Light Reflection in a Mirror Cube

• Ravalanche
In both cases the light is travelling at 300,000 km/s. However, the microwave oven is much more efficient at reducing the intensity of the light.f

Ravalanche

not sure if this is posted,

so imagine i have a 3- dimensional cube. let's say its about 10m x 10m x 10m in volume. this cube is empty on the inside but its inner surface is covered with mirrors. normal plane mirrors all around the walls, like tiles but instead mirrors. so what happens if i switch on a light bulb inside this box and immediately switch it off. will the light escape when i can ensure that there is not a single hole / leak available for the light to escape. or in other words. will the light be still bouncing around in the cube even after the light source is switched off?

Since there is no such thing as a perfect mirror the light, being adsorped by the mirrors, will be gone in a few nanosecsonds.

Will there not be some absorption at each single reflection?

And light travels so fast that the number of absorptions will be so high!

So the energy is absorbed so fast.

mirrors absorb light? that's new to me.
well then, is there any surface that totally reflects light?

mirrors absorb light? that's new to me.

there you go learn something new every day. :)
Only a percentage of the light is reflected maybe as low as 50% maybe as high as 90%

zoom onto the surface of the mirror and at the microscopic level there's likely to be significant spaces between the "blobs' of reflecting material deposited on the glass/other material. All those spaces/gaps are places where the incident light is going to get through and either be absorbed by the glass or "transmitted" through and out the other side.

well then, is there any surface that totally reflects light?

not that I'm personally aware of.
some of the best mirrors are produced for astronomical telescopes and they are often ~ 90%
its all depends on wavelength as well
have a look at this wiki entry... http://en.wikipedia.org/wiki/Optical_coating" [Broken]

cheers
Dave

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alright, thanks a lot for the info and the link too of course.

cheers.

How would the reflective ability of a pool of mercury compair to a mirror?

Ratchettrack

Clean mercury was what they used to use for mirrors in the old days (on the back of the glass with paint behind it) . So it obviously works tolerably well. But it does rely on the mercury surface being uncontaminated. One reason they don't use it now is that it's toxic but the main reason is that there are better coatings. Real mercury mirrors always look a weird colour to me - but the ones you can see are all antiques. The reflectivity to optical light is only about 75%, apparently.

A Google search for “high reflectivity first surface mirrors” brought many results. Here are some specifications from one manufacturer:

Dielectric High Reflection Front Surface Mirrors
Highest reflection in the visible range
Reflection Properties
Ravg > 99% at 425-680nm at 45°
Ravg > 97% bei 400-690nm at 45°
http://www.pgo-online.com/intl/katalog/shr_dielectric_mirror.html

So even if you built your mirror box using these (imperfect) mirrors, the light would rapidly diminish in amplitude and would disappear very soon.

Cheers,
Bobbywhy

If one assumes 99% reflectivity then that's 100 reflections to reduce by a factor of 1/e in intensity. In a 10x10x10 meter cube, something like 1 km total distance. At 300,000 km/second, around 3 microseconds.

That's about 30 times faster than a good quality photographic strobe.

zoom onto the surface of the mirror and at the microscopic level there's likely to be significant spaces between the "blobs' of reflecting material deposited on the glass/other material. All those spaces/gaps are places where the incident light is going to get through and either be absorbed by the glass or "transmitted" through and out the other side.

Dave
The structure of a mirror surface might reflect light as well if the wavelengt is considerably longer than the distance between the spaces in the structure. You can use a metal grid to reflect radiowaves pretty well, even if there is visible small holes in that reflector.

In your microwave oven you have a grid like that on the door window. The microwaves do almost not escape from that door even if 50% of the door is open holes. Visible light pass throug because the wavelengt is much shorter. (What I have learned)

That said, there is no such thing as a perfect mirror - yet.

Vidar

If one assumes 99% reflectivity then that's 100 reflections to reduce by a factor of 1/e in intensity. In a 10x10x10 meter cube, something like 1 km total distance. At 300,000 km/second, around 3 microseconds.

That's about 30 times faster than a good quality photographic strobe.
It's interesting to compare signal attenuation in an optical fibre which carries light over thousands of km, involving multiple TIR reflections- plus all that glass.

It's interesting to compare signal attenuation in an optical fibre which carries light over thousands of km, involving multiple TIR reflections- plus all that glass.

Don't they have to amplify the signal on long runs to maintain it? I would think that the lights energy loss would be less because its reflection off the inside of the fiber is more of a ricochet.I also would think that all that glass doesn't matter too much because none of the light seems to escape out the side of the fiber. That makes me think that all the glass is doing is slowing down the light by increasing its wavelength and leaving some energy behind. The extra energy must go into the fiber in the form of heat. Does that then mean that the amount of heat build up in the fiber is equal to the lights energy loss or is there something else picking up some of the energy.

Ratchettrack

Don't they have to amplify the signal on long runs to maintain it? I would think that the lights energy loss would be less because its reflection off the inside of the fiber is more of a ricochet.I also would think that all that glass doesn't matter too much because none of the light seems to escape out the side of the fiber. That makes me think that all the glass is doing is slowing down the light by increasing its wavelength and leaving some energy behind. The extra energy must go into the fiber in the form of heat. Does that then mean that the amount of heat build up in the fiber is equal to the lights energy loss or is there something else picking up some of the energy.

Ratchettrack

I think you should read a bit about optical fibres before trying to make a contribution about something so sophisticated.
What's a "ricochet" when applied to total internal reflection? Do you know anything about Total Internal Reflection?

Dear Ravalanche,

The idea of enclosing light in a cavity with reflecting walls is very old. In fact there is intense search for high-Q cavities. Q or the quality factor is a measurement of the goodness of a resonator, and it is related to the decay time (dacay being cause due to losses). Optical fibers already mentioned are one way of making such a device. Another one is using photonic crystals. However, the best resonators are so called whispering gallery resonators (named so after their acoustic analogs in the form of some architectural constructions) whose Q-s are much above those of any other device.

Just to give you an idea of their construction:

That said, there is no such thing as a perfect mirror - yet.

Vidar

But is there a [conservation] law in nature that says an ideal mirror is not allowed?

Even if such a reflective surface is permitted by nature, the Compton Effect would cause the wavelength of the light bulb's emitted light to eventually increase to ∞...

But is there a [conservation] law in nature that says an ideal mirror is not allowed?

Even if such a reflective surface is permitted by nature, the Compton Effect would cause the wavelength of the light bulb's emitted light to eventually increase to ∞...
Different metals reflect light differently. A reflective surface for visible light might absorb most of light with shorter or longer wavelength. Dont know about any law except it would be impossible with more than 100% reflective.

If the mirrors in the cube really was reflecting 100% the energy will stop increasing as soon as the bulb shuts off. Then the cube will act as a capacitor which stores the electromagnetic radiation inside it.

It's interesting to compare signal attenuation in an optical fibre which carries light over thousands of km, involving multiple TIR reflections- plus all that glass.

not really ;)

in reality, for practical data transmissions, distances between repeaters is quite low ( depending on the data transmission speed)
Optical fibre systems I was installing with Telecom in New Zealand, had repeaters at 10km separation
Have a look here for some Bit Error Rate and attenuation calculations

Dave

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I think you should read a bit about optical fibres before trying to make a contribution about something so sophisticated.
What's a "ricochet" when applied to total internal reflection? Do you know anything about Total Internal Reflection?

Well, I didn't state things as a fact but instead as what i was thinking. I'm looking for insite here. Are you inferring that you should only post on this forum if you know all the answers already? If everyone who post already knows all the facts and are not allowed to inquire, then what is the reason for the forum. I joined this forum to learn something but your indicating that is unacceptable. My post wasn't to make a contribution other then stimulating though. I don't claim to have the answers. Besides a contribution is worthless unless there is someone on the other end to benifit from it in some way. As far as this thread goes, I'm the one on the other end. I'm not in the position to act like I know everything and figure if someone else doesn't know something, their an idiot. Ricochet is when something strikes a boundry at an angle and rebounds back off the boundry without going thru it. Isn't that kind of what the light is doing in the fiber?

Ratchettrack

....Ricochet is when something strikes a boundry at an angle and rebounds back off the boundry without going thru it. Isn't that kind of what the light is doing in the fiber?

Ratchettrack

yes it is, but its just not a term used to describe the interaction of light with other objects :)
In this instance, reflection is the term more commonly used

yup its all live and learn, and compared to some of the "heavyweights of physics" on this forum, I'm just a young lad haha

Dave

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not really ;)

in reality, for practical data transmissions, distances between repeaters is quite low ( depending on the data transmission speed)
Optical fibre systems I was installing with Telecom in New Zealand, had repeaters at 10km separation
Have a look here for some Bit Error Rate and attenuation calculations

Dave
I understand that 100km is the sort of spacing, using the newer (0.2dB/km) fibres. That is fairly 'stunning', in my estimation. Obviously, attenuation is not the only problem and bandwidth also needs to be considered - but we were only discussing losses in this thread.

Well, I didn't state things as a fact but instead as what i was thinking. I'm looking for insite here. Are you inferring that you should only post on this forum if you know all the answers already? If everyone who post already knows all the facts and are not allowed to inquire, then what is the reason for the forum. I joined this forum to learn something but your indicating that is unacceptable. My post wasn't to make a contribution other then stimulating though. I don't claim to have the answers. Besides a contribution is worthless unless there is someone on the other end to benifit from it in some way. As far as this thread goes, I'm the one on the other end. I'm not in the position to act like I know everything and figure if someone else doesn't know something, their an idiot. Ricochet is when something strikes a boundry at an angle and rebounds back off the boundry without going thru it. Isn't that kind of what the light is doing in the fiber?

Ratchettrack

Perhaps I made a slightly grumpy reply - sorry. But there were two 'new' ideas in my post. (Clearly new to you, at any rate) I used the term TIR and the notion of a very long path length absorption. Presumably you were sitting with your browser, reading that post and a 'considered' response would surely require a bit of a look elsewhere to see what TIR is and to look at light losses in glass. Wiki is a good first go for most questions (instant expert). Or at least, you could have asked what I meant by TIR (which, of course, relies entirely on oblique incidence on the surface - look it up). If you have ever looked at the edge of a sheet of ordinary glass you must have noticed that even a few cm thickness reduces the transmitted light level so these fibres are very impressively low loss in comparison.

Also, the light doesn't "slow up" on the way through and its wavelength remains the same. Yes, the lost light energy will turn up as heat - but spread over many km of fibre so it would be hard to detect.

Also, the light doesn't "slow up" on the way through and its wavelength remains the same. Yes, the lost light energy will turn up as heat - but spread over many km of fibre so it would be hard to detect.

Thanks, I'm a high school guy that has worked a lot with physics. Have several patents and have worked out a complex problem that all the engineers of my competors said couldn't be done. Most people that know me think I'm an engineer. Got them fooled. I'm from a family of engineers. That said, I'm not in the class of most on here and don't have the formal training. As a result, I don't know the proper terms but I do have a pretty good sense for the concepts. You might say I'm a jack of many areas but a master of none. I'm also sorry for retaliating as aggressively as I did. Guess that comes from my insecure years. Anyway I want to thank all of you for your guidance on this Light Box thread. I gained a lot of insite.

If anyone is interested in the complex problem I worked out, just say so and I post a new thread presenting the problem.

To your final statements, if heat is produced but the light doesn't slow down and its wavelength remains the same, where does the heat energy then come form?

Ratchettrack

The heat comes from photons that were absorbed. The light intensity drops

The short answer is the box will get slightly warmer, as the optical energy is reduced
to longer and longer IR wavelengths.
At some point the components of the mirror become a black body, and it all turns to heat.
The same question could be stated for computers, what percentage of each watt going into a computer turns to heat, effectively 100%, as the computer does no real work.