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Optical Fiber, beam path

  • Thread starter C4l1
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  • #1
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Homework Statement


An optical fiber is one way to guide light efficiently from one point to an other. It is currently used for data communication: it offers low loss and very high bandwidth, ideal for the requirements of the internet. Generally, we can describe an optical fiber as a medium with an index of refraction n which depends only on the distance from its axis for example Ox. If we assume an incoming beam in the plane Oxyat start crossing Ox with an angle θ0,
show that, by solving the Euler-Lagrange equations, the equation for the beam path can be written as:

##n \sin(i) = a##

with a a constant to determine and i the angle that the ray makes with the normal. You will have to keep some approximation on the angle

Homework Equations


Euler lagrange equations:
##\frac{d}{dt}\frac{\partial\mathcal{L}}{\partial x'}-\frac{\partial\mathcal{L}}{\partial x}=0 ##

##\frac{d}{dt}\frac{\partial\mathcal{L}}{\partial y'}-\frac{\partial\mathcal{L}}{\partial y}=0 ##

##\frac{d}{dt}\frac{\partial\mathcal{L}}{\partial z'}-\frac{\partial\mathcal{L}}{\partial z}=0 ##

The Attempt at a Solution


I've tried rewriting the Euler-Lagrange equations using the Principle of Fermat, that light will travel the path between A and b that is shortest, in order to get a differential equation:

##\frac{d}{ds}\left(n\frac{d\vec{r}}{ds}\right)={\vec\nabla}n##

with s the distance travelled by the light and r the position in space.
From this point I did not get any further on relating this equations to this particular beam path length in an optical fiber proposed, especially on relating it to the angle i.
 

Answers and Replies

  • #2
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Are you excluding the only thing that keeps the light in the fiber is the angle of refraction provided by the difference in R(i) of two mediums glass n1 and n2 (cladding,air,etc..)?
 

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