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+### 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 $n$.
+
+\begin{center}
+ \Large \textbf{Solution}
+\end{center}
+
+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 and its wave length.\\
+
+Turning to the moving frame we have that the velocity of the wave inside the dielectric is $\frac{c}{n}$, turning back to the earth frame the velocity will be:
+
+\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 pointing 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:
+
+\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}
+
+#### Answer
+
+[Insert a concise answer or boxed result]