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I am confused with the realtionship between intensity and amplitude of wave. My understanding is that energy in a wave is proportional to its intensity; which is proportional to the square of the maximum height of the wave. is that a correct understanding.? If that understanding is correct I have a red right and I increase the brightness what happens.? Do I increase the amplitude of the wave.?

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The maximum height of a wave is also referred to as it's amplitude, and yes, you are correct. The energy of a wave is proportional to the square of it's amplitude, and by consequence, it's intensity depends on the square of $A$ as well. If the intensity were to decay as $1/r^2$ then it's amplitude would decay as $1/r$ as well. Increasing the brightness of your light would give it a larger amplitude and thus a larger intensity as well.

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If you insist on taking a wave, your statements are correct. The energy of the wave IS proportional to the intensity, which is in turn proportional to the square of amplitude of vibrations produced by the wave(in this case being the vibrations of electric and magnetic fields). When you increase the brightness, the amplitude of vibrations of the electric and magnetic fields increase, leading to an increase in the energy carried by the wave.

If, on the other hand, you consider photons, intensity is the number of photons reaching your eye per unit area per unit time. This again relates directly to energy; more the number of photons, more the energy. When you increase the brightness, you are actually increasing the number of photons. I do admit, though, that I can't bring an amplitude into this photon picture. Maybe someone else can help.

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  • $\begingroup$ At photon explanation, doesn't the frequency of photons play a role in the intensity? $\endgroup$ – physicsguy19 Oct 26 '18 at 17:21
  • $\begingroup$ Nope. The frequency determines the energy of each photon. The intensity is simply what it always is: number per unit area per unit time. You are trying to relate intensity with energy and hence with frequency. But in the photon picture, the energy of the radiation arises from two INDEPENDENT factors : energy of each photon which depends on the frequency, and the intensity. $\endgroup$ – Abhirup Mukherjee Oct 26 '18 at 20:24
  • $\begingroup$ Actually sir, you are wrong (I am majoring in quantum optics). Intensity is proportional to the energy flux, no matter the wavelength. Meaning for the same intensity, there are twice the amount of 700nm photons than 350nm photons. Furthermore, photons are just terms "counting" the energy in an electromagnetic wave, and classical light sources (e.g. lasers) emit a superposition of photon number, so increasing the wave amplitude is, in a way, more accurate of counting photons $\endgroup$ – Ofek Gillon Nov 2 '18 at 14:53

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