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en/12.1.23.md
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| + | ### Statement | ||
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| + | $12.1.23.$ [Insert the problem statement] | ||
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| + | ### Solution | ||
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| + | For a perfect reflector, the radiation pressure on the surface is $p = \dfrac{2I}{c}\cos^2\theta$, where I is the energy flux density, c is the speed of light, and $\theta$ is the angle of incidence measured with respect to the normal. | ||
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| + | With $ I = 600\ \text{W/m}^2 and c = 3.0\times10^8\ \text{m/s}$ | ||
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| + | $\frac{I}{c} = \frac{600}{3.0\times10^8} = 2.0\times10^{-6}\ \text{Pa}$ | ||
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| + | Normal incidence ($\theta = 0^\circ$): | ||
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| + | $p = 2 \times 2.0\times10^{-6} = 4.0\times10^{-6}\ \text{Pa}$ | ||
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| + | Incidence at $30^\circ$ (with respect to the normal): | ||
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| + | $p = 2 \times 2.0\times10^{-6} \times \cos^2 30^\circ = 4.0\times10^{-6} \times \left(\frac{\sqrt{3}}{2}\right)^2 = 4.0\times10^{-6} \times 0.75 = 3.0\times10^{-6}\ \text{Pa}$ | ||
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| + | If $30^\$circ were interpreted as the angle with respect to the surface, it would be $60^\$circ with respect to the normal and the pressure would be $1.0\times10^{-6}\ \text{Pa}$. | ||
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| + | $\boxed{p_{\text{normal}} = 4\times10^{-6}\ \text{Pa},\qquad p_{30^\circ} = 3\times10^{-6}\ \text{Pa}}$. | ||
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| + | #### Answer | ||
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| + | [Insert a concise answer or boxed result] | ||
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| ### Statement | |||
| $12.1.23.$ [Insert the problem statement] | |||
| ### Solution | |||
| For a perfect reflector, the radiation pressure on the surface is $p = \dfrac{2I}{c}\cos^2\theta$, where I is the energy flux density, c is the speed of light, and $\theta$ is the angle of incidence measured with respect to the normal. | |||
| With $ I = 600\ \text{W/m}^2 and c = 3.0\times10^8\ \text{m/s}$ | |||
| $\frac{I}{c} = \frac{600}{3.0\times10^8} = 2.0\times10^{-6}\ \text{Pa}$ | |||
| Normal incidence ($\theta = 0^\circ$): | |||
| $p = 2 \times 2.0\times10^{-6} = 4.0\times10^{-6}\ \text{Pa}$ | |||
| Incidence at $30^\circ$ (with respect to the normal): | |||
| $p = 2 \times 2.0\times10^{-6} \times \cos^2 30^\circ = 4.0\times10^{-6} \times \left(\frac{\sqrt{3}}{2}\right)^2 = 4.0\times10^{-6} \times 0.75 = 3.0\times10^{-6}\ \text{Pa}$ | |||
| If $30^\$circ were interpreted as the angle with respect to the surface, it would be $60^\$circ with respect to the normal and the pressure would be $1.0\times10^{-6}\ \text{Pa}$. | |||
| $\boxed{p_{\text{normal}} = 4\times10^{-6}\ \text{Pa},\qquad p_{30^\circ} = 3\times10^{-6}\ \text{Pa}}$. | |||
| #### Answer | |||
| [Insert a concise answer or boxed result] | |||