Imagine a flat panel covered with tall thin fiberoptic cones packed close together preferably submerged in a clear material with a refractive index similar to air. Light shining on the panel will shine through the first material and strike the surface of the cones. Since the cables have a larger refractive index, some would be bent in and some would be reflected off. The light reflected off would then strike another cable, etc. This could absorb a large portion of the incoming light. Once in the cables the light would be contained as long as the angles are kept large enough to prevent escape. It could then be culminated together by merging fibers. These panels could form roofs that would absorb a large portion of the light falling on them and channel it into fiberoptic cables used for lighting or energy production.-- MisterQED, Jan 31 2011 Wikipedia - Optical Fiber http://en.wikipedia.../wiki/Optical_fiber [MisterQED, Jan 31 2011] The thought that lead to this Light tube accumulatorC Mr's anno. QED [pashute, Jul 01 2013] I think to get any useful bend with refraction, you're going to have to suffer some loss just like you would with reflection. Air's refraction index is really low, almost identical to a vacuum which is the base you work from being an index 1. Air 1.0008, Water 1.330, Glass, 1.510. I'm pretty sure the higher the refraction, the more it impacts the light and "bends" it.
Whenever you put a straight line of light into a curving path you're going to have it "ricochet" around loosing energy with every bounce and it will never actually straighten out. You can get it from point a to point b eventually but not without cumulative loss. Fiber optics work because the laser light is many times more powerful, concentrated and coherent than sunlight so it's no big deal, you basically get most of what you put in out at the other end, at least enough to use as a medium to transfer data. You just need to be able to tell whether the light is on, off or where it is in between.
To get useful light for illumination you're going to need to get a significant portion into the house through your scheme and whether using reflection or refraction you're going to get attenuation of your original source at every turn.
So it would work, but would you get enough light to make it worth while? That's the question.-- doctorremulac3, Jan 31 2011 There are no losses with reflection or refraction. You are right about the Index changing the amount of bend you get. Losses come from lack of clarity of the glass and light escaping from the system. Once any kind of light is in a fiber, it can only escape by hitting a refraction gradient at less than it's critical angle of reflection which is based on the gradient and the frequency of the light. As long as the light impinges on the gradient at greater than that angle it is reflected. You can test this by looking at the side of a fish tank.
You can send any kind of light into a fiber and get it out the other end as long as its frequency and angle of insertion are within limits. The reason you don't use incoherent light for communications is that the different frequencies end up bouncing differently and one end of the band will end up bouncing at higher angles and thus travelling farther before exiting out the other end of the fiber. This muddles the signal, but outside the losses due to lack of clarity, all that went in comes out. (link)-- MisterQED, Jan 31 2011 //Losses come from lack of clarity of the glass and light escaping from the system.//
//but outside the losses due to lack of clarity, all that went in comes out.//
You're assuming that you get 100% reflection off any given surface which you don't and the material you refract through it reducing some of the light going through it as well. Light doesn't pass through or bounce off any material without some loss.
If you don't think the reflection or refraction effects attenuation take an optical fiber, put a light through it, measure it, then wrap the fiber around a pencil and measure how much light's coming out now. It'll be quite a bit less because you've increased the number of times the light hits the walls of the fiber. This can't be explained by the light simply traveling through more glass due to the increased number of angles because the coil still would attenuate more than that same fiber perfectly straight hundreds of times longer.-- doctorremulac3, Feb 01 2011 Thanks-- pashute, Feb 04 2011 //You're assuming that you get 100% reflection off any given surface which you don't//
actually, yes you do... in certain circumstances. It's called total internal reflection (TIR).-- afinehowdoyoudo, Feb 06 2011 I asked my physics lecturer that question. He answered that in the real world, total internal reflection is always less than 1, i.e. there is some loss each time. I had suspected that to be the case, which is why I asked.-- spidermother, Feb 07 2011 There is also the problem that unless you are tracking the sun a large majority of sunlight hitting the fiber is not going to be parallel to the length of the fiber. It will, therefore, hit the first reflection at steeper than the critical angle and you won't get TIR in the first palce.-- MechE, Feb 07 2011 //He answered that in the real world, total internal reflection is always less than 1, i.e. there is some loss each time.//
Originally fibers were created of single materials and reflections would happen at the edge of the fiber. It is tough to make a fiber optically smooth so light would be lost due to imperfections. Modern fiber uses materials whose index varies radially, so light does not follow a sawtooth path but a shorter sinusoidal path which shouldn't intersect the sides of the fiber.This allows for nearly total internal reflection. Nothing is perfect, but it is close.-- MisterQED, Feb 07 2011 //There is also the problem that unless you are tracking the sun a large majority of sunlight hitting the fiber is not going to be parallel to the length of the fiber. It will, therefore, hit the first reflection at steeper than the critical angle and you won't get TIR in the first place.//
Getting TIR without tracking is the whole point of the conical end sections. It does it thru multiple reflections.-- MisterQED, Feb 07 2011 //actually, yes you do... in certain circumstances. It's called total internal reflection (TIR).//
...actually, no you don't. Total internal reflection is a term referring to how light is redirected, not a statement that 100% of that light is reflected.-- doctorremulac3, Feb 08 2011 If I understand this correctly, you are merging the optic fibers at an angle each into the next, and receiving the same result as my proposed (but flawed) mutiple L's idea. Correct?
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dr.m says that you may loose a lot from the extra refractions. maybe, but its still worth trying, because if it works, you have three benefits:
A flat collection system (rather than a parabolic mirror).
Extensible - you can add units to it at any size (unless there is a limit - which I'm not aware of)
Light energy be lead to a distance at a small cost.-- pashute, Jul 01 2013 The bottom line is - can you imagine something made of transparent materials, with any combination of refractive indices, which looks profoundly black? It may be possible, but my guess is that physics will jump up and bite you. Everything transparent that has a refractive index higher than air, that I can think of, either reflects light, or scatters it, or does both.
A lot of effort and thought has gone into coupling light into fibreoptics (for instance, coupling a freespace laser beam into a fibre), and in general the only way to do it efficiently is to start with well-conditioned light and do lots of things to it very accurately.
So, I'm neither convinced nor anticonvinced, but very skeptical.-- MaxwellBuchanan, Jul 01 2013 random, halfbakery