Правка разделов «Statement», «Answer»

naz правка от 2 июл. 2026 г., 20:36

правка #19588 предыдущая #19587 ← раньше

en/13.2.8.md +5 −2

		@@ -1,6 +1,7 @@
			### Statement

		−	$13.2.8.$ ~~[Insert the problem statement]~~
		+	$13.2.8.$
		+	A cone with an apex angle $2\alpha$ is made of Plexiglas. A parallel beam of light falls onto the base of the cone. Describe the behavior of the light. The refractive index of Plexiglas is $n = 1.5$.

			### Solution

			The problem concerns the phenomenon of total internal reflection, i.e., when the refracted ray emerges parallel to the surface.

			Consider a ray from the beam; it strikes the surface of the cone at an angle $\pi/2 - \alpha$ to the surface. The condition for total internal reflection is derived from Snell's law:

			\begin{equation}
			n \cos(\alpha) = 1
			\end{equation}

			\begin{equation}
			\alpha = \arccos(2/3)
			\end{equation}

			For angles larger than the critical one, it is clear that the rays escape immediately after the first intersection. However, the case of smaller angles is more interesting. Consider the second intersection after the first reflection. The angle at which the ray crosses the cone surface for the second time is found from the triangle:

			\begin{equation}
			\alpha' = \beta + 2\alpha
			\end{equation}

			As we can see, the angle of incidence decreases by $2\alpha$ with each reflection after the first, until it once again becomes smaller than the critical angle.
		@@ -26,4 +27,6 @@Solution

			#### Answer

		−	~~[Insert a concise answer or boxed result]~~
		+	For $\alpha > \arccos(2/3)$, the light will partially escape from the cone already at the first reflection.
		+
		+	For $\alpha < \arccos(2/3)$, the light will first be totally reflected from the lateral surface. After each reflection, the angle between the ray and the normal to the cone surface decreases by $2\alpha$, and after several reflections, the light will again leave the cone through its surface.

	@@ -1,6 +1,7 @@
	### Statement		### Statement

	$13.2.8.$ ~~[Insert the problem statement]~~		$13.2.8.$
			A cone with an apex angle $2\alpha$ is made of Plexiglas. A parallel beam of light falls onto the base of the cone. Describe the behavior of the light. The refractive index of Plexiglas is $n = 1.5$.

	### Solution		### Solution

	The problem concerns the phenomenon of total internal reflection, i.e., when the refracted ray emerges parallel to the surface.		The problem concerns the phenomenon of total internal reflection, i.e., when the refracted ray emerges parallel to the surface.

	Consider a ray from the beam; it strikes the surface of the cone at an angle $\pi/2 - \alpha$ to the surface. The condition for total internal reflection is derived from Snell's law:		Consider a ray from the beam; it strikes the surface of the cone at an angle $\pi/2 - \alpha$ to the surface. The condition for total internal reflection is derived from Snell's law:

	\begin{equation}		\begin{equation}
	n \cos(\alpha) = 1		n \cos(\alpha) = 1
	\end{equation}		\end{equation}

	\begin{equation}		\begin{equation}
	\alpha = \arccos(2/3)		\alpha = \arccos(2/3)
	\end{equation}		\end{equation}

	For angles larger than the critical one, it is clear that the rays escape immediately after the first intersection. However, the case of smaller angles is more interesting. Consider the second intersection after the first reflection. The angle at which the ray crosses the cone surface for the second time is found from the triangle:		For angles larger than the critical one, it is clear that the rays escape immediately after the first intersection. However, the case of smaller angles is more interesting. Consider the second intersection after the first reflection. The angle at which the ray crosses the cone surface for the second time is found from the triangle:

	\begin{equation}		\begin{equation}
	\alpha' = \beta + 2\alpha		\alpha' = \beta + 2\alpha
	\end{equation}		\end{equation}

	As we can see, the angle of incidence decreases by $2\alpha$ with each reflection after the first, until it once again becomes smaller than the critical angle.		As we can see, the angle of incidence decreases by $2\alpha$ with each reflection after the first, until it once again becomes smaller than the critical angle.
	@@ -26,4 +27,6 @@Solution

	#### Answer		#### Answer

	~~[Insert a concise answer or boxed result]~~		For $\alpha > \arccos(2/3)$, the light will partially escape from the cone already at the first reflection.

			For $\alpha < \arccos(2/3)$, the light will first be totally reflected from the lateral surface. After each reflection, the angle between the ray and the normal to the cone surface decreases by $2\alpha$, and after several reflections, the light will again leave the cone through its surface.