The Wavefunction
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Conservation of energy states that energy can never be created or destroyed etc
So, when I turn off a light switch, what happens to the light energy?
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Gogregg
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#2
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It dissipates, I believe.
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Zarek
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#3
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Turns in to heat, at a guess..
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aamirac
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Energy can be transferred
Light to electricity with wasted heat energy
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The Wavefunction
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So dissipated as heat is the general consensus. Although it seems odd to me at the speed at which it transfers. Think there might be more to it, but dissipated as heat will do for me, now.
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Smilin’ Knight
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#6
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when you turn off the light the process of changing electricity into light stops. any light in the room at the time is absorbed into the walls/other stuff and as it does that it is converted into heat
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General Josh
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It does not dissipate, it is absorbed by it's surroundings. I'll keep it simple, but basically if you want to be able to see light, you have to destroy it. It you know anything about light and wave particle duality, then you'll know that light is an electromagnetic wave which is made up of particles called photons. When you destroy a photon, you can see it. That's how you can see, basically. You destroy photons when they enter your eye, and an energy transfer takes place. So when the light collides, it's photons are destroyed, and the energy is transferred to it's surroundings.
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Kallisto
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(Original post by The Wavefunction)
Conservation of energy states that energy can never be created or destroyed etc
So, when I turn off a light switch, what happens to the light energy?
Conservation of energy does not mean that it can never be created or destroyed only, but also transformed in other energy forms.

In terms of your example with a switch - and a light bulb - electricity is transformed to light. After that light in form of rays disperse around the room, as light has a polarization character, so it can be everywhere. When the switch was turned off, the light source is not in existence any longer, thus no light rays disoperse in the room. But the light is not 'gone'. When the switch is turn on again, electricity flows through the light bulb and it can be transformed to light again.
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mik1a
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The light hits the wall, some of it is absorbed by the wall: it excites the atomic structure of the paint/plaster/wallpaper, causes thermal vibrations which are dissipated via (1) thermal conduction into the wall, (2) via thermal conduction into the adjacent air molecules, after which it is convected around the room, and (3) re-radiated at a thermal frequency, typically much lower wavelength than visible light. Any light that is not absorbed at first contact with the wall is reflected or scattered as visible light away from the wall, only to hit another wall to undergo the same process. Each time the light hits a wall, some percentage of it is absorbed, and it doesn't take long for the amount of ambient radiation to decrease in the room.
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General Josh
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(Original post by mik1a)
The light hits the wall, some of it is absorbed by the wall: it excites the atomic structure of the paint/plaster/wallpaper, causes thermal vibrations which are dissipated via (1) thermal conduction into the wall, (2) via thermal conduction into the adjacent air molecules, after which it is convected around the room, and (3) re-radiated at a thermal frequency, typically much lower wavelength than visible light. Any light that is not absorbed at first contact with the wall is reflected or scattered as visible light away from the wall, only to hit another wall to undergo the same process. Each time the light hits a wall, some percentage of it is absorbed, and it doesn't take long for the amount of ambient radiation to decrease in the room.
I think you are just listing a bunch of terms without understanding them properly. Of course you're right, and the thermal energy is conducted into the wall, but you're going a bit overboard. The "thermal frequency" will of course have to be lower than that of visible light because it came from visible light, lost energy, and therefore must have a lower frequency (wave packet/photon lost energy, e=hf taking h to be Planck's constant, so f=e/h, meaning that the frequency must decrease).
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mik1a
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(Original post by General Josh)
I think you are just listing a bunch of terms without understanding them properly.
With respect, the more likely explanation is that you don't understand me.


(Original post by General Josh)
Of course you're right, and the thermal energy is conducted into the wall, but you're going a bit overboard. The "thermal frequency" will of course have to be lower than that of visible light because it came from visible light, lost energy, and therefore must have a lower frequency (wave packet/photon lost energy, e=hf taking h to be Planck's constant, so f=e/h, meaning that the frequency must decrease).
Thermal frequency refers to blackbody emission, i.e. emission in a broad spectrum that is primarily governed by the emitter's temperature due (fundamentally) to vibrations of the lattice structure. Everything radiates like this, and the wall is able to lower the frequency of light by absorbing in the visible and emitting in the infra-red by this mechanism.

This is entirely separate from scattering, wherein the photon hits an electron, causes it to jump to a higher energy level, then it falls back to where it started and radiates another photon of exactly the same frequency back out (but not necessarily in the same direction).

Both these mechanisms are interesting, but very different.
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