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 330 nm is approximately

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

Answer 1

Answer: (D)

50.5 A

From Rayleigh's law of scattering, intensity
$I\propto \frac{1}{{\lambda }^4}$
$\therefore \frac{I_1}{I_2}={\left(\frac{{\lambda }_2}{{\lambda }_1}\right)}^4$
$\Rightarrow \frac{I_1}{I_2}={\left(\frac{330}{880}\right)}^4={\left(\frac{3}{8}\right)}^4=\frac{81}{4096}$
$I_2=\frac{4096}{81}A=\left(50.557\right)A$