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| ### Solution | | ### Solution |
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| We’ll model the interaction of light with fog particles. We'll assume that the light intensity decreases due to absorption by the droplets. We'll neglect scattering effects, which is acceptable if the droplet size is larger than the wavelength of light. | | We’ll model the interaction of light with fog particles. We'll assume that the light intensity decreases due to absorption by the droplets. We'll neglect scattering effects, which is acceptable if the droplet size is larger than the wavelength of light. |
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| We treat the fog droplets as ideal spheres with cross-sectional area \(S = \pi r^2\). When a photon encounters a droplet, we assume it is absorbed and its energy is converted entirely into internal energy (heat). A droplet can absorb any number of photons. | | We treat the fog droplets as ideal spheres with cross-sectional area \(S = \pi r^2\). When a photon encounters a droplet, we assume it is absorbed and its energy is converted entirely into internal energy (heat). A droplet can absorb any number of photons. |
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| Now consider a thin volume element between two parallel planes separated by a small thickness \(dx\). Light travels perpendicular to these planes. Suppose there are \(N\) droplets in volume \(V\). If a photon enters a sub-volume \(V_k = \pi r^2 dx\) that contains a droplet, it will be absorbed. | | Now consider a thin volume element between two parallel planes separated by a small thickness \(dx\). Light travels perpendicular to these planes. Suppose there are \(N\) droplets in volume \(V\). If a photon enters a sub-volume \(V_k = \pi r^2 dx\) that contains a droplet, it will be absorbed. |
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| The probability \(p\) that a photon will be absorbed is: | | The probability \(p\) that a photon will be absorbed is: |
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| $$p = \frac{N \pi r^2 dx}{V} \tag{1}$$ | | $$p = \frac{N \pi r^2 dx}{V} \tag{1}$$ |
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| On the other hand, if the incoming light intensity is \(I\), and the outgoing is \(I - dI\), the change \(-dI\) reflects the absorbed photons. So the absorption probability is also: | | On the other hand, if the incoming light intensity is \(I\), and the outgoing is \(I - dI\), the change \(-dI\) reflects the absorbed photons. So the absorption probability is also: |
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| $$p = \frac{-dI}{I} \tag{2}$$ | | $$p = \frac{-dI}{I} \tag{2}$$ |
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| Equating (1) and (2): | | Equating (1) and (2): |
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| $$-\frac{dI}{I} = \frac{N \pi r^2 dx}{V}$$ | | $$-\frac{dI}{I} = \frac{N \pi r^2 dx}{V}$$ |
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| Integrate this differential equation from \(I_0\) to \(0.9 I_0\) (since intensity drops by 10%) and from \(x = 0\) to \(x = L\): | | Integrate this differential equation from \(I_0\) to \(0.9 I_0\) (since intensity drops by 10%) and from \(x = 0\) to \(x = L\): |
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| $$\int_{I_0}^{0.9 I_0} \frac{dI}{I} = -\frac{N \pi r^2}{V} \int_0^L dx$$ | | $$\int_{I_0}^{0.9 I_0} \frac{dI}{I} = -\frac{N \pi r^2}{V} \int_0^L dx$$ |
| $$\ln \frac{0.9 I_0}{I_0} = -\frac{N \pi r^2}{V} L$$ | | $$\ln \frac{0.9 I_0}{I_0} = -\frac{N \pi r^2}{V} L$$ |
| $$\ln 0.9 = -\frac{N}{V} \pi r^2 L$$ | | $$\ln 0.9 = -\frac{N}{V} \pi r^2 L$$ |
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| Let \(n = \frac{N}{V}\), the number of droplets per unit volume: | | Let \(n = \frac{N}{V}\), the number of droplets per unit volume: |
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| $$n = \frac{-\ln 0.9}{\pi r^2 L}$$ | | $$n = \frac{-\ln 0.9}{\pi r^2 L}$$ |
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| Using the approximation: | | Using the approximation: |
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| $$-\ln 0.9 \approx 0.1 = \frac{1}{10}$$ | | $$-\ln 0.9 \approx 0.1 = \frac{1}{10}$$ |
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| We get the final estimate: | | We get the final estimate: |
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| $$n \approx \frac{1}{10 \pi r^2 L} = \left(10 \pi r^2 L\right)^{-1}$$ | | $$n \approx \frac{1}{10 \pi r^2 L} = \left(10 \pi r^2 L\right)^{-1}$$ |
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