Question

When sunlight is scattered by atmospheric atoms and molecules the amount of scattering of light of wavelength $880nm$ is $A$ . Then, the amount of scattering of light of wavelength $330nm$ is approximately

• A. $10A$
• B. $20A$
• C. $40A$
• D. $50.5A$

Answer 1

Answer: (D)

We know from Rayleigh's law of scattering
$I \propto \frac{1}{\lambda ^{4}}$
So, $\frac{I_{1}}{I_{2}}=\left(\frac{\left(\lambda \right)_{2}}{\left(\lambda \right)_{1}}\right)^{4}$
$\Rightarrow \, \frac{A}{I_{2}}=\left(\frac{330}{880}\right)^{4}=\left(\frac{3}{8}\right)^{4}=\frac{81}{4096}$
$I_{2}=\frac{4096}{81} \, A=\left(50.57\right)A$