Question

What is the relative efficiency of air molecules to scatter blue light ($\lambda = 470$ nm) as compared to red light ($\lambda = 640$ nm)?

• A. 3.45
• B. 1.83
• C. 0.54
• D. 0.29

Hint:

The Rayleigh Law ($\frac{1}{\lambda^4}$) is why the sky is blue! Shorter wavelengths (blue) scatter much more efficiently than longer wavelengths (red).

Answer 1

The Correct Option is A

Concept: Rayleigh scattering describes the scattering of light by particles much smaller than the wavelength of the light (such as air molecules). The intensity of scattered light ($I$) is inversely proportional to the fourth power of the wavelength ($\lambda$): \[ I \propto \frac{1}{\lambda^4} \]

Step 1: Setting up the ratio.
To find the relative efficiency of blue light ($I_B$) compared to red light ($I_R$), we use the formula: \[ \text{Relative Efficiency} = \frac{I_B}{I_R} = \left( \frac{\lambda_{\text{red}}}{\lambda_{\text{blue}}} \right)^4 \]

Step 2: Perform the calculation.
Given $\lambda_{\text{blue}} = 470$ nm and $\lambda_{\text{red}} = 640$ nm: \[ \text{Ratio} = \left( \frac{640}{470} \right)^4 \approx (1.3617)^4 \] \[ \text{Ratio} \approx 3.435 \]

Step 3: Conclusion.
The calculated value of approximately 3.44 correlates most closely with the provided option 3.45.