Light in a mirror box

Ravalanche

not sure if this is posted,

so imagine i have a 3- dimensional cube. lets say its about 10m x 10m x 10m in volume. this cube is empty on the inside but its inner surface is covered with mirrors. normal plane mirrors all around the walls, like tiles but instead mirrors. so what happens if i switch on a light bulb inside this box and immediately switch it off. will the light escape when i can ensure that there is not a single hole / leak available for the light to escape. or in other words. will the light be still bouncing around in the cube even after the light source is switched off?

Integral

Since there is no such thing as a perfect mirror the light, being adsorped by the mirrors, will be gone in a few nanosecsonds.

grzz

Will there not be some absorption at each single reflection?

And light travels so fast that the number of absorptions will be so high!

So the energy is absorbed so fast.

Ravalanche

mirrors absorb light? thats new to me.
well then, is there any surface that totally reflects light?

davenn

there ya go learn something new every day. :)
Only a percentage of the light is reflected maybe as low as 50% maybe as high as 90%

zoom onto the surface of the mirror and at the microscopic level there's likely to be significant spaces between the "blobs' of reflecting material deposited on the glass/other material. All those spaces/gaps are places where the incident light is going to get through and either be absorbed by the glass or "transmitted" through and out the other side.

not that I'm personally aware of.
some of the best mirrors are produced for astronomical telescopes and they are often ~ 90%
its all depends on wavelength as well
have a look at this wiki entry.... http://en.wikipedia.org/wiki/Optical_coating

cheers
Dave

Ravalanche

alright, thanks alot for the info and the link too of course.

cheers.

ratchettrack

How would the reflective ability of a pool of mercury compair to a mirror?

Ratchettrack

sophiecentaur

Clean mercury was what they used to use for mirrors in the old days (on the back of the glass with paint behind it) . So it obviously works tolerably well. But it does rely on the mercury surface being uncontaminated. One reason they don't use it now is that it's toxic but the main reason is that there are better coatings. Real mercury mirrors always look a wierd colour to me - but the ones you can see are all antiques. The reflectivity to optical light is only about 75%, apparently.

Bobbywhy

A Google search for “high reflectivity first surface mirrors” brought many results. Here are some specifications from one manufacturer:

Dielectric High Reflection Front Surface Mirrors
Highest reflection in the visible range
Reflection Properties
Ravg > 99% at 425-680nm at 45°
Ravg > 97% bei 400-690nm at 45°
http://www.pgo-online.com/intl/katal...ic_mirror.html

So even if you built your mirror box using these (imperfect) mirrors, the light would rapidly diminish in amplitude and would disappear very soon.

Cheers,
Bobbywhy

jbriggs444

If one assumes 99% reflectivity then that's 100 reflections to reduce by a factor of 1/e in intensity. In a 10x10x10 meter cube, something like 1 km total distance. At 300,000 km/second, around 3 microseconds.

That's about 30 times faster than a good quality photographic strobe.

Low-Q

The structure of a mirror surface might reflect light as well if the wavelengt is considerably longer than the distance between the spaces in the structure. You can use a metal grid to reflect radiowaves pretty well, even if there is visible small holes in that reflector.

In your microwave oven you have a grid like that on the door window. The microwaves do almost not escape from that door even if 50% of the door is open holes. Visible light pass throug because the wavelengt is much shorter. (What I have learned)

That said, there is no such thing as a perfect mirror - yet.

Vidar

sophiecentaur

It's interesting to compare signal attenuation in an optical fibre which carries light over thousands of km, involving multiple TIR reflections- plus all that glass.

ratchettrack

Don't they have to amplify the signal on long runs to maintain it? I would think that the lights energy loss would be less because its reflection off the inside of the fiber is more of a ricochet.I also would think that all that glass doesn't matter too much because none of the light seems to escape out the side of the fiber. That makes me think that all the glass is doing is slowing down the light by increasing its wavelength and leaving some energy behind. The extra energy must go into the fiber in the form of heat. Does that then mean that the amount of heat build up in the fiber is equal to the lights energy loss or is there something else picking up some of the energy.

Ratchettrack

sophiecentaur

I think you should read a bit about optical fibres before trying to make a contribution about something so sophisticated.
What's a "ricochet" when applied to total internal reflection? Do you know anything about Total Internal Reflection?

Alkim

Dear Ravalanche,

The idea of enclosing light in a cavity with reflecting walls is very old. In fact there is intense search for high-Q cavities. Q or the quality factor is a measurement of the goodness of a resonator, and it is related to the decay time (dacay being cause due to losses). Optical fibers already mentioned are one way of making such a device. Another one is using photonic crystals. However, the best resonators are so called whispering gallery resonators (named so after their acoustic analogs in the form of some architectural constructions) whose Q-s are much above those of any other device.

Just to give you an idea of their construction:

http://www.google.es/search?hl=es&q=...w=1366&bih=653

2112rush2112

But is there a [conservation] law in nature that says an ideal mirror is not allowed?

Even if such a reflective surface is permitted by nature, the Compton Effect would cause the wavelength of the light bulb's emitted light to eventually increase to ∞...

Low-Q

Different metals reflect light differently. A reflective surface for visible light might absorb most of light with shorter or longer wavelength. Dont know about any law except it would be impossible with more than 100% reflective.

If the mirrors in the cube really was reflecting 100% the energy will stop increasing as soon as the bulb shuts off. Then the cube will act as a capacitor which stores the electromagnetic radiation inside it.