Discussion Forum

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

Answer:

Explanation:

From Rayleigh's law of scattering, intensity

$I\propto\frac{1}{\lambda^{4}}$

 $\therefore$    $\frac{I_{1}}{I_{2}}\propto\left(\frac{\lambda_{2}}{\lambda_{1}}\right)^{4}$

$\Rightarrow\frac{I_{1}}{I_{2}}=\left(\frac{330}{880}\right)^{4}=\left(\frac{3}{8}\right)^{4}=\frac{81}{4096}$

${I_{2}}=(\frac{4096}{81})$Å =(50.557)Å