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naz правка от 2 июл. 2026 г., 20:31

правка #19551 предыдущая #19549 ← раньше позже →

en/13.2.8.md +29 −0

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		+	### Statement
		+
		+	$13.2.8.$ [Insert the problem statement]
		+
		+	### 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.
		+
		+	#### Answer
		+
		+	[Insert a concise answer or boxed result]

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			### Statement

			$13.2.8.$ [Insert the problem statement]

			### 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.

			#### Answer

			[Insert a concise answer or boxed result]