Правка разделов «Statement», «Answer»
en/12.1.26.md
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| @@ -1,7 +1,11 @@ | |||
| ### Statement | |||
| − | $12.1.26.$ | ||
| + | $12.1.26.$ | ||
| + | The average pressure of a plane sinusoidal wave incident at an angle α on | ||
| + | a metal surface is equal to P. Determine the amplitude of the electric field | ||
| + | strength of this wave. | ||
| + | |||
| ### Solution | |||
| The average radiation pressure on a perfect mirror for a sinusoidal plane wave incident at an angle$ \alpha $(measured with respect to the normal) is: | |||
| $P = \frac{2I}{c}\cos^2\alpha$ | |||
| where I is the average intensity of the incident wave. For a plane electromagnetic wave in vacuum | |||
| $ I = \dfrac{1}{2}\varepsilon_0 c E_0^2 $ | |||
| where E_0 is the amplitude of the electric field. Substituting and solving for$ E_0$: | |||
| $P = \varepsilon_0 E_0^2 \cos^2\alpha | |||
| \quad\Longrightarrow\quad | |||
| \boxed{E_0 = \frac{1}{\cos\alpha}\sqrt{\frac{P}{\varepsilon_0}}}$ | |||
| If the angle were measured with respect to the surface,$ \cos\alpha$ is replaced by$ \sin\alpha$. In the CGS system, the equivalent expression is \displaystyle $E_0 = \frac{\sqrt{4\pi P}}{\cos\alpha}$ | |||
| (But this is only for knowledge ) | |||
| @@ -23,4 +27,4 @@Solution | |||
| #### Answer | |||
| − | [Insert a concise answer or boxed result] | ||
| + | $\boxed{E_0 = \frac{1}{\cos\alpha}\sqrt{\frac{P}{\varepsilon_0}}}$ | ||
| @@ -1,7 +1,11 @@ | |||
| ### Statement | ### Statement | ||
| $12.1.26.$ |
$12.1.26.$ | ||
| The average pressure of a plane sinusoidal wave incident at an angle α on | |||
| a metal surface is equal to P. Determine the amplitude of the electric field | |||
| strength of this wave. | |||
| ### Solution | ### Solution | ||
| The average radiation pressure on a perfect mirror for a sinusoidal plane wave incident at an angle$ \alpha $(measured with respect to the normal) is: | The average radiation pressure on a perfect mirror for a sinusoidal plane wave incident at an angle$ \alpha $(measured with respect to the normal) is: | ||
| $P = \frac{2I}{c}\cos^2\alpha$ | $P = \frac{2I}{c}\cos^2\alpha$ | ||
| where I is the average intensity of the incident wave. For a plane electromagnetic wave in vacuum | where I is the average intensity of the incident wave. For a plane electromagnetic wave in vacuum | ||
| $ I = \dfrac{1}{2}\varepsilon_0 c E_0^2 $ | $ I = \dfrac{1}{2}\varepsilon_0 c E_0^2 $ | ||
| where E_0 is the amplitude of the electric field. Substituting and solving for$ E_0$: | where E_0 is the amplitude of the electric field. Substituting and solving for$ E_0$: | ||
| $P = \varepsilon_0 E_0^2 \cos^2\alpha | $P = \varepsilon_0 E_0^2 \cos^2\alpha | ||
| \quad\Longrightarrow\quad | \quad\Longrightarrow\quad | ||
| \boxed{E_0 = \frac{1}{\cos\alpha}\sqrt{\frac{P}{\varepsilon_0}}}$ | \boxed{E_0 = \frac{1}{\cos\alpha}\sqrt{\frac{P}{\varepsilon_0}}}$ | ||
| If the angle were measured with respect to the surface,$ \cos\alpha$ is replaced by$ \sin\alpha$. In the CGS system, the equivalent expression is \displaystyle $E_0 = \frac{\sqrt{4\pi P}}{\cos\alpha}$ | If the angle were measured with respect to the surface,$ \cos\alpha$ is replaced by$ \sin\alpha$. In the CGS system, the equivalent expression is \displaystyle $E_0 = \frac{\sqrt{4\pi P}}{\cos\alpha}$ | ||
| (But this is only for knowledge ) | (But this is only for knowledge ) | ||
| @@ -23,4 +27,4 @@Solution | |||
| #### Answer | #### Answer | ||
| [Insert a concise answer or boxed result] | $\boxed{E_0 = \frac{1}{\cos\alpha}\sqrt{\frac{P}{\varepsilon_0}}}$ | ||