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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?

newzad
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    Possible duplicates: http://physics.stackexchange.com/q/8347/2451 , http://physics.stackexchange.com/q/12417/2451 and links therein. – Qmechanic May 03 '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: http://physics.stackexchange.com/q/55768/16660

     – Wouter May 04 '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? – newzad May 04 '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 04 '13 at 14:15

2 Answers2

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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.

Wouter
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Short answer: No.

You have to remember that light is very fast. The law of specular reflection states that an angle made with a mirror and incoming beam will be the same angle the outgoing beam and the mirror make.

If you were able to put light in completely perpendicular to the surface of lets say a half-a-meter radius mirror(insides are mirrors) sphere then it would go in and bounce out. If you put it in at some angle, lets say just enough so that it would go in and bounce around, it would be going so fast that eventually it would reach the same hole it came in at. There is a particular Euler problem at http://projecteuler.net/problem=202 involving sending a beam of light into a triangle mirror system that asks "how many ways can a laser beam enter at vertex C, bounce off 12017639147 surfaces, then exit through the same vertex?".

Euler

You can imagine our sphere will have a similar situation in which even after a billion or so bounces the beam of light will eventually escape from the same hole we used to send the beam in.

3*10^8 m/s is a blistering fast speed to conceptualize.

EDIT: User changed the question as I was typing a response! I will let other people attempt a response.

John M
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