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Why do we need to keep lights on?

by GarageDweller
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GarageDweller
#1
Mar5-14, 04:47 AM
P: 104
When we turn a light on a wave of high frequency EM waves are emitted, however as soon as we turn the light off the light fades. Why is this?
If we were to do the same thing in a perfectly reflecting room (perhaps a spherical room with mirrors on the walls) will the same occur?
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adjacent
#2
Mar5-14, 04:55 AM
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If we were to do the same thing in a perfectly reflecting room (perhaps a spherical room with mirrors on the walls) will the same occur?
I posted a similar question some months ago: Here
Maybe it can offer some help

Look towards the later posts.
GarageDweller
#3
Mar5-14, 05:06 AM
P: 104
My question is why light seems to fade so fast.
Where does it go?


edit: After reading the link you provided, I guess the waves are absorbed by the surrounding?

CWatters
#4
Mar5-14, 05:20 AM
P: 3,085
Why do we need to keep lights on?

It's absorbed by it's surroundings. Even mirrors aren't 100% efficient. Google says a mirror typically reflects 80-90% of the incident light. The rest is absorbed and turned into heat.

So after around 20-30 reflections only 1% is left. If your mirrored room was 10m in diameter the light would be down to 1% after travelling just 200-300m. Since light travels at 3*10^8m/s it won't take very long for 99% to be absorbed.
abitslow
#5
Mar5-14, 09:31 AM
P: 140
According to wikipedia, the best narrow wavelength mirrors reflect ~ 99.999%. Some simple math: F=0.99999
hence F^500 000 ~0.0067...meaning 500,000 reflections is enough to reduce 100% to 0.7%. Light travels at 300 000 000 m/s. Say in a room 3 m across. 3*500,000 300 000 000 = 0.005 seconds. In 0.005 seconds the light will be "gone". With more realistic broad wavelength mirrors, reflection is more like 95% and so 100 reflections would get you to 0.6%. Air is composed of electric dipoles we call atoms and molecules. Each of those can interact with light in several ways, look up Rayleigh Scattering. If that weren't true, day and night would be very different. Anyway, even if the mirrors were "perfect", air would absorb the light, eventually. Take out the air, and you'd have more important things to worry about. All matter emits light, the most general is black body radiation. The implication of this is that theoretically, we can see in the dark; all we need is to cool our eyeballs down to absolute zero. As they warm up from zero, they begin to emit light (em radiation) and so begin to compete with the light from our surroundings. So, only bright objects are visible becuz its the difference in light intensity that matters to us.
danshawen
#6
May10-14, 05:12 PM
P: 12
One of the Nobel prizes awarded for physics for 2012 included a pair of physicists who had observed the decay of photons being reflected inside of a superconducting cavity that was theoretically capable of reflecting the energy losslessly. Even this idea is not as easy as it sounds, because such a cavity also has reflective properties that may be influenced by things such as external magnetic fields. Also, it is not uncommon for superconducting material domains to be less than optical or electrical perfection, to say the least. As such, the abstract of the experiment does not make for easy reading.

Of course, this question is one of the quantum physics conundrums (like Schroedinger's cat) as well, because in order to determine that the photon(s) are still inside the cavity, you need to observe them in a way that does not absorb its energy prematurely, and this is even more of an experimental challenge.
Vanadium 50
#7
May10-14, 07:22 PM
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Danshawen, what does this have to do with the question?
danshawen
#8
May10-14, 09:55 PM
P: 12
"perfectly reflecting room (perhaps a spherical room with mirrors on the walls"

-- does not necessarily need to be a mirror, does it? Or is there a PF sanction against improving the question?
CWatters
#9
May11-14, 06:12 AM
P: 3,085
Quote Quote by danshawen View Post
One of the Nobel prizes awarded for physics for 2012 included a pair of physicists who had observed the decay of photons being reflected inside of a superconducting cavity that was theoretically capable of reflecting the energy losslessly. Even this idea is not as easy as it sounds, because such a cavity also has reflective properties that may be influenced by things such as external magnetic fields. Also, it is not uncommon for superconducting material domains to be less than optical or electrical perfection, to say the least. As such, the abstract of the experiment does not make for easy reading.

Of course, this question is one of the quantum physics conundrums (like Schroedinger's cat) as well, because in order to determine that the photon(s) are still inside the cavity, you need to observe them in a way that does not absorb its energy prematurely, and this is even more of an experimental challenge.
Thanks for the interesting post.


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