Solution

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\Large \textbf{Statement}
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$14.2.6$: 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 $n$.

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\Large \textbf{Solution}
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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}