Правка разделов «Statement», «Solution», «Answer»

JAMF правка от 13 июн. 2026 г., 15:36

правка #18716 предыдущая #18715 ← раньше

en/14.2.6.md +8 −16

		@@ -1,37 +1,27 @@
			### Statement

		−	$14.2.6.$ [Insert the problem statement]
		−
		−	### Solution
		−
		−	\begin{center}
		−	\Large \textbf{Statement}
		−	\end{center}
		−
			$14.2.6$: Determine the difference between the frequencies of a plane wave outside and
			inside the dielectric, the plane boundary of which is moving towards the wave
			at a speed $\beta c$. The frequency of the wave outside the dielectric is $\nu$, the
		−	refractive index of the wave in the dielectric is ~~$n$~~.
		+	refractive index of the wave in the dielectric is n.

		−	\begin{center}
		−	~~\Large \textbf{~~Solution}
		−	\end{center}
		+	### Solution

			The new frequency can be calculated using the idea that:

			\begin{equation}
			c = \lambda \nu \rightarrow \frac{c}{\lambda} = \nu
			\end{equation}

		−	~~For to~~ calculate the frequency of the electromagnetic wave inside the moving dielectric need calculate the velocity of the wave in the ~~frame of the earth a~~nd its wave length.\\
		+	To calculate the frequency of the electromagnetic wave inside the moving dielectric, we need to calculate the velocity of the wave in the Earth frame and its wave length.

		−	~~Turning to t~~he moving frame ~~we have that th~~e velocity of the wave inside the dielectric is $\frac{c}{n}$~~, t~~urning back to the earth frame the velocity ~~will be~~:
		+	In the moving frame, the velocity of the wave inside the dielectric is $\frac{c}{n}$. Turning back to the Earth frame, the velocity becomes:

			\begin{equation}
			c_1 = \frac{c/n - \beta c}{1 - \frac{\beta}{n}} = c \frac{1 - n \beta}{n - \beta}
			\end{equation}

		−	The negative sign is because the ~~velocity of the dielectric is~~ point~~ing~~ to the left in the earth frame. Also, the wave length in the earth frame is $\frac{\lambda}{n}$, then the frequency of the electromagnetic wave inside of the moving dielectric is:
		+	The negative sign is because the dielectric's velocity points to the left in the Earth frame. Also, the wave length in the Earth frame is $\frac{\lambda}{n}$; then the frequency of the electromagnetic wave inside the moving dielectric is:

			\begin{equation}
			\nu_1 = \frac{c_1}{\lambda/n} = \nu \frac{1 - n \beta}{n - \beta}
			\end{equation}

			Then the difference is:

			\begin{equation}
			\Delta \nu = \nu_1 - \nu = \nu \left(\frac{1 - n \beta}{n - \beta} - 1\right)
			\end{equation}
		@@ -45,4 +35,6 @@Solution

			#### Answer

		−	[Insert a concise answer or boxed result]
		+	\begin{equation}
		+	\Delta \nu = \nu_1 - \nu = \nu \left(\frac{1 - n \beta}{n - \beta} - 1\right)
		+	\end{equation}

	@@ -1,37 +1,27 @@
	### Statement		### Statement

	$14.2.6.$ [Insert the problem statement]

	### Solution

	\begin{center}
	\Large \textbf{Statement}
	\end{center}

	$14.2.6$: Determine the difference between the frequencies of a plane wave outside and		$14.2.6$: Determine the difference between the frequencies of a plane wave outside and
	inside the dielectric, the plane boundary of which is moving towards the wave		inside the dielectric, the plane boundary of which is moving towards the wave
	at a speed $\beta c$. The frequency of the wave outside the dielectric is $\nu$, the		at a speed $\beta c$. The frequency of the wave outside the dielectric is $\nu$, the
	refractive index of the wave in the dielectric is ~~$n$~~.		refractive index of the wave in the dielectric is n.

	\begin{center}		### Solution
	~~\Large \textbf{~~Solution}
	\end{center}

	The new frequency can be calculated using the idea that:		The new frequency can be calculated using the idea that:

	\begin{equation}		\begin{equation}
	c = \lambda \nu \rightarrow \frac{c}{\lambda} = \nu		c = \lambda \nu \rightarrow \frac{c}{\lambda} = \nu
	\end{equation}		\end{equation}

	~~For to~~ calculate the frequency of the electromagnetic wave inside the moving dielectric need calculate the velocity of the wave in the ~~frame of the earth a~~nd its wave length.\\		To calculate the frequency of the electromagnetic wave inside the moving dielectric, we need to calculate the velocity of the wave in the Earth frame and its wave length.

	~~Turning to t~~he moving frame ~~we have that th~~e velocity of the wave inside the dielectric is $\frac{c}{n}$~~, t~~urning back to the earth frame the velocity ~~will be~~:		In the moving frame, the velocity of the wave inside the dielectric is $\frac{c}{n}$. Turning back to the Earth frame, the velocity becomes:

	\begin{equation}		\begin{equation}
	c_1 = \frac{c/n - \beta c}{1 - \frac{\beta}{n}} = c \frac{1 - n \beta}{n - \beta}		c_1 = \frac{c/n - \beta c}{1 - \frac{\beta}{n}} = c \frac{1 - n \beta}{n - \beta}
	\end{equation}		\end{equation}

	The negative sign is because the ~~velocity of the dielectric is~~ point~~ing~~ to the left in the earth frame. Also, the wave length in the earth frame is $\frac{\lambda}{n}$, then the frequency of the electromagnetic wave inside of the moving dielectric is:		The negative sign is because the dielectric's velocity points to the left in the Earth frame. Also, the wave length in the Earth frame is $\frac{\lambda}{n}$; then the frequency of the electromagnetic wave inside the moving dielectric is:

	\begin{equation}		\begin{equation}
	\nu_1 = \frac{c_1}{\lambda/n} = \nu \frac{1 - n \beta}{n - \beta}		\nu_1 = \frac{c_1}{\lambda/n} = \nu \frac{1 - n \beta}{n - \beta}
	\end{equation}		\end{equation}

	Then the difference is:		Then the difference is:

	\begin{equation}		\begin{equation}
	\Delta \nu = \nu_1 - \nu = \nu \left(\frac{1 - n \beta}{n - \beta} - 1\right)		\Delta \nu = \nu_1 - \nu = \nu \left(\frac{1 - n \beta}{n - \beta} - 1\right)
	\end{equation}		\end{equation}
	@@ -45,4 +35,6 @@Solution

	#### Answer		#### Answer

	[Insert a concise answer or boxed result]		\begin{equation}
			\Delta \nu = \nu_1 - \nu = \nu \left(\frac{1 - n \beta}{n - \beta} - 1\right)
			\end{equation}