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][1]. **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.
 




  


  [1]: http://mathoverflow.net/questions/70421/light-rays-bouncing-in-twisted-tubes