Consider a corner reflector with angle $\alpha$ between its semi-planes:

reflector with angle shown as α

Let a plane wave come from the bottom into this reflector (possible at an angle). The objective is to find the total wave including this incoming one and all the reflections — so as to satisfy some boundary conditions on the reflecting surfaces, e.g. homogeneous Dirichlet boundary condtions.

Physical intuition (educated by quantum mechanics) suggests: try viewing the wave as a particle with momentum proportional to wave vector $\vec k$, and taking into account all the possible reflections of such a particle. Then represent all the straight parts of the trajectory with plane waves, find amplitudes of these waves such that will make each of waves satisfy boundary conditions at the surface of (single!) reflection, and combine all the waves obtained with corresponding amplitudes accumulated in multiple reflections.

Although one might have some objections to the suggestion above, this appears to work out nicely for $\alpha=\frac\pi n$ where $n\in\mathbb N$ (which corresponds to the case when the reflector becomes a reversing or non-reversing mirror, depending on parity of $n$). E.g. for $\alpha=\frac\pi4$ we have the following possible reflections (red color marks incoming "wave"):

paths of possible reflections

Adding all the plane waves with the wave vectors matching directions of these reflections (skipping duplicate ones), we get for the example incoming wave $u_0$ defined as


the final incoming+reflected wave:

$$\begin{align} u_{f}(x,y) &= \exp\left( i \left(\frac3{10} x + y\right)\right) + \exp\left(-i \left(\frac3{10} x + y\right)\right) +\\ &+ \exp\left( i \left(x - \frac3{10} y\right)\right) + \exp\left(-i \left(x - \frac3{10} y\right)\right) -\\ &- \exp\left( i \frac{7x + 13y}{10 \sqrt2}\right) - \exp\left(-i \frac{7x + 13y}{10 \sqrt2}\right) -\\ &- \exp\left( i \frac{13x - 7y}{10 \sqrt2}\right) - \exp\left(-i \frac{13x - 7y}{10 \sqrt2}\right), \end{align} $$

which simplifies to a real-valued standing wave (because this mirror is non-reversing) containing only $\cos$ terms. Here's what it looks like (green lines show the reflector boundaries):

final wave for α=π/4

The problem is that this simple intuition doesn't work out for angles not of the form $\frac\pi n$, when outgoing wavevectors are not parallel. We can still find the solution as a finite superposition of plane waves in the case of $\alpha=\frac mn \pi$ (with $m\in\mathbb N$ and $n>m$) by finding the solution for the case $\alpha=\frac \pi n$ and rotating it so as to match the zeros of actual boundaries. But the intuition of reflections appears broken here. E.g. when $\alpha=\frac34\pi$, we have only one reflection per side:

one reflection per side

But the solution satisfying the boundary conditions requires two additional, "virtual", reflections from the $\alpha=\frac\pi4$ reflector (i.e. the solution for $\alpha=\frac34\pi$ is the same as for $\alpha=\frac\pi4$).

My question is: what is the intuition for these "virtual" reflections in the case of $\alpha=\frac mn \pi$?

  • $\begingroup$ May need a bounty--too complex. $\endgroup$
    – user45664
    Feb 3, 2019 at 17:43

1 Answer 1


The construction with virtual reflections is simply wrong. It doesn't solve the problem of scattering by a non-$\frac\pi n$-angle reflector, although it is a solution of the wave equation.

Any time a corner reflector has opening angle other than $\frac\pi n$, i.e. when we can't simply superpose plane waves corresponding to the classically-intuitive wave vectors to satisfy the boundary conditions, the actual solution is non-classical. In particular, solutions of the scattering problem for a reflector with general rational opening angle $\frac pq\pi$ (with $p,q\in\mathbb{N}$) contain, aside from the incident and reflected plane waves, a cylindrical wave, which is a diffracted wave.

Below is an animation demonstrating this wave. Its mathematical description is discussed in this Math.SE question.


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