physics question

[cjolley]

To simplify things lets assume the mirror is perfect and the other walls absorb all of the light that falls on them.

The fact is that the average flux will be higher.
why because 1/4 of the light will have to travel farther before it encounters a surface that does not reflect it.

Every thing else is a detail or a percentage and a complex calculation.
But the answer will always be somewhere between the result with the original four walls and my most extreme example.

[Brian Ellis]

Put it this way. Consider the radiative energy over the visible spectrum.

Take a largish room and paint the six surfaces matt black. Hang a 15 W bulb from the ceiling in the middle. Can you read a newspaper anywhere in the room? Of course not, because the energy has been absorbed by the walls. Now, put a metal reflector above the bulb. It will become possible to read the paper under the bulb, because the energy which was directed upwards has now been partially directed downwards and added to the ordinary downwards radiation. Now paint the six surfaces white and it becomes even easier to read the paper, because even more of the energy is reflected onto your newspaper, instead of being dissipated as heat in the black paint. If the 6 surfaces were all mirrors, then little energy would be converted to heat, except where there is absorption by the presence of the newspaper, your body and clothes and the chair you are esconced upon as you read about the latest bombings in Israel/Iraq/Pakistan/East Timor/London/Glasgow/Russia etc.

Of course, there will be a difference between speculum reflection (essentially a mirror is close to a perfect speculum) and reflection from structured surfaces.

Now here's a purely hypothetical question. Imagine a six-surfaced room consisting of 100% speculum surfaces, with no air or other absorbing molecules. Somehow or another you introduce a light into that room and switch it off. There is nothing to absorb the light energy and convert it into heat. Does the light stay on for ever, bouncing back and forth off the walls?

[Dr Mordrid]

No. At the QM level light is a resonant coupling between the electron shells of the source (lamp) and those of the reflecting and/or absorbing surfaces (walls, mirrors, meter etc.).

In reflection the electron shells in these surfaces move into a higher energy wave state and effectively become the light source. They then couple to the next surface and so on. Lose the original source and the coupling chain disappears.

Back to darkness.

Of course no energy transfer is 100% efficient so there are loses all along the way due to absorption.

[VJ]

Thanks guys!

This sort of matches my intuition but when answering it to my friend, he found it odd. I have directed him to this thread, hopefully it will make things clearer for him as well.

Jörg

[az]

Also very simple: There is a reason lamps have reflectors.

[Dr Mordrid]

One more clarification about the end of the coupling.

Resonant couplings also occur between distant objects like a star and a telescopes CCD detector. Now imagine that somehow that star, lets say it's 5 light years distant, winked out; how soon would that information become apparent to the CCD? 5 years because information also cannot exceed C.

The implication is that the light in the box doesn't go out instantly either; it persists just long enough for all it to be absorbed, which is a vanishing short time.

[az]

Yes, but if all surfaces in the room were perfect reflectors (which don't exist, of course), light would bounce around the room infinitely, no? Of course, the moment you'd measure it, it would vanish into your sensor

[Dr Mordrid]

Damned sensor