Physics Friday 19

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 (xf,yf), xf≥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



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:

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

And so

Now, we note that . We have



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).


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One Response to “Physics Friday 19”

  1. Physics Friday 106 « Twisted One 151's Weblog Says:

    […] 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). […]

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