Consider a material of a refractive index linearly increasing with depth from the surface: we have *n*=1+*ky*, *y*>0, for some positive constant *k*, with *y* being depth from the surface. What will be the trajectory of a light ray entering the surface *y*=0 at an angle *θ*>0 from the surface normal.

Let x be the direction along the surface, with the x component of the initial direction being positive for *θ*>0. We choose our origin to be the point the ray meets the surface.

Using Fermat’s Principle, we know that the optical path length must be extremal.

For a path from (0,0) to (*x*_{f},*y*_{f}), *x*_{f}≥0, with a path described by , we obtain an optical path length of:

We want to find the extrema, so we use the calculus of variations, and the Euler-Lagrange equation:

with , we have

Now:

,

,

,

So:

This is a second-order non-linear differential equation, with no explicit dependence on the independent variable *x*. So, we make the transformation , (see case II in this Monday Math post), and get:

This is separable:

This, in turn, is separable:

Rearranging,

,

and using *y*(0)=0, we find:

And so

Now, we note that . We have

so

Thus:

And plugging this back into our equation for *y*(x), we get:

And thus the path of a light ray within the material is a segment of a catenary (see also Physics Friday 6).

### Like this:

Like Loading...

*Related*

Tags: Calculus of Variations, Catenary, Euler-Lagrange, Fermat's Principle, Friday Physics, Optics, physics, Refraction

This entry was posted on May 9, 2008 at 12:01 am and is filed under Math/Science. You can follow any responses to this entry through the RSS 2.0 feed.
You can leave a response, or trackback from your own site.

January 29, 2010 at 12:11 am |

[…] fitting the above, and thus use the calculus of variations, as I did in Physics Friday 6 and Physics Friday 19. The surface area of the surface of revolution is given by the integral (see equations 1-3 here). […]