Light transmission question

Is there a material that would allow light in but not out. such as a polymer or something? Or a coating that you could put on a polymer such a silicone tubing?
Or a polymer with a very high transparency, higher than glass for example?

Originally posted by mlprior
Is there a material that would allow light in but not out. such as a polymer or something? Or a coating that you could put on a polymer such a silicone tubing?
Or a polymer with a very high transparency, higher than glass for example?

http://web.mit.edu/newsoffice/1998/mirror.html


What's this all about? 😛

Originally posted by leisurelysloth
http://web.mit.edu/newsoffice/1998/mirror.html


What's this all about? 😛

Yes, smarty pants....it's a mirror.

OK, I'm not talking about reflecting light, I'm talking about absorbing and then trapping it.

Oh, never mind already!

😠

Originally posted by mlprior
Yes, smarty pants....it's a mirror.

OK, I'm not talking about reflecting light, I'm talking about absorbing and then trapping it.

Oh, never mind already!

😠

There are mirror arrays that can trap light but there is no reflecting surface, no matter what kind that is 100% reflective, even the multilayer additive reflection coatings are say 99.9% reflective. So that means that after 1000 such reflection, there will be no more light left but you can trap light for that time/distance, say a series of mirrors to slightly deflect a beam eventually bringing it around to the original mirror, which would have a hole in it for the entering beam but enough area around it to keep the beam going in a daisy chain kind of thing, so if one trip around took (picking a # out of mid air) say, 20 mirrors are arranged to reflect a beam around a circumferance of 10 meters, say, well light goes about 300 million meters per second so one meter takes light about 3 nanoseconds in space, a bit slower in air and a lot slower in water but you get the idea, so in our example, 99.9% reflective mirrors, 20 of them sending light on a series of reflections that ends up being a total of ten meters would lose 20/1000ths of its energy on each trip around the circle (2 percent) So it would lose all of its energy after 50 circular trips, supposing you had a VERY fast way to switch on and off the light beam, so you make a light beam that is 10 meters long, or about 30 nanoseconds of light and let that into the mirror system, the leading edge of the beam would go round the circle just in time to see the tail end of the beam and you would get 50 such trips or a trail of light that would follow its tail so to speak, for about 1500 nanoseconds or 1.5 microseconds for a total beam path of about 1500 meters before it poops out. Now if the beam was very well collimated, say a nice tight laser beam, and the mirrors were in space with no air you could separate the mirrors to give say, 1000 meters circular light path and the beam would last 100 times longer or 150 microseconds, still not a whole lot of time! It takes 1000 microseconds to equal one millisecond so you can see that would be about 0.15 milliseconds, the light would still be dead before you can blink. So the storage part won't fly. Absorbing on the other hand, there are carbon nanotube absorbers now that absorb something like 99.9999% of light hitting such a surface, the blackest black of all times but it would in turn heat up the surface, turn it into infrared energy and re-emit it as heat radiation as well. So the trapping part can be done but what do you do with it after that? Now its heat. Be a great solar heating collector, it absorbs light better than anything else ever. Outside of that application though, not sure what you would do with it. I guess you could use it for those black velvet Elvis paintings you get in Tijuana🙂

Originally posted by sonhouse
There are mirror arrays that can trap light but there is no reflecting surface, no matter what kind that is 100% reflective, even the multilayer additive reflection coatings are say 99.9% reflective. So that means that after 1000 such reflection, there will be no more light left ...

I don't mean to be the jerk "corrector", but a point of interest:

If a surface were 99.9% reflective, there would be approx. 36.7% of the light left after 1000 reflections. (.999^1000)

Originally posted by forkedknight
I don't mean to be the jerk "corrector", but a point of interest:

If a surface were 99.9% reflective, there would be approx. 36.7% of the light left after 1000 reflections. (.999^1000)

Well thanks for that, I assumed that if you reflected 99.9% you would lose one part in a thousand after each reflection. What is wrong with that logic? I did your calculation with my casio and it indeed comes out at 36% but why? Oh, wait, if you lose one part in a thousand after each reflection, then the second reflection loses one part in one thousand of what is left not the original #. Aha, I see it now. Thanks for pointing that out and you are not a jerk sir! I think we need more digits to do the whole calculation then to see how long a light path I set up would actually last before it poops out. I think I will do it again on my trusty rusty HP48🙂

Originally posted by mlprior
Is there a material that would allow light in but not out. such as a polymer or something? Or a coating that you could put on a polymer such a silicone tubing?
Or a polymer with a very high transparency, higher than glass for example?

The higher than glass transparency thing was something I studied here in the photonics industry, I was thinking for instance, that the optical fibers we use that transmits our optical signals for 80 Km before needing amplification, that would be the same as a piece of glass 80 Km thick. But what is happening is the light is guided by total internal reflections and if you had a piece of the plastic that makes up the fiber actually 80 Km thick, it would be pretty opaque I think so I was comparing apples and oranges at that point. You would have to compare the difference in transparency of glass to perfection by just showing how much it would attenuate compared to vacuum, total vacuum that is, no atoms in the way to muck things up, scattering light, etc. Anyway thats one way to look at it.

Originally posted by sonhouse
Well thanks for that, I assumed that if you reflected 99.9% you would lose one part in a thousand after each reflection. What is wrong with that logic? I did your calculation with my casio and it indeed comes out at 36% but why? Oh, wait, if you lose one part in a thousand after each reflection, then the second reflection loses one part in one thousand of w ...[text shortened]... up would actually last before it poops out. I think I will do it again on my trusty rusty HP48🙂

Mathmatically, it will never "poop out"; it will, instead, asymtotically approach 0.

The amount of light will be attenuated by
.9 after 105 reflections
.5 after 692 reflections
.1 after 2301
.01 after 4602
.001 after 6904
etc.

In practice, the field just accepts a certain level of attenuation that is considered to be "enough", whether that be at 10%, 1%, .1% or some other level.

For electrical signals, that point is at 3dB down, giving the attenuated signal half of the power of the original, which equates to .707 attenuation, because signal power is proportional to V^2

Originally posted by forkedknight
Mathmatically, it will never "poop out"; it will, instead, asymtotically approach 0.

The amount of light will be attenuated by
.9 after 105 reflections
.5 after 692 reflections
.1 after 2301
.01 after 4602
.001 after 6904
etc.

In practice, the field just accepts a certain level of attenuation that is considered to be "enough", whether that ...[text shortened]... the original, which equates to .707 attenuation, because signal power is proportional to V^2

Yep, that I knew, I am into amateur radio, my call is AI3N, live in Pa. So how many trips around my hypothetical 20 mirror reflector series would it take to make the light go down to 1% of its starting value?
That has implications for quantum computers because they can use light paths like that for information storage. I used to work on the Apollo project in Apollo tracking and timing and one of the devices way back then was a delay line, an audio delay line that they used a transducer to put in information and to sonically vibrate a long wire in a spiral shape factor. They did not have our neat digital delay lines such as I use in my audio recording studio, Cakewalk Sonar. So they had to use analog means to store information, it was very complex for sure, especially for 1969!
The same kind of thing can be used for optical information storage. So wonder what the time would be where it would be below detectability.
We can also just use fiber optic networks too, because the devices we make here at Cyoptics can send signals down the fiber for 50 miles without a repeater, so that is another technique possible for data storage without resorting to electronics. BTW, we are working on a laser modulator for fibers that can stuff one TERAbit per second down a single fiber for 50 miles without the need for amplification!

Originally posted by forkedknight
Mathmatically, it will never "poop out"; it will, instead, asymtotically approach 0.

Not exactly, since light power is not a continuous variable. A photon has a finite amount of energy, so when the light power has reached the order of that of a few photons, it may "poop out".