Curious question about fiber optic sights, how they work?

[nabiul]

Could someone with real scientific knowledge answer this?

I have thought about it and still can't seem to figure this out. How do those nifty plastic rods channel the light entering from the sides to the ends? All of the theory I know suggests that the light striking the sides would pass right through after being displaced from refraction, UNLESS the sides of the plastic object were slanted in a prism like structure which would make it a 'one-way' light channel. But the plastic rods used in gun and bow sights are just regular cylindrical rods from what I can tell.

Any ideas?

 

[XXXXX]

The sight is a glass or plastic cylinder, so light is passing from air into plastic/glass.

When light enters a material that is more dense than the material it is leaving, it is refracted like this:

So the direction of the ray of light changes towards the other end of the cylinder. This allows a lot of light to be funneled into a relatively small area.

Except in this case, the sides of the fiber optic cylinder is clad with a reflective coating, which does not allow the light to readily escape. In fact, in high end fiber optics, it is totally reflected off the walls back into the cylinder, like this:

So, all that light is funneled in the front using the concept of "refraction" and then it "reflects" off the walls of the cylinder towards the shooter's eye.

That's all fine and dandy, but to make the other end glow bright and evenly, many sights/fiber optics use a rougher (think sandblasting/hazed over) exit path for the light which makes the light scatter evenly and brightly.

Think of sandblasted glass and how it "glows" when light is on the other side of it, because of all the different reflection and refraction angles.

Some exit paths are slightly concave as well, to focus the light down the line of sight.

Anyways, I hope I explained it well.

 

[nabiul]

I know that, but that does not explain how light from the sides leaves through the ends. The refracted light is just displaced and bends back toward the original beam direction when it leaves the material; light from more dense to less dense bends back in the other direction, away from the normal of the material surface.

ex:

In fiberoptics, a cladding isn't needed either, only a less optically dense material on the outside and air is almost the best thing for that. Since light leaving a more dense material bends towards the surface of the material, there is a critical angle where the light leaving becomes paralell to the surface of the material, and after that reflects completely back inside the material. This is the principle that real fiber optics uses.

ex:

So by conventional theory, for the light from the sides to be leaving the ends, it would have to strike the material surface at such an angle that the refracted ray strikes the opposite surface of the material at an angle greater than the critical angle, at which point the light would be completely reflected towards the end of the material. However this is impossible, as I have already said the critical angle is when the path of the light beam is refracted 90 degrees to be paralell to the surface of the material, so the only light that can be reflected has to come from the inside of the material.
ex:

Now the one way I have come up with is slanting the material surfaces so that the refracted ray does strike with an angle greater than the critical angle.
ex:

>.> So I'm still stuck on how this actually works, if any one out there knows exactly how please tell me, and don't spare the technical details.

 

[Grtwhthntr]

###XX response pretty much summed it up for me, interesting.

 

[nabiul]

It didn't for me, his diagram shows light entering from the sides bending away from the end of the tube, not towards.

I've taken physics in highschool and first year university, none of the basic theory (snells law, refraction/reflection, wave front interference) covers how it works.

If it was explainable in the way ###xx mentioned, any optical material would have glowing sides since all materials are more optically dense than air, but they don't.

 

[TDC]

Total internal reflection...

TDC

 

[Ratzilla]

First off, I'd like to say that I am by no means an expert in the subject. I major in Engineering Physics and have taken a 2nd year university course in optics as well as a 4th year level course. I have learned a few things so all I can offer is my scientific explanation.

First thing to understand are the coefficients in between reflection and transmittance. The coefficient of reflection and transmittance are related through stoke's relation http ://en.wikipedia.org/wiki/Stokes_relations.

Think of the fiber optic like a cavity. The principles suggest that light will enter and get "trapped" by being continuing bounced off the inside. http ://en.wikipedia.org/wiki/Optical_cavity . Although the fiber optic doesn't use the same optical cavities that are required for lasing, it still gathers light. By trapping light, I mean that light will be repeatedly bounce off the inside edges. What happens if light comes at any angle? Eventually, a good portion of that light will reach the ends.

I think the mistake is thinking of the fiber like an optical waveguide. Any light that enters into the rod from one end will propagate until it reaches the end. In addition to that, light can still enter from the sides. Your misunderstanding comes from the part where you have deduced that light has to come from "inside" the medium. It does not. You have already deemed that it is impossible for that scenario to occur for the picture with the critical angle and the beam thats "impossible". The critical angle is not important in whether or not this works. It tells you which are the allowable angles for total internal reflection to occur. Light can enter through the ends and propagate to the other end but this does not explain why it's BRIGHTER.

If you point your sight at something bright, it won't appear brighter. Let the light source shine on the sides and you'll see it very bright. Also try this. Cover up the front of the sight and you'll notice that as long as you let light hit the sides, the sights will still appear just as bright.

I am also a little confused why you need it to "channel light in one direction". Light will exit on both ends of the optic fiber. And this is mainly due to the fact that "less light" exits through the sides than enters through the sides.

I'm not sure if I answered your question but I don't really understand what you mean by

only a less optically dense material on the outside and air is almost the best thing

. Do you mean that a light in a material that is optically more denser than air?. Also, In fiber optics, the critical angle is not the angle at which they transmit. They do not want the optic to travel parallel in the direction of the material.

Tell me what you think. It's really late so I don't know if I have made my points clearly.

 

[nabiul]

The less optically dense comment was in response to ###xx's statement that fiber optics needs a cladding to work, it doesn't. Just the interface of the exiting ray has to go to from a higher refractive index to lower, thats all I meant.

Ok I see, you are talking about the natural reflection you get along with the refraction, I forgot that part. But only a certain % of the light is reflected everytime the beam encounters the material/air interface, meaning that within a few reflections, most of the light will have exited through the sides. Does some property exist that allows materials to reflect a higher % of light?

I'm not sure what stokes relation has to do with this, I couldn't understand the wiki page, but from what I read it's just talking about the 180 degree phase shift that occurs when reflecting from a material with a higher refractive index?

Ok there is one part of theory that I'm not so clear on, when you have a material with X wavelengths of thickness of the incoming light as to cause destructive interference from the two reflections, does the light remain 'trapped' in the material? I mean when you 'cancel out' two EM waves, the energy has to go somewhere. This is in relation to a not so clear question that was in our text, having to do with the reflection losses of solar panels and applying a coating to cause thin film interference which would reduce the 'loss of light'.


I see what you are saying, but I'm still not clear on what exact property that these sights have from regular transparent materials that make them act as an optical cavity.

Also going by this theory, there would be an ideal length of 'fiber' which would have the maximum possible glow from the reflections and anything longer would have no effect on the brightness since there would be an insignificant amount of light left after a certain number of reflections. However manufacturers of fiber-optic telescopic sights use several windings of the stuff?

 

[nabiul]

I've found part of the answer, apparently some types of fiber have a flourescent dye inside them so part of the light actually does come from inside. Still not there at the 100% though.