1
$\begingroup$

I want to channel light from an LED indicator through some clear, circular acrylic rod. I want a rough-ish idea of the smallest bend radius through which the acrylic can be bent and maintain TIR. I will be experimenting, and would like an equation to guide the design.

The surrounding medium will be air. I don't know if there are different acrylic grades that need to be factored in, hopefully not because retailers don't seem to give further information.

The wavelength will vary, but only visible light is of interest. So I'd like an equation to give minimum bend radius, given a particular wavelength and rod diameter.

$\endgroup$
3
$\begingroup$

You can figure this out from first principles. Here's how:

First, you will need the critical angle for the acrylic material. This is the steepest angle that light can strike the acrylic/air interface and still be totally reflected, and it follows from the basic laws of refraction.

sin(t) = n1/n2

Where n1 and n2 are the refractive indices of the air and rod. Then, you have a geometry problem to work out how that fits in with the curve of the rod. Here's a diagram; the edges of the rod are black, the ray of light is red, the blue and green are constructions to help solve it.

diagram of light reflecting in rod

The rod of diameter d is bent around a curve of radius r. The light beam is tangent to the inner edge at B, and strikes the outer edge at angle t - so it stays in, but only just. Angle OBA is between a radius and a tangent, so is a right angle.

We can then express t using some trigonometry as

sin(t) = r/(r+d)

Serendipity! we already know sin(t) in terms of refractive indices, so we can write:

n1/n2 = r/(r+d)

And re-arrange for r:

r = d/(n2/n1 -1)

If we use the refractive index of air (n1=1) and acrylic (n2=1.49) then this simplifies to

r > 2.04d

Now, there may be slight variations in refractive index for different acrylic grades, and for different wavelengths, so you'll probably want to keep to a bend radius a bit larger than that, but it should give a good starting point.

| improve this answer | |
$\endgroup$
1
$\begingroup$

This is a well-known and multiply-engineered problem, as you might imagine. The simplest fix, at least conceptually, is to coat the exterior with a metallic layer so that you no longer depend on internal reflection & Snell's Law. At this point, your don't need the core (glass) any more, just a shiny tube.

But for situations (glass, or cladded optical fibers) where you do depend on total internal reflection, you want a material with the highest possible index of refraction. Follow the equations in other answers to see why. The usual problem is that high-index materials tend to have a higher absorption coefficient as well. One way or the other you lose some of your signal.

Finally, if you have money, you might be interested in "holey fibers"

| improve this answer | |
$\endgroup$
1
$\begingroup$

I've realised something which Jack B's answer didn't mention.

Say you have a very long, straight rod section, of which one end has light entering, at all angles.

  • Any beams of light which are beyond the critical angle immediately escape the rod.
  • the remaining rays continue bouncing indefinitely.

But:

  • The moment there is any bend of any kind, it is inevitable that some light will exit.
| improve this answer | |
$\endgroup$
  • $\begingroup$ In addition, after multiple reflections, it gets worse. $\endgroup$ – Carl Witthoft Jul 5 '17 at 15:33
  • $\begingroup$ Indeed. My answer applies to light which enters straight (or very nearly straight) into the rod. You might wish to make sure most of the light does this by carefully positioning the rod and lamp, shaping the end of the rod, or adding additional optical elements to collimate the light. $\endgroup$ – Jack B Jul 5 '17 at 16:03

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.