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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. (Note that I want an optic fiber to be bidirectional.)

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 other words, is it always possible so slice an optic fibers by planes which cut the walls at right angle?

In particular,

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

• 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.
• I realized that if the walls are only piecewise smooth then one can make an optic fiber with a pair of equidecomposable figures at the ends. (The construction is the same, but one splits tube into few on the way and then rearrange them back together.)
• An extract from the answer of Marcos Cossarini: Note that if one can cut an optic fiber in two pieces in such a way that almost all rays pass the cut at most once then the cut has to be flat and orthogonal to the boundary. After such cut, one gets two optical fibers. Applying a bit of differential geometry the problem can be reformulated in an equivalent way: is it true that any optic fiber admits such a cut.
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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. (Note that I want an optic fiber to be bidirectional.)

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 other words, is it always possible so slice an optic fibers by planes which cut the walls at right angle?

In particular,

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

• 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.
• I realized that if the walls are only piecewise smooth then one can make an optic fiber with a pair of equidecomposable figures at the ends. (The construction is the same, but one splits tube into few on the way and then rearrange them back together.)
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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 other words, is it always possible so slice an optic fibers by planes which cut the walls at right angle?

In particular,

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