Light in a mirror box

davenn

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

ratchettrack

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

davenn

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

sophiecentaur

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.

sophiecentaur

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.

ratchettrack

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

Khashishi

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

johnbbahm

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.