons per second of that type of light. So if you want to compare intensity of similar types of light, the amplitude is the variable of choice.
When light waves interfere with each other, areas of greater intensity result when photons pile on top of each other, and this is measured by the greater amplitude.
Ernie
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