Let $\gamma\colon[a,b]\to \mathbb R^3$ be a smooth curve with curvature $< 1$. Consider a tube, formed by one parameter family of unit circles with center at $\gamma(t)$ in the plane orthogonal to $\dot\gamma(t)$.

A light ray which is comming into tube from one end and bouncing with perfect reflection from the interior walls will emerge from the other end with probability 1; see this question. Let us call a tube with this property an optic fiber.

One can construct an optic fiber along the same lines using any simple close smooth plane curve $(x(\theta),y(\theta))$ instead of circle. To do this one has to choose a parallel normal frame $e_1,e_2$ along $\gamma$ (i.e., such that $\dot e_i(t)\parallel\dot\gamma(t)$ for all $t$) and consider the tube $[a,b]\times\mathbb S^1\hookrightarrow\mathbb R^3$ defined as $$(t,\theta)\mapsto \gamma(t)+x(\theta){\cdot}e_1(t)+y(\theta){\cdot}e_2(t)$$ (The condition that the frame is parallel implies that any normal plane to $\gamma$ cuts tube at right angle.) This way we get an optic fiber with congruent ends.

Question 1. Are there any constructions of optic fibers different from the one described above?

In particular,

Question 2. Is there an optic fiber with noncongruent ends?

Comments

• I feel that the answer is "NO", but have no idea "WHY".
• From Liouville's theorem, it is clear that the ends must have the same area.