Where Does Light Go When We Turn Off the Switch?

[Mozart]

What happens to the light in my room when I turn off the light switch? Where does it go?

All I can think of is that it gets absorbed by all the different coloured things in my room, but then I think about a white room, and wonder why the light still disappears if an object that is white relfects all light.

Another thing I don't quite understand is how I can be in a dark room and be looking out of my window at night and see my neighbours rooms all lit up. It makes me wonder why the gap between us is still dark if the light should be traveling through it to my eyes.

Thanks

 

[Homer Simpson]

Thats an interesting idea! I don't know much about it. Seems to me that if the light (photons ?) was indeed reflecting, it would quickly dampen out much like an echo, except way faster due to the speed of light. As for the neigbours light, if its foggy out the fog would be illuminated. Light needs somthing to bounce off to be visible.

edit: To add my own little related question in: what types of radiation would reflect well in a mirror?? would gamma, alpha, beta?? Microwaves? Neutrons?? If so, why? And if not, what is the difference?? Mozarts question has twisted my brain.

 

[rbj]

a better scenario would be a room with all walls as mirrors. but there would still be absorbtion. but, hey, it would sure as hell be bright when the light's on.

 

[-Job-]

I don't think you can have a room whose walls reflect 100% of the light. Even if the walls reflect 99.99999% of incident light, considering how fast light propagates and how many times it hits the walls it probably wouldn't take very long for all the light to be absorved. Are we assuming there is an atmosphere inside the room? If there is then nothing is preventing the light from interacting with the particles in the atmosphere and be converted into heat for example. I think this question was asked before.

 

[Dr.Brain]

While studying light in your room , photonic-nature of light should be stressed-upon since the dimensions of the objects in your room are larger than wavelength of light. All objects in your room are partial absorbers , with exceptions (mirrors) , when you put on the light, photons of variable energy-values are ejected in all directions , some of them being absorbed by the objects while some reflect back, thus forming an image on your retina. Becaus ethe photons are being lost in some or the other way , the tube-light continuous its continuous supply , when you switch it off , the supply is cut off , and the initially ejected photons terminate in some way or the other.

Regarding your second concern , the gap between you and your neighbour's windows is NOT dark , its just that you are not getting any light reflected into your eyes from that region.Only source you are getting light from is your neighbour's room, and if someone blocks the path of light between you and the neighbour's room , the light is blocked.Remember darkness is "absence of light" .

BJ

 

[Mozart]

Thank you all, but I am still wondering how the light completely disappear s in any room not just a completely white room or a fully mirrored room.

-job- mentioned above that:
" Are we assuming there is an atmosphere inside the room? If there is then nothing is preventing the light from interacting with the particles in the atmosphere and be converted into heat for example."

Is the electromagnetic energy just changing to a different frequency, and the energy doesn't actually disappear?

What if we took out all the gas from a white room (with just walls, a ceiling, and a floor) and turned on a switch that emitted photons. After we shut the switch off I would think that the photons would keep on bouncing off the walls, ceiling and floor never being absorbed. So would the room always be bright until say a human went to observe the room but by doing this would provide the photons a place to be absorbed and turned into heat for example?

 

[-Job-]

Even a white wall will absorbe some radiation. If we shine a very strong light against a white wall does it not get a little warmer? I don't believe any surface is guaranteed to reflect 100% of the light.
Consider a photon that is emitted in the scenario you just described. Suppose when this photon hits a surface (wall, ceiling ...) it is reflected with probability k and absorved with probability 1-k. Suppose that it is reflected, then it will hit another wall eventually and may be reflected again or may not. Consider that light travels 300,000 Km/sec, how many times can it bounce off the walls of a typical room in half a second? Many, many times, so on average even if the walls are very reflective chances are all the light will be absorbed very quickly, and if they're not then even faster. The probability that enough photons stick around to be noticeable is very very low.

 

[Mozart]

Great points.

 

[Ubern0va]

I'm pretty sure that light energy is absorbed, similar to heat. With a constant light source, the amount of reflection is constant (more or less), but, when you turn off the light, there are no more new photons to reflect. It's the constant light source that makes you perceive a problem with all the light suddenly disappearing because it gives the illusion that the room used to be 'filled' with light (when in fact it's not).

 

[Mk]

Goddam its the third time I lost this post

What happens to the light in my room when I turn off the light switch? Where does it go?

All I can think of is that it gets absorbed by all the different coloured things in my room, but then I think about a white room, and wonder why the light still disappears if an object that is white relfects all light.

Answer:

When light hits a metallic surface, the conduction electrons actualy ABSORBS the photons, causing a transition from one conduction band to another (there's a whole spiel here that I'm skipping about the transition is actually between two points between two different band separated by a reciprocal lattice vector). But this doesn't last and the conduction electrons will drop back to the original band and re-emit this photon, preserving the transverse momentum but with a 180 phase difference. Classically, the conduction electron gains energy from the oscillating E-field in the photon, and re-radiate this field. But since the electrons have negative charge, the field shift phase by 180 deg.

(WTF happened to what I typed here? I forgot what I did. Sorry. Something about... mirrors and white.)

All I can think of is that it gets absorbed by all the different coloured things in my room, but then I think about a white room, and wonder why the light still disappears if an object that is white relfects all light.

There's where you make the incorrect assumption. White does not reflect all light, otherwise it would look like a mirror, which reflects almost all light. Since the white object is white, it is obviously not looking like a mirror. You're probably confused now, or maybe you've are enlightened (haha I'm so funny). White is a color that is composed of all wavelengths. If color is wavelengths that are reflected off an object and enter the eye, what differentiates between a perfect mirror and a perfectly white object? White requires equal proportions of each wavelength. Seen in the color wheel, white requires equal proportions of perfect red, perfect green, and perfect blue.
----
The Color Wheel: http://en.wikipedia.org/wiki/Image:AdditiveColorMixing.png
See that red, green, and blue mix to make WHITE.
Red and green make yellow
Red and blue make magenta
Blue and green make cyan.
See that yellow, magenta, and cyan mix to make BLACK
----

Consider a photon that is emitted in the scenario you just described. Suppose when this photon hits a surface (wall, ceiling ...) it is reflected with probability k and absorved with probability 1-k. Suppose that it is reflected, then it will hit another wall eventually and may be reflected again or may not. Consider that light travels 300,000 Km/sec, how many times can it bounce off the walls of a typical room in half a second? Many, many times, so on average even if the walls are very reflective chances are all the light will be absorbed very quickly, and if they're not then even faster. The probability that enough photons stick around to be noticeable is very very low.

Actually if you do the math out...

OK, for the best dielectric mirrors around, I would estimate from what I've read that one can expect losses on the order of one part per million, i.e. 99.9999 percent reflectivity.

source: http://www.rp-photonics.com/encyclop...k_supermirrors

(and I'm open for better estimates).

With this ultra-high efficiency (I would assume the bandwidth over which this level of refelctivity would be maintained would be very low, requiring the light being reflected to be a laser pulse) the light could reflect about 693,000 times before it halved in intensity. If you know the mean free path between reflections (this would be proportional to the radius of the sphere, you can figure out the time it would require for the pulse to halve in power.

Let's say that the [mirrored] sphere was room-sized, and that the mean free path was 10 meters, the intensity would halve every .02 seconds. If you have a huge room, 100 meters, you could get that up to .2 seconds.

If we shine a very strong light against a white wall does it not get a little warmer?

Actually, if you only aim one photon against the white wall it will get warmer.

Have I answered all your questions?

*Very* interesting related threads, I suggest you read.
https://www.physicsforums.com/showthread.php?t=90267
https://www.physicsforums.com/showthread.php?t=99218

 

[SpaceTiger]

There's where you make the incorrect assumption. White does not reflect all light, otherwise it would look like a mirror, which reflects almost all light.

I was under the impression that the difference between a mirror and a white surface had to do with the direction of the reflection. Imagine I have a hypothetical wave front that is parallel to the mirror or white surface and moving towards it. I would think that a white surface would scatter the light in all directions, while an ideal mirror would reflect the entire wavefront (or the part that hits the mirror) back in the opposite direction.

 

[Mk]

I was under the impression that the difference between a mirror and a white surface had to do with the direction of the reflection. Imagine I have a hypothetical wave front that is parallel to the mirror or white surface and moving towards it. I would think that a white surface would scatter the light in all directions, while an ideal mirror would reflect the entire wavefront (or the part that hits the mirror) back in the opposite direction.

Mmmm, maybe you're right. Are you sure? I'm not. Sorry, I may have sounded like I was totally sure.

White is a color that is composed of all wavelengths. If color is wavelengths that are reflected off an object and enter the eye, what differentiates between a perfect mirror and a perfectly white object? White requires equal proportions of each wavelength. Seen in the color wheel, white requires equal proportions of perfect red, perfect green, and perfect blue.

After reading it again, I still like my hypothesis, with a scattering part added—mirrors reflect back, white just scatters it out. Anybody know who is right?

[edit: Yeah, I meant the white scatters light out, while reflecting all colors equally, and the mirror reflects all color back the direction it came]

 

[krab]

No. When we say a wall is white, we mean it reflects ALL visible colours equally. Just like a mirror, except the mirror retains the direction of the incident light and doesn't scatter the light (like spacetiger said). Shine a red light on a white wall, and the wall suddenly looks as though it is red. That's how a projection screen works. Projection screens are white.

 

[spacetime]

Light gets absorbed in the atoms of the wall. Whatever be the material of the wall, there is a probability of photons getting absorbed by it, and the process happens quickly because of the fast speed of light. So the energy goes into increasing the thermal energy of the atoms of the wall.

I believe in the hypothetical situation when the walls are perfectly reflecting, you would continue to see the walls even when you switch off the light, but then again, whatever light falls on your eyes or on you would be absorbed

http://www.geocities.com/physics_all"

 

[krab]

Also, if the walls are perfectly reflecting and you don't switch off the light, you are pumping energy in continuously, it doesn't get dissipated, and the temperature in the room increases to infinity. Its in fact an optical cavity with a quality factor Q of infinity.

 

[Mk]

Also, if the walls are perfectly reflecting and you don't switch off the light, you are pumping energy in continuously, it doesn't get dissipated, and the temperature in the room increases to infinity. Its in fact an optical cavity with a quality factor Q of infinity.

How are you pumping in energy continuously? Its just one flash of light, and if its a perfect mirror, in perfect vacuum, the light would bounce around for eternity.

 

[krab]

No, the OP question was

What happens to the light in my room when I turn off the light switch? Where does it go?

i.e. not discussing a flash of light here, but a case where light is already on. I'm pointing out that if the walls are perfectly white and you don't switch off the light, the room will get monotonically hotter. So that not being the case, you know that the light is being continuously absorbed. That means that switching off the light will eliminate the light in the time it takes for the light to make a few passes across the room. (For reflectivity being less than say 90%) To be specific, the room being say 3 metres on a side, the light would disappear in a few tens of nanoseconds. This is in fact much shorter than the cool-down time of the bulb after electricity is interrupted.

 

[Mk]

Yes, here is a post by pervect.

for the best dielectric mirrors around, I would estimate from what I've read that one can expect losses on the order of one part per million, i.e. 99.9999 percent reflectivity.

source: http://www.rp-photonics.com/encyclop...k_supermirrors

(and I'm open for better estimates).

With this ultra-high efficiency (I would assume the bandwidth over which this level of refelctivity would be maintained would be very low, requiring the light being reflected to be a laser pulse) the light could reflect about 693,000 times before it halved in intensity. If you know the mean free path between reflections (this would be proportional to the radius of the sphere, you can figure out the time it would require for the pulse to halve in power.

Let's say that the [mirrored] sphere was room-sized, and that the mean free path was 10 meters, the intensity would halve every .02 seconds. If you have a huge room, 100 meters, you could get that up to .2 seconds.