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en/12.1.26.md
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| + | ### Statement | ||
| + | |||
| + | $12.1.26.$ [Insert the problem statement] | ||
| + | |||
| + | ### 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 ) | ||
| + | |||
| + | #### Answer | ||
| + | |||
| + | [Insert a concise answer or boxed result] | ||
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| ### Statement | |||
| $12.1.26.$ [Insert the problem statement] | |||
| ### 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 ) | |||
| #### Answer | |||
| [Insert a concise answer or boxed result] | |||