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en/14.3.16.md
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| + | ### Statement | ||
| + | |||
| + | $14.3.16.$ [Insert the problem statement] | ||
| + | |||
| + | ### Solution | ||
| + | |||
| + | $14.3.16$ Solve problem $14.3.15$ for plane wave propagation in a dielectric medium with | ||
| + | refractive index $n$. | ||
| + | |||
| + | $14.3.15$ How many times will the amplitude of a plane electromagnetic wave change | ||
| + | when it passes into a coordinate system moving at a speed $\beta c$ in the direction | ||
| + | of wave propagation? | ||
| + | |||
| + | To solve this problem we need to use the Lorentz transformation for fields, taking into account that the fields of the electromagnetic wave are | ||
| + | perpendicular to the velocity of the moving frame. Also, in each frame, the relation $\vec{E} = [\vec{v} \times \vec{B}]$ holds, where $\vec{v}$ is the | ||
| + | velocity vector of the electromagnetic wave; in this case $\vec{v} = \frac{c}{n}$. | ||
| + | |||
| + | \begin{equation} | ||
| + | E' = \frac{E - \beta c B}{\sqrt{1-\beta^2}} = E \frac{1-\beta n}{\sqrt{1-\beta^2}} | ||
| + | \end{equation} | ||
| + | |||
| + | And using the relation between fields: | ||
| + | |||
| + | \begin{equation} | ||
| + | B' = B \frac{1-\beta n}{\sqrt{1-\beta^2}} | ||
| + | \end{equation} | ||
| + | |||
| + | So, the amplitude becomes $\frac{1-\beta n}{\sqrt{1-\beta^2}}$ times smaller. | ||
| + | |||
| + | #### Answer | ||
| + | |||
| + | [Insert a concise answer or boxed result] | ||
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| ### Statement | |||
| $14.3.16.$ [Insert the problem statement] | |||
| ### Solution | |||
| $14.3.16$ Solve problem $14.3.15$ for plane wave propagation in a dielectric medium with | |||
| refractive index $n$. | |||
| $14.3.15$ How many times will the amplitude of a plane electromagnetic wave change | |||
| when it passes into a coordinate system moving at a speed $\beta c$ in the direction | |||
| of wave propagation? | |||
| To solve this problem we need to use the Lorentz transformation for fields, taking into account that the fields of the electromagnetic wave are | |||
| perpendicular to the velocity of the moving frame. Also, in each frame, the relation $\vec{E} = [\vec{v} \times \vec{B}]$ holds, where $\vec{v}$ is the | |||
| velocity vector of the electromagnetic wave; in this case $\vec{v} = \frac{c}{n}$. | |||
| \begin{equation} | |||
| E' = \frac{E - \beta c B}{\sqrt{1-\beta^2}} = E \frac{1-\beta n}{\sqrt{1-\beta^2}} | |||
| \end{equation} | |||
| And using the relation between fields: | |||
| \begin{equation} | |||
| B' = B \frac{1-\beta n}{\sqrt{1-\beta^2}} | |||
| \end{equation} | |||
| So, the amplitude becomes $\frac{1-\beta n}{\sqrt{1-\beta^2}}$ times smaller. | |||
| #### Answer | |||
| [Insert a concise answer or boxed result] | |||