Expert: Daniel Mazur - 8/9/2006

Question
I am a new grad student in biomedical engineering.  The subject of diffraction interests me beyond what my textbook explains.  One part has me very confused.

1.  Do you normally (in an ordinary thin pipe with collimated light parallel to it going through it) lose more and more light the longer the pipe is due to diffraction (or does it just affect you at the entrance and exit of non TIR pipes?

2. Does TIR, even in a very, very thin fiberoptic, make diffraction "not happen" in the middle of the fiberoptic?  

3. Does light diffract as light exits the fiberoptic?  I know it diverges (since light exiting isn't collimated) but is some of that divergence due to diffraction?  (i.e. if the light were collimated would it still diffract at the fiberoptic exit?)  It would seem that it might not diffract at the exit if the boundary at the exit was TIR.

4. My textbook just shows that wavelength/diameter determines the diffraction  (kind of a ratio) -- but how would you calculate
  a. the maximum degree of divergence of collimated light leaving a pipe?
  b. the amount of light lost at a given angle of that divergence?  For example, with a pipe of diameter D and light of wavelength L, what percentage of the light would be lost being difracted 10 degrees off the central axis? 5 degrees etc.?

This is a very interesting thing but my textbook says nothing about it.

Thank you very much for whatever light you can shed!

John

Thank you very much.


Answer
Hello John,

your question is beyond my curriculum and knowledge, I am afraid. Let me just answer what I know. I learned something at school and don't know, if it is 100% valid or relevant.

Fiberoptic cable is a bundle of optical fibers, where the number of fibers is N times the number of signals transmitted together. This N>1 because individual fibers can easily become "blind", so that some redundancy is necessary. Each optical fibre is usually coated with high-reflectivity metal to minimize losses en-route, but that's not all. Each optical fibre has a refined profile of the material's refraction index n across the diameter, so that non-axial waves do not get delayed with respect to the axial waves. Thanks to this, a square pulse sent in on one end does not smear into a gaussian-like one, by the time it reaches the end of the fibre. The refraction index profile further reduces the amount of losses by reducing the power density reaching the walls of the fibre.

1. Now, because the losses are non-zero in every practical application, an optical cable cannot be infinitely long. You need an optical amlifier every 500 meters or so to recover the signal to its original strength. However, if you're talking about really short cables, there the losses at the ends of the fibre may be larger that the losses along the cable.

2. I don't know enough to answer this. It seems to me that there is no need for diffraction to happen until the light reaches the fibre end. Scattering on impurities, yes, or if your fibre is designed with periodically varying refractive index along its length, yes indeed. I apologize, I cannot help more.

3. The end of a fibre should, really, be where some diffraction happens. Popular experiments with laser shining at a small opening (slit or circular hole) show us all. You can treat signal in the oprtical fibre as a planar wawe, to some extent, and as such, when it reaches a point of abrupt change in boundary conditions, it will tend to diffract. The spreading will increase with decreasing fibre diameter as ~Lambda/D.

4. I don't know any better advice than look it up in a general physics textbook, or maybe one of the Electromagnetism and Optics. Provided diffraction is the main reason for light divergence at optical fibre ends, you will find diffraction patterns (angular intensity profiles) there and you can derive answers to both your questions there.
a) As far as I remember, Lambda/D is the angular width of the zeroth maximum in the diffraction pattern of a collimated light beam (quasi-planar wave). This also may be the value you are asking for, but first consult the Optics/Diffraction. Even Wikipedia http://en.wikipedia.org/wiki/Diffraction may give you some idea.
b) When you get the expression of angular dependence of intensity in the diffraction pattern, just integrate it from zero to finite angle and that's the answer.

Note: While the diffraction on a narrow rectangular slit has a simple enough solution, diffraction on a round hole (which is the case of fiber end) has solutions only in Bessel's functions, which are well know, but not so easy to operate with. So, if you need precise enough expressions in4a) and 4b), you will need to employ computational software like Stephen Wolfram's Mathematica.

I hope my answer has helped a bit and wish you good luck with your studies.
Daniel