After a long night, complete with many ales, my brother and I found ourselves in a Physics/Optics quagmire.
Please bear with me as I attempt to word this question correctly;

Would it be possible:
For a beam of light to travel through a physical barrier (physical to solid matter,) not liquid or gas (clearly,) through one way, but if reversed would not travel the same path?
--for instance through a mesh of mirrors.

regardless of what end result would achieve, would it be possible perhaps through refraction?

it is possible (depending on what you are asking exactly), but only in a very limited way.

Assuming we are talking about a "black box", where we shine light in one end, and detect it coming out the other end, and we time how long this takes (not really caring about the internal structure of the box). We also stipulate that the box must be prepared in such a way that you don't know which end we will shine the light into (simple example: no cheating and moving mirrors inside the box between tests).

Classically I can't think of a way to do it. The temptation is to use a one way mirror but those actually won't help you here.

At a more fundamental level, individual photons do not behave deterministically. So it would be possible to run a single photon through one way, and a single one through the opposite way and get a different result. However I think if you took an average of a large number of trials you would find you got the same average for both ways. Worse still, you could run a photon through the normal way, then wait, and run another photon through the normal way and get a different result from the first! So does that even satisfy your goal? Hard to say...

I don't think this is possible for actual transmission, but you could in theory create a "one-way filter". So light would shine through one way, but not the other. One example could be using polarized filter. So a setup like
input -> polarizing crystal or mirror (parallel to filter) -> polarized filter -> polarizing crystal or mirror (perpendicular to filter) -> output

So going from input to output, your light will be polarized parallel to the filter and will pass through, but going the other way it will be perpendicular and will be filtered out. I guess technically that's "different paths".

not quite sure what you are saying, but if somewhere in your system you have 2 filters that are perpendicular to each other---with nothing in-between them---there won't be any output from either direction.

there's only 1 filter there

Polarizer are filters.

It is of course possible to do if you allow a power source. The simplest example would just be a photon detector on one end, and a flash light on the other... Light coming in one end hits the detector, the detector turns on the flash light, and light comes out the other end. Photons going the other way hit the flashlight and nothing happens. There are much more complicated systems involving stuff like Faraday rotators, but they all have this quality, they need power to get the desired effect.

2 polarizers and 1 filter. The idea is to polarize light going one way so it passes through the filter, but to polarize light going the other way so it doesn't.

What do you mean by "polarizers are filters"? A mirror can polarize light, but it won't filter it.

This post was edited by russian on Apr 10 2014 11:51am

Actually, scratch all that. How about this:



We have the incident I, reflected R and transmitted T. Let's ignore T for now and focus on R. If we shine the exact same ray back along the R vector, wouldn't we get a new ray T2, which would be the transmitted component of reversed R and would be at the same theta T angle to the normal (going towards the upper-right corner of the picture)?

BTW, I'm glad we finally got an interesting topic in here!

Ok here is what I see: The input to your black box is I, the output is R (from your picture). If we then do it the opposite way, input light along R, we get light out at I. But if you use say I as the input and T as the output, to go the opposite way we'll have to feed light in on the T path, and again we'll get light at I.

I'm kind of cheating here because I realized a while ago that what you are trying to construct is a Maxwell demon. The challenging part is to find the problem with each suggestion.