Reflection of light

Question

There is a beam of light which strikes $BC$ at the point given in the diagram at an angle $\alpha = 19.94^{\circ}$ with $BC$ and reflects at an equal angle. The reflections continue following the law:

$$\text{angle of incidence} = \text{angle of reflection}$$

The question asks me to find out how many times will the ray get reflected. I analysed its path and came to the inclusion that the direction of light ray will reverse, i.e. it will start coming out of the angle but not following its initial path, after the $36$-th reflection, which occurs at the surface $AB$. But I am unable to analyse further.

Any help would be appreciated. Thanks in advance.

Answer 1

Write a rule that tells you the angle at next reflection based on the current one. Then you have a recursive definition of a sequence which you can analyze.

Call the second reflection point $D$. Starting with $\alpha_n$, the angle $\angle BCD=\alpha_n$, and $\angle CDB=180^\circ-\beta-\alpha_n$. The angle of incidence at the second reflection point is therefore $\alpha_{n+1}=\angle ADC=\alpha_n+\beta$.

This tells you directly, that $\beta$ is added at each reflection:

$$\alpha_n=\alpha+n\beta$$

if $n$ is the number of reflections.

What exactly happens depends on $\beta$, which isn't given.

However, the "stopping condition" is, that the ray will not only reverse, but have an angle that will shoot between both legs of the angle (and therefore never touch any surface again). Think about what it means that angle is greater than $90^\circ$, greater than $180^\circ$ and smaller than $0$, and which angle is the one that never gets a reflection again (it depends on $\beta$). Also sketch and figure out if you have to re-derive the recursion if the ray goes the other way.

Answer 2

As I mentioned in my comment, an often-useful trick for analyzing problems that involve reflections is to “unfold” the path into a straight line. For this problem, this involves successive reflections of the region between the two mirror rays, like so:

Examining this diagram, we can see that the incident ray will escape once it and the bounding ray of one of these successive reflections no longer converge. Thus, we seek the least positive integer $n$ for which $\pi+\alpha \ge 2\pi-n\beta$, from which $$n = \left\lceil {\pi-\alpha \over \beta} \right\rceil.$$ This is also the number of reflections before escape. (This assumes that the first reflecting surface struck by the ray is $\overrightarrow{BC}$.)