Question

The ratio of scattering efficiency of red light of wavelength 0.65 µm to blue light of wavelength 0.45 µm by air molecules in the atmosphere is _________ (rounded off to two decimal places).

Hint:

Rayleigh scattering efficiency is inversely proportional to the fourth power of the wavelength, leading to stronger scattering for shorter wavelengths.

Answer 1

Correct Answer: 0.22

The scattering efficiency of light by molecules in the atmosphere follows Rayleigh scattering, which is inversely proportional to the fourth power of the wavelength (\( \lambda \)): \[ \text{Scattering Efficiency} \propto \frac{1}{\lambda^4} \] The ratio of scattering efficiencies for red and blue light is: \[ \text{Ratio} = \frac{\left( \frac{1}{\lambda_{\text{red}}^4} \right)}{\left( \frac{1}{\lambda_{\text{blue}}^4} \right)} = \left( \frac{\lambda_{\text{blue}}}{\lambda_{\text{red}}} \right)^4 \] Substituting the values for \( \lambda_{\text{red}} = 0.65 \ \mu m \) and \( \lambda_{\text{blue}} = 0.45 \ \mu m \): \[ \text{Ratio} = \left( \frac{0.45}{0.65} \right)^4 = \left( 0.6923 \right)^4 = 0.2295 \] Thus, the ratio of scattering efficiencies is: \[ \boxed{0.23} \]