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I recalled the fact that a perfect concave mirror has not a spherical surface, but a parabolic one.

I now wonder what the surface would be for a perfect planoconvex and biconvex lens (same surface mirror imaged on both sides), that can focus all monochromatic rays that pass through it into one spot. I took my pen and paper and tried to derive a formula for the surface using Snell's Law, but I can't seem to get anywhere.

I'd be grateful if someone could lead me along the right path.

Pritt Balagopal
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A parabolic mirror is perfect only for bringing rays parallel to the axis to a focus at a single point. Are you content with this limited sort of perfection for a lens? If you are, then you can find the required shape of curved surface for plano-convex lens by applying Fermat's Principle. For details, Google "Perfect plan-convex lens" and go to the University of Arizona site. This site also finds the shape by applying Snell's law, but this method is much more complicated.

For the parallel beam landing normally on the plane face of the lens, the wavefronts are parallel to that face. Each wavefront will arrive all at the same time at that face and will go on to arrive – all parts of it – at the principal focus of the lens simultaneously. Therefore each point on the wavefront takes the same time to reach the focus. The time taken is given by $$t = \text{time travelling through glass + time travelling through air}$$ $$\text{So}\ \ \ t=\frac{d_{glass}}{c/n} + \frac{d_{air}}{c}\ \ =\ \ \frac{1}{c} \left(n\ d_{glass} + d_{air}\right)$$ The shape of the convex surface must therefore be such that $\left(n\ d_{glass} + d_{air}\right)$ is the same for all routes from their entry points on the plane face of the lens to the principal focus.

Fermat's principle is that in going from A to B, light takes the path for which its time of travel is a minimum (or has a stationary value). In the case we're considering, point A would be the point at infinity from which the parallel rays originate and point B would be the principal focus. Since all paths are equally possible, all must take the same time.

Philip Wood
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