I am failing to explain why light won't remain inside the wooden box in the following situation. I considered a wooden box closed from all the sides, with a bulb inside it. If we switch on the bulb, light will be emitted from the bulb inside the box, if we switch off the bulb, I assumed the light to disappear, which I practically assume to be true. If I am wrong here, please explain. So, practically I assumed light to disappear if we switch off the bulb, after being switched on for a while. I got a question here, why light has to disappear the moment we switch off the bulb?

Thinking about the above situation, I considered the case of bullet being fired into the wooden face from the inside of the same box. No doubt, bullet smashes the wooden face into pieces and comes out. Here bullet has high speed and even has mass, so that, it disappears out of the wooden box. I considered the photons now, I thought photons have high speed much greater than bullet, but they have very less mass (I am not sure with rest or relativistic), so I thought that, photons should remain into the wooden box after colliding the wooden face. If they would had, I thought that there should have been light inside the box even after the bulb was switched off, but it is not. Is it that photons are piercing into the voids between the atoms with high speed to get out of wooden box?

If we say photons are getting out of the wooden box piercing into the voids, why won't it be in total internal reflection phenomenon. If we assume light to be incident at the interface from the denser medium to the rarer medium, at an angle greater than critical angle, photons should even then pierce into the voids. But, my teacher has taught me that, in total internal reflection there would be no refraction. By this even my assumption of photon getting out of the box from the voids also falls down. As radiation is a form of energy, it can't be lost with out being utilised or else it can't be lost without getting emitted out of the box so as to avoid violation of conservation of energy. Is here, the light absorbed inside the box by the medium? I don't know whether I have misunderstood here or whether we could explain from wave nature of light, so as to why light disappears from the box?


2 Answers 2


The (massless) photons are absorbed by the wood, giving the atoms in the wood energy. If you turn the lightbulb on inside the box, then turn it off and feel the inside of the box, you'll notice that it's warm: there are your "missing" photons.

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    $\begingroup$ Note that this heating is smaller than what most household thermometers are capable of detecting. You can't burn your house down by turning the lights on and off. (or by just leaving them on) $\endgroup$ Dec 20, 2013 at 15:35
  • $\begingroup$ Thank you for the answer Sir. Would there be such a fast dissipation of energy even if we turn on once and switch off instantaneously (thus avoiding box to be warm)? $\endgroup$
    – Sensebe
    Dec 20, 2013 at 15:36
  • $\begingroup$ @VINAY: Yes, the box would still absorb the photons emitted by the light bulb. $\endgroup$
    – Kyle Kanos
    Dec 20, 2013 at 15:48

Most of the light will be absorbed to increase the molecular vibrations of wood (i.e heat) and some will be reflected depending upon wavelength of photons. This heat will increase the temperature of inner boundary of wooden box. If the wooden box has no air in it then due to temp. diff. the molecular vibrations will flow at the rate of speed of sound(perhaps) towards the outer boundary. After some time the whole wooden box will acquire a constant temp. in the steady state. At the outer boundary of box there will be a temp. diff. this will also cause the flow of heat from box to the enviournment . If the env. is also vacuuo then the wooden box will emit the photons from its outer boundary and box will slowly go to 0K which takes infinite time hence the box in practice can never reach 0K..Usually the env. is air in experiments and most of the energy is transmitted as molecular vibrations to the air's molecules.

"why light has to disappear the moment we switch off the bulb?"

To show you why in practice light disappear i am going to model the current situation very grossly.
Let's assume
(0) The source of photons(e.g. a laser) produces photons of a single frequence $\nu$.
(1) when any photon strikes the wooden box all the molecules of wooden box are in ground states. Now when the photon strikes the surface it interacts with the particular molecule. Since the molecule is in its ground state so stimulated emission cannot happen.
(2) suppose there are two types of molecules in the material $A$ and $B$..The ground energy levels of $A$ is $T_1$ and of $B$ is $T_2$. The conentration of $A$ and $B$ is 50-50.
(3) The photon strikes either $A$ or $B$. The probability of striking $A$ and $B$ are $\frac{1}{2}$ and $\frac{1}{2}$ respectively.
(4) $T_1$ and $T_2$ are of such kind that the photon of energy $h\nu$ is able to cause excitation of only $A$ kind of molecules.
(5) The average time b/w any two succesive reflections of photon is $\frac{1}{3\times 10^{8}}$ sec.
(5) The probability that the electrons is absorbed by $A$ is 1 and the probability that the photon will be scattered away by $B$ is 1.

Let's find the probbility of finding the photon in the box after 0.01 sec we switch off the source .

At time $ T< \frac{1}{3\times 10^{8}}$ sec the probability of finding the photon in the box is $1$.
suddenly after $T = \frac{1}{3\times 10^{8}}$ sec the photon will make its first collision with the surface of box. There are 50-50 chances for it to strike $A$ or $B$.
i.e. The probability of finding the photon in the box after first collision is $\frac{1}{2}$.
The probability of finding the electron after its second collision is $\frac{1}{2} \times \frac{1}{2}$
Similarily the probability of finding the electron after its nth collision is $$\frac{1}{{2}^n}$$
The time $T$ after n collisions is $n \times \frac{1}{3\times 10^{8}}$ sec.
In 0.01 sec the number of collisions is $0.01 \times 3 \times 10^{8}= 3 \times 10^{6}$
The probability of finding the electron in box after $3 \times 10^{6}$ collisions is
So the probability of finding a photon after 0.01 sec you switched off the source is nearly $0$.

I think we don't need precise quantum machenical calculation to answer why why light has to disappear the moment we switch off the bulb?

  • $\begingroup$ If there occurs total internal reflection enumerable times i.e photons keep reflecting then light will not disappear!. $\endgroup$
    – user31782
    Dec 20, 2013 at 15:57
  • $\begingroup$ If it is photons we are talking about, i.e. quantum mechanical framework, there can be no total reflection, there will always be a probability of scattering with the surface electrons of the mirror and loosing energy. Infinity is a big concept. $\endgroup$
    – anna v
    Dec 20, 2013 at 16:15
  • $\begingroup$ @annav i am just trying to explain as par my understanding of the concept involved. if my answer has some technical inaccuracies then i accept them. Please give your answer with some rigrous quantum mechanics so that I and some other beginners in physics could learn something interesting about photon. $\endgroup$
    – user31782
    Dec 20, 2013 at 16:28
  • $\begingroup$ I was replying to your comment , not the answer. Have a look about the photon elementary particle crossections nicadd.niu.edu/~piot/phys_630/Lesson17.pdf $\endgroup$
    – anna v
    Dec 20, 2013 at 17:20

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