My initial query was to find the distance, denoted $a$ from the centre, which a light source radiating in all directions (see below, where centre is represented by a black dot and light source by an empty circle), would result in the highest intensity of light crossing the horizontal line of length 2d. The constraints were that a ray can only bounce once off the circle and once it crosses the horizontal line it is within the tube beneath and cannot escape. The more rays that cross the horizontal line the more intense the light is within the tube.
I approached the problem by thinking of flux into the tube and finding the position to be along the centre of the circle. Following on from this I created a series of trigonometric problems which concluded with the most efficient place would be when $\phi_1$ and $\phi_2$ are maximised as a function of $a$. I believe this to be the way to approach the problem and seem to find the algebra very difficult. If anybody has any insight into where to begin that would be greatly appreciated. Of course any questions please feel free to comment.
I want to find the position of reflection point $P$ for which the ray is reflected on endpoint $B$ of horizontal segment $AB$ (see diagram below, where I set for simplicity $R=1$). From the sine rule applied to triangle $CSP$ one has $\sin\alpha/a=\sin(\alpha+\theta)$, that is: $$ \sin\alpha=a\sin\alpha\cos\theta+a\cos\alpha\sin\theta. $$ On the other hand, one also has $$ \cot\alpha=\tan(\pi/2-\alpha)={PH\over AH}= {\sqrt{1-d^2}+\cos\theta\over d+\sin\theta}. $$ Substituting this relation into the previous one we obtain the equation: $$ d+\left(1-a\sqrt{1-d^2}\right)\sin\theta=a(d+2\sin\theta)\cos\theta. $$ Squaring both sides leads to a quartic equation for $t=\sin\theta$, which has however the spurious solution $t=-d$. Factoring that out we are then left with a cubic equation:
$$ 4 a^2 t^3+\left(1-3 a^2-2 a \sqrt{1-d^2}\right)t+\left(1-a^2\right) d=0, $$ which has the solutions: $$ t_1=-2A-2B,\quad t_2=\left(1+i \sqrt{3}\right)A+\left(1-i \sqrt{3}\right)B,\quad t_3=\left(1-i \sqrt{3}\right)A+\left(1+i \sqrt{3}\right)B, $$ where: $$ \begin{align} A=&\frac{3 a^2+2 a \sqrt{1-d^2}-1}{12a \sqrt[3]{ad(1-a^2) +\sqrt{a^2d^2(1-a^2)^2-{1\over27}\left(3 a^2+2a \sqrt{1-d^2}-1\right)^3}}}\\ \\ B=&\frac{\sqrt[3]{ad(1-a^2)+\sqrt{a^2d^2(1-a^2)^2-{1\over27} \left(3 a^2+2 a \sqrt{1-d^2}-1\right)^3}}}{4a}\\ \end{align} $$
Solution $t_1$ is always real and negative, while $t_2$ and $t_3$ are real only for some values of $a$ and $d$. By symmetry, $-t_1$ relates to the case of the reflected ray arriving at $A$ (see diagram below).
The other two solution $\theta=\arcsin t_2$ and $\theta=\arcsin t_3$ relate to the cases shown in the first diagram and in the following one:
Now I think you have enough results to complete the computation by yourself.