Правка разделов «Problem 13.2.20», «Solution», «(1) Finding the boundary points A and B»

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−### Statement
−$13.2.20.$ [Insert the problem statement]
+#### Problem 13.2.20
+Light rays lying in a plane perpendicular to the axis of a glass semicylinder fall onto its flat surface at an angle of $45^{\circ}$. From which part of the lateral surface of the semicylinder will the light rays emerge? The refractive index of glass is $n$.
### Solution
−![For problem $13.2.20$ |1240x981, 31%](../../img/13.2.20/scan-0.png)
+Before solving the problem, we need to understand its essence.
+To do this, let us rephrase the question: "What is the range of the transmitting region on the curved surface of the cylinder from which light emerges?"
+Now we understand that we are required to find the angle between the two boundary points (by boundary points I mean the points on the cylinder where rays will not pass through). We will consider the semicircle of the cylinder — that is, a top view.
−#### Answer
+#### (1) Finding the boundary points A and B.
−[Insert a concise answer or boxed result]
+Suppose we have a ray emerging from the interior of the semicircle to the outside. Then consider the projection onto the axis perpendicular to the tangent — i.e., onto the radius of the semicircle — and write Snell's law for this axis.
+
+\[
+n \sin(\omega) = \sin(\sigma)
+\]
+
+Since the rays emerge when $\sigma \leq \frac{\pi}{2}$, we have two boundary points A and B with the same equation. Thus we get:
+
+\[
+\begin{cases}
+n \sin(\rho_{1}) = \sin(\sigma) \\
+n \sin(\rho_{2}) = \sin(\sigma)
+\end{cases}
+\]
+
+\[
+\Rightarrow \rho_{1} = \arcsin\left(\frac{1}{n}\right) = \rho_{2} \quad \text{or} \quad \rho_{1} = \rho_{2} \quad \text{(see Fig. 1).}
+\]
+
+\[
+\boxed{\rho_{1} = \rho_{2} = \rho = \arcsin\left(\frac{1}{n}\right)}
+\]
+![$Fig.1 $|1240x981, 50%](../../img/13.2.20/7309.png)
+
+
+#### (2) We have found the main boundary points A and B. Now, through geometry, we need to find $\phi$.
+
+From points D and C, draw rays from the external medium to points A and B respectively.
+Then we have triangles $\triangle DOA$ and $\triangle CBO$, where the angles $\rho_{1}$ and $\rho_{2}$ are known, and the angles ODA and OCB are equal to $\frac{\pi}{2} - \gamma$, where $\gamma$ is the angle of the refracted ray from the external medium into the cylinder (see Fig. 2).
+
+\[
+\sin(\beta) = n \sin(\gamma) \quad \Rightarrow \quad \gamma = \arcsin\left(\frac{1}{n\sqrt{2}}\right)
+\]
+($\beta$ from the condition is $45^{\circ}$).
+
+Then from geometry, we find:
+
+\[
+\begin{cases}
+\angle DOA = \dfrac{\pi}{2} - \rho + \gamma \\\\
+\angle COB = \dfrac{\pi}{2} - \rho - \gamma
+\end{cases}
+\]
+
+![$Fig.2$|1168x894, 50%](../../img/13.2.20/7306.png)
+
+#### (3) Now we find $\phi$:
+
+\[
+\phi = \pi - (\angle DOA + \angle COB) = \pi - \left( \frac{\pi}{2} - \rho + \gamma + \frac{\pi}{2} - \rho - \gamma \right) = 2\rho = 2\arcsin\left(\frac{1}{n}\right)
+\]
+
+
+### Answer :
+\[
+\boxed{\phi = 2\arcsin\left(\frac{1}{n}\right)}
+\]