# Keep the light beam in a closed room, is it possible? [duplicate]

This question already has an answer here:

I mean if I am in a room totally closed to light. If I switch on a torch for a second then switch it off. So will the inside of room be always bright?

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## marked as duplicate by John Rennie, Waffle's Crazy Peanut, Qmechanic♦May 4 '13 at 17:01

Possible duplicates: physics.stackexchange.com/q/8347/2451 , physics.stackexchange.com/q/12417/2451 and links therein. – Qmechanic May 3 '13 at 18:56
No mirror can be perfectly reflective due to quantum tunneling so that already answers your question. But even if it could be done, you would never be able to check the situation because when you look inside, the light almost instantly leaves through the peephole. Also, related question: physics.stackexchange.com/q/55768/16660 – Wouter May 4 '13 at 12:55
@Wouter excuse me for my lack of knowledge. As I have understood if the room is covered by perfect mirrors, the room will be bright for a little (how little?) of time, but not forever because of the quantum tunelling. Am I right? – nikamed May 4 '13 at 13:36
I've poured my previous comment into an answer together with what I hope answers your comment as well. – Wouter May 4 '13 at 14:15

No mirror can be perfectly reflective due to quantum tunneling so that already answers your question. But even if it could be done, you would never be able to check the situation because when you look inside, the light almost instantly leaves through the peephole. This also poses a problem for your initiation method, which John M already touched on: you need to be very, very quick to insert the light beam and remove your insertion device/hole before it can reabsorb the light.

In any case, let's consider what would happen if you had been able to pull of this insertion of a light beam. The light would definitely be inside for a short but finite period of time. The problem is that even classically perfect mirrors are impossible with current technology; I believe the highest reflectivity that we're able to achieve is about 99.999% if the wavelength is just right. Alas, even this kind of amazing reflectivity means that a light beam will lose 90% of its intensity in less than a second in a spherical container with a diameter of $1\,\mathrm{km}$!

You can do the maths of this yourself: use that the reflectivity $R$ is the factor by which the intensity $I$ of the light beam is reduced when it reflects off a wall. So

$$I_{after} = R I_{before}$$

Now give the intensity of the light an index corresponding to the number of interactions with the wall. So the original beam has intensity $I_0$. After one reflection, it still has $I_1 = RI_0$ left. After two reflections, it has $I_2 = RI_1 = R^2I_0$ left. There is a pattern here. After $n$ reflections, it has $I_n = R^nI_0$ left. Now, you want to find out for which $n$ the intensity has dropped to 10 percent of the original $I_0$. So the question is: when is $R^n < 0.1$ (with $n$ a natural number)?

You can probably work that out by yourself. Now consider a spherical container of diameter $1\,\mathrm{km}$. Then the light will always have to travel $1\,\mathrm{km}$ or less before it hits the wall again. In other words, it has to travel (at the highest) $n$ times $1\,\mathrm{km}$ before it has lost 90 percent of its original intensity. Using that the speed of light is approximately $300000\,\mathrm{km/s}$ you should find that it only takes the light about $0.77\,\mathrm{s}$ to do this.

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