Let $M^2$ be an orientable surface with boundary endowed with a Riemannian metric $g$. We know that if we put in the manifold $M \times \mathbb{R}$ the product metric $\overline{g} = g + dt^2$, then the slices $M_t = M \times \{t\}$ are totally geodesic and are free-boundary surfaces, which means that their boundary meet the boundary $\partial (M \times \mathbb{R}) = \partial M \times \mathbb{R}$ orthogonally.

Suppose now that we choose an angle $\theta \in (0, \pi)$. Is there a "natural" Riemannian metric $g_\theta$ on $M \times \mathbb{R}$ such that the slices $M_t$ are totally geodesic, meet the boundary of $M \times \mathbb{R}$ at a constant angle $\theta$ and $g_{\pi/2}$ agrees with the product metric $\overline{g}$? Is there an explicit formula?