It is night. I’ve a fine LED lighting the room. I switch off the power and the room is immediately dark. Where does the light in the room go to when I switch off the power?
It’s absorbed by all the surfaces.
And, even if you trapped a photon in a perfectly reflecting container, as soon as it entered your eye (so that you could see it), the box would be dark.
So what happens in a large reflective container with a person inside it? Does the light just slowly dim after you turn it off?
Light is really, really fast.
If you had a room that was 10’ across, with mirrored walls, light would travel from one wall to another in 10 nanoseconds (10 billionths of a second). A typical mirror reflects something like 96% of the light that falls on it. So, in one microsecond, the light has made 100 trips, and 100 reflections. If the original intensity was 1, after 1 microsecond, the intensity would be .96^100 or 1.7%, after 2 microseconds, the intensity would be .02% of the original.
So, there’s probably no way to arrange a room that would allow you to see the light fading out - the human eye isn’t that great at seeing things go from bright to dark rapidly (think about the afterimages resulting from the lights going out suddenly).
I suspect you could make a sensor that would display the fade-out on an oscilloscope, but that’s not quite the same thing.
Also, remember that any photons that hit you will be absorbed pretty much entirely, so it’s actually harder to do than I described above.
Light bounces back and forth until it is absorbed by the person inside, or by the wall. There’s no such thing as a 100% reflective surface.
Let’s say the container is 300 m (1000 ft) in size - so light takes about 1 microsecond to travel the length of the container. If the walls are 99.9% reflective (only achieved in mirrors for very special scientific applications), in 1 millisecond (0.001 second) the light would have reflected 1000 times, and brightness has decreased to less than half. In 0.005 seconds, 99.3% of the light has been absorbed.
And just to point out the obvious, the answer to where the light “goes to” is the same while the light is on, not just when you turn it off. The bulb must continually emit new light because all the old light is being absorbed almost instantaneously. If it wasn’t being absorbed as quickly as it was produced, then the total light level of the room would increase over time.
I’m not sure I understand your question here is at all. If the conductivity is not closed there would be no electrical conductivity in wires.
It same way if you cut the return wire the light would go off.
I’m not sure what you are doing with LED lighting in your room. Is there short some where or incomplete wiring going to and from the light to power source.
I have never had this problem. So I’m not sure if you asking why there is no power unless you have short some where or incomplete wiring.
Everyone else understood it.
He’s asking where did the light go after it was cut off. The light that existed already before.
The number of bounces light makes in a box with walls that reflect a proportion of the light delta is on the order of (delta)/(1 - delta). If your reflectivity is 0.999999 having N nines, then the number of bounces is about 10[sup]N[/sup] bounces. So for N = 1, you get 10 bounces. for N = 3, you get about 1000 bounces.
Let’s say your room is 10 meters on a side. If your walls were immaculate mirrors with N = 6, then you’d get a million bounces. Light travels at 10[sup]8[/sup] meters per second The amount of time needed to travel that distance is 10 m * 10[sup]6[/sup]/10[sup]8[/sup] m/sec = 0.1 sec. So you’re not going to see tapering off.
Worse, you’re not going to see reflectivities anywhere near that high. Ultra-high mirrors are less than N = 3 (0.999) reflective. Spectrolon broad-band diffuse scattering material, used in integrating spheres, where high reflectivity is desired, is just over 0.99 (N = 2). Extreme cases of special-case high reflectivity at one wavelength don’t even get to N = 5. For the common “best” case of N = 3 you’d get 1000 bounces, and that would take 0.0001 sec = 100 microseconds. You won’t see that, although I suppose with a good fast detector you could measure it. With an ordinary wall you’d be at less than 0.9, or N less than 1, and it’d be over in a microsecond.
This question is actually better suited for GQ. Thread relocated from IMHO.
Light becomes heat energy … you’ll need a thermometer to see the “light” after the power is turned off.
You could just as well ask where the ripples in a pond go after you stop tossing stones into it. Light, as we currently understand it, is made up of ripples in a field called the photon field; we call the well-defined examples of these ripples “photons” but, really, in modern physics, the field is primary. Anyway, the ripples are caused by, and carry, some energy which was put into the field, and they can transfer that energy to other things (electrons, atomic nuclei, etc.) which are “coupled to” the photon field. So the energy in the field moves around and eventually is entirely transferred to things which don’t excite the photon field in ways your eyes can perceive.
This can be an interesting entry point to the concept of a conserved quantity: Energy is conserved. The number of photons, which isn’t even necessarily well-defined, is most certainly not conserved. The amount of visible light is not conserved, either. Learning which quantities are conserved is an important part of figuring out the physics of a situation.
Thank you, this is an explanation that makes sense to me. But then what is meant by “light is a wave and a particle”? It doesn’t mean a phyiscal particle, does it, or where do those particles go?
I can pretty much guarantee that in the responses that follow, you’re not going to get an answer to that question that satisfies you.
Yes, light is a physical particle. A particle of light is called a photon. When a photon interacts with an electron in an atom, many things can happen. Sometimes the photon transfers its energy to the electron. The electron jumps to a higher energy level, and the photon disappears. Where did it go? It disappeared. That’s it. In the quantum world, unlike in the familiar macroscopic world, particles are not conserved.
I suggest reading Feynman’s “QED: A Strange Theory of Light and Matter” for a good introduction to the topic.
–Mark
Help me understand this. If this is true then why can’t we think of the room acting like a bucket into which light is being poured? Unless the light is being absorbed at exactly the same rate as its being emitted then the light level will either rise or decrease, and of course we never see that,
If you’re fast enough, you can hit the switch, leap into bed and get under the covers before the light disappears.
You never see that because it is true that light is being absorbed at the same rate that it is emitted. As was mentioned above, even if the room is a .9999 perfect mirror all the light will be absorbed in a few microseconds.
And when you first turn on the light, for a tiny fraction of a second, it is true that the rates aren’t the same, and so the light level is rising. It very quickly gets to a point where the light levels are high enough that enough is absorbed, and then the rates are the same.
I honestly flashed on this thread when I saw OP:
**The door is closed. Is the refrigerator light on? **The door is closed. Is the refrigerator light on? [Firefox memory usage question] - Factual Questions - Straight Dope Message Board,
and couldn’t remember if that old one was on a similar topic, as it sounded.
And then I saw that a mod, sadly, had edited the title. Different topic.