Question

6.24Consider a medium in which the refractive index n is inversely proportional tor2: that is, n a/r2, where r is the distance from the origin. Use Fermat's principle, that the integral (6.3) is stationary, to find the path of a ray of light travelling in a plane containing the origin. [Hint: Use two- dimensional polar coordinates and write the path as φ = φ(r). The Fermat integral should have the form .ff(ф, ф'.r) dr, where f(d, ф. r) is actually independent of ф. The Euler-Lagrange equation therefore reduces to af/ao'-const. You can solve this for φ, and then integrate to give φ as a function of r. Rewrite this to give r as a function of φ and show that the resulting path is a circle through the origin. Discuss the progress of the light around the circle.] (time of traveldt- 1 U is a minimum. If n is constant, then it can be taken outside the integral and the problem reduces to finding the shortest path between points 1 and 2 (and the answer is, of course, a straight line). In general, the refractive index can vary, n = n(x, y), and our problem is to find the path y(x) for which the integral (6.3)

Answer 1

Solution: Fermat's principle states that n(r, φ)ds is a minimum. In polar coor-dinates ue have which leads to Hence we need to minimize which gives This is indeed independent of φ. Euler-Lagrange then simplifies to įfod-const or 20 This gives C2

Take a path in which φ 2ncreases with r. Integration gives Ст arcsin(-) + D φ where D is another constant.Inverting this equation gives sin(0-D) sin(-D) What is the equation for a circle through the origin in polar coordinates? supposes the origin of the circle is at (0,R) and the angle θ measures the angle along the circle, starting at the bottom. We then havie r-Rsin(0) y-R- R cos() This gives 2 2R(1 - cos()) 22in2 sim 2 2

which has the form ue have with ф-30. This is indeed the Telation between ф and . So our path is part of a circle through the origin. Evaluation The first thing to notice is that the actual path of the light will never go through the origin, because that would take an infinite time. At the origin the integrand diverges like 1/r2, which cannot be integrated So if we take two points, we construct a circle through these two points and the origin, and the path of light takes the part of the circle not going through the origin. Second, if light goes from a region with a lower indear of refraction to one with a higher one, it curves towards the normal. In our erample the normals are lines coming out of the origin, and a circle through the origin has indeed the correct form, bending towards the origiin at very small distance. The curvature of the path is correct.