A geometry puzzle

Here’s a fun problem I came across when trying to analyse some data, which I thought I’d write up to illustrate the kind of interesting puzzles I get paid to solve as a physicist. Perhaps have a go yourselves and let me know if there are more intuitive solutions.

During the course of our experiments, we occasionally use scintillating crystals to detect high energy particles. As the particle passes through the crystal, it can deposit some energy which is converted to visible light by the crystal. The light then travels in a straight line out of the crystal to a detector of some kind.

The problem is, the light can bounce around from the edges of the crystal before it reaches the detector, which reduces the amount of light reaching the detector. It would be nice to calculate the expected fraction of light leaving the crystal, as a function of emission angle, for a given crystal size and reflection coefficient R. In terms of the figure below, how much light leaves the left face of the crystal (blue), and in what direction?

CrystalSetup.png
Here, the crystal is blue, the light power is P, and the reflection coefficient is R. The ratio of the horizontal to vertical dimension of the crystal is w.

Lets start simply – assume the light travels at \theta=0 towards the right face with power P = 1. Upon hitting the right face a fraction R is reflected back into the crystal, and a fraction (1-R) transmitted.

The light then hits the left side of the crystal, transmitting R\cdot(1-R) and reflecting R^2. As you can imagine, this process repeats, giving the infinite series

P_R = (1-R) + R^2(1-R) + R^4(1-R) + \dots

P_L = R(1-R) + R^3(1-R) + R^5(1-R) + \dots

Doing the sum (see geometric series) we have

P_R = \frac{1-R}{1 - R^2} = \frac{1}{1+R}

P_L = \frac{R(1-R)}{1 - R^2} = \frac{R}{1+R}

The total power P = P_L + P_R = 1 as is required from energy conservation. As R \rightarrow 1, P_L \rightarrow P_R \rightarrow 1/2 and as R \rightarrow 0, P_L \rightarrow 0 and P_R \rightarrow 1.

This makes sense  – when there are lots of bounces inside the crystal, it shouldn’t matter in which direction the light was initially moving, and equal power is emitted from both sides. When there are very few bounces, because the light is mostly transmitted, the right face emits the majority of the power.

Things get a bit more complicated when the light moves at an angle – the animation below shows what happens for light emitted at different angles from the centre of the crystal:

R06IJ.gif
Here R = 0.6 and w = 3.

And for different reflection coefficients:

R01-1IJ.gif
Here \theta = 0.1 rads.

… and different aspect ratios:

w1-5.gif
Here R = 0.6.

… and a range of angles

AngleRange.png

 

It is clear that the emitted light varies rapidly with crystal properties and emission angle, and it would be nice to have some kind of general model for the angular distribution of the emitted power.

We’ll use the same method of counting light bounces to generalise the 1D model above. For a particular face of the crystal, in this case the left face, there must still be 2 bounces from a vertical face between events where light escapes, just like the above. However if the light is moving at a nonzero angle, there are also bounces from the upper and lower faces to consider.

Each time the light has moved 2w horizontally, and therefore escapes the crystal, it has also moved 2w\tan\theta vertically. Given that we have defined the vertical dimension of the crystal to be 1, this is also the additional number of bounces.

In analogy to the above then, the infinite series becomes

P_L = R(1-R)(1 + R^2R^{2w\tan\theta} + R^4R^{4w\tan\theta} + \dots)

which for \theta = 0 becomes equivalent to the 1D case. Taking into account the initial bounces before the first emission from the left side, and some technicalities depending on the definition of \tan\theta, the summed series looks like:

P_L(\theta) = \frac{(1-R)R^{w|\tan\theta|/2}}{1 - R^{2 + 2w|\tan\theta|}} \times \left\{ \begin{array}{ll} R^{1 + w|\tan\theta|}&|\theta|<\pi/2\\ 1&\text{otherwise} \end{array} \right.

This looks a bit nasty, but how does it compare to a simulation of this process? It turns out, pretty well:

ModelComparison.png
The power transmitted through the other faces comes about from a simple redefinition of \theta, and taking w\rightarrow 1/w as appropriate. Here R = 0.5, w = 3.

 

Fixing w and varying R from 0.1 to 0.9 gives the following plots:

ModelvsR.gif
Here w = 3.

Interestingly, at low R, i.e. few bounces, P(\theta) becomes step-like, changing whenever there is a new bounce. At large R there are many bounces, and P(\theta) becomes much smoother and matches the model very well.

We can also do some sanity checks and see how the total fractional flux from each face, i.e.

F_L(w, R) = \frac{1}{2\pi}\int_0^{2\pi}P_L(\theta)\,d\theta

varies with reflection coefficient and aspect ratio:

rvsw
Integrated over all angles, the L/R and U/D pairs are equivalent.

For a square crystal w = 1, the fractional flux is 1/4 from each face irrespective of R, which makes sense. As the reflection coefficient goes to zero, the fraction leaving each face just becomes the fractional angle subtended by each face.

More interesting is the limit R \rightarrow 1, as it’s not obvious a priori how the flux will be distributed among the faces. Let R = 1 - \epsilon where \epsilon is small, then substituting into the above:

P_L(\theta) = \frac{1}{2 + 2w|\tan\theta|} \times \left\{ \begin{array}{ll} 1 - 3/2\cdot w|\tan\theta|\epsilon & |\theta| < \pi/2\\ 1 - 1/2\cdot w|\tan\theta|\epsilon & |\theta| > \pi/2 \end{array} \right.

Letting \epsilon = 0, the integral becomes relatively simple and

F_L(w) = \frac{1}{2\pi}\int_0^{2\pi} \frac{1}{2 + 2w|\tan\theta|}\,d\theta = \frac{1 + \frac{4}{\pi}w\log w }{2 + 2w^2}

This is a fun result! It somehow involves a logarithm, arising from light bouncing around inside a rectangle. Plotting below, it also has a maximum around w = 1 .87 which is a non-obvious result, and decays like \log w / w for large w.

FLw.png

It would be tempting to conclude that in order to maximise light output from one of these crystals, you’d pick this aspect ratio. This would be a nice result, but this analysis is too simplistic for that. A large simplifying assumption is that the reflection coefficient is independent of incidence angle, which is obviously not the case. We have also assumed that all of the light originates from the centre of the crystal, which will not be the case in practice.

Nevertheless this was a fun puzzle to work out over a recent plane ride, and might be useful experimentally in the future.

You can find the (bad) notebook here:

https://github.com/jasmcole/Blog/tree/master/Geometry%20Puzzle

There are two functions, simulateCrystal and simulateCrystalFast, which do the numerical simulations. The former is based on simple stepping, the latter checks for intersections with crystal faces and propagates one bounce at a time.

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