Solution

The main idea of the method is that the phase of a plane wave $\Phi = \omega t - kx$ is a Lorentz invariant. It follows that the frequency $\nu$ and the wavenumber $k = \nu/c$ transform during transitions between inertial reference frames according to the same Lorentz transformation formulas as time $t$ and coordinate $x$.

The solution algorithm for both problems is the same:

1. Transition from the laboratory frame ($K$) to the rest frame of the moving object ($K'$).
2. Analysis of the process (reflection or refraction) in frame $K'$, where the boundary is stationary.
3. Inverse transition from $K'$ to the laboratory frame $K$ for the modified wave.

Solution to problem 14.2.5 (Reflection from a mirror)

Let the incident wave travel along the $x$-axis. Its frequency is $\nu$, and the wavenumber is $k = \nu/c$. The mirror moves towards the wave with a velocity $v = \beta c$. The projection of the mirror's velocity in frame $K$ is $V_x = -\beta c$.

Step 1. Transition to the mirror frame $K'$.
In the rest frame of the mirror $K'$, the frequency of the incident wave $\nu'$ is determined by the formula of the relativistic Doppler effect for an approaching source (which directly follows from the Lorentz transformations):
$$\nu' = \nu \sqrt{\frac{1 + \beta}{1 - \beta}}$$

Step 2. Reflection in frame $K'$.
In frame $K'$, the mirror is stationary. Upon normal reflection from a stationary ideal mirror, the wave frequency is conserved, and the direction of propagation is reversed.
Reflected frequency: $\nu'_{ref} = \nu'$.
Reflected wavenumber (taking into account the direction along the $-x$ axis): $k'_{ref} = -\frac{\nu'}{c}$.

Step 3. Return to the laboratory frame $K$.
Frame $K'$ (the mirror) moves relative to $K$ with a velocity $V = -\beta c$.
We apply the inverse Lorentz transformation for the frequency of the reflected wave:
$$\nu_{ref} = \gamma (\nu'_{ref} + V k'_{ref}) = \frac{1}{\sqrt{1 - \beta^2}} \left( \nu' + (-\beta c) \left(-\frac{\nu'}{c}\right) \right)$$
$$\nu_{ref} = \frac{1}{\sqrt{1 - \beta^2}} \nu' (1 + \beta) = \nu' \sqrt{\frac{1 + \beta}{1 - \beta}}$$
Substituting $\nu'$ from Step 1:
$$\nu_{ref} = \left( \nu \sqrt{\frac{1 + \beta}{1 - \beta}} \right) \sqrt{\frac{1 + \beta}{1 - \beta}} = \nu \frac{1 + \beta}{1 - \beta}$$
The change in wave frequency $\Delta \nu$:
$$\Delta \nu = \nu_{ref} - \nu = \nu \frac{1 + \beta}{1 - \beta} - \nu = \nu \frac{2\beta}{1 - \beta}$$

Solution to problem 14.2.6 (Wave inside a dielectric)

Step 1. Transition to the dielectric frame $K'$.
Similarly to the first problem, in the rest frame of the dielectric $K'$, the frequency of the incident wave is:
$$\nu' = \nu \sqrt{\frac{1 + \beta}{1 - \beta}}$$

Step 2. Entering the dielectric in frame $K'$.
In frame $K'$, the boundary of the dielectric is stationary. When passing through a stationary boundary between media, the wave frequency does not change:
$$\nu'_{in} = \nu'$$
The wave continues to propagate in the same direction (along $+x$). The phase velocity in the dielectric is $c/n$, so the wavenumber inside is:
$$k'_{in} = \frac{\nu'_{in}}{c/n} = n \frac{\nu'}{c}$$

Step 3. Return to the laboratory frame $K$.
The velocity of frame $K'$ relative to $K$ is still $V = -\beta c$.
We apply the inverse Lorentz transformation for the frequency of the transmitted wave:
$$\nu_{in} = \gamma (\nu'_{in} + V k'_{in}) = \frac{1}{\sqrt{1 - \beta^2}} \left( \nu' + (-\beta c) \left(n \frac{\nu'}{c}\right) \right)$$
$$\nu_{in} = \frac{1}{\sqrt{1 - \beta^2}} \nu' (1 - n\beta)$$
Substituting the expression for $\nu'$:
$$\nu_{in} = \frac{1}{\sqrt{1 - \beta^2}} \left( \nu \sqrt{\frac{1 + \beta}{1 - \beta}} \right) (1 - n\beta) = \nu \frac{1 - n\beta}{1 - \beta}$$
Let us find the difference in frequencies $\Delta \nu$ of the plane wave outside ($\nu$) and inside ($\nu_{in}$) the dielectric:
$$\Delta \nu = \nu - \nu_{in} = \nu - \nu \frac{1 - n\beta}{1 - \beta} = \nu \frac{1 - \beta - 1 + n\beta}{1 - \beta} = \nu \beta \frac{n - 1}{1 - \beta}$$