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+### Statement
+
+$13.1.12.$ [Insert the problem statement]
+
+### Solution
+
+The ring is cut from the base of a hollow cone of height h and semi-angle$ \alpha \ll 1$. Its radius is approximately $R = h \tan\alpha \approx h\alpha. $
+When placed with the wide part facing the beam, light hits the inner conical surface parallel to the axis.
+
+Each ray arrives with an angle of incidence$ \alpha$ with respect to the normal to the surface. Reflection deflects the ray by an angle$ 2\alpha $towards the axis. The distance f from the point of incidence to the focus on the axis satisfies$ R = f \tan 2\alpha$
+Therefore,
+
+$f = \frac{R}{\tan 2\alpha} = \frac{h\tan\alpha}{\tan 2\alpha} = \frac{h}{2}\bigl(1 - \tan^2\alpha\bigr)$
+
+Since the angle at the vertex is small, $\tan^2\alpha \ll 1$, and the focus is located at a distance
+
+$\boxed{\dfrac{h}{2}}$
+
+from the plane of the ring.
+
+
+#### Answer
+
+[Insert a concise answer or boxed result]