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

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

Tardigrade Admin · Accepted Answer

We know from Rayleigh's law of scattering I propto (1/λ 4) So, (I1/I2)=(((λ )2/(λ )1))4 ⇒ (A/I2)=((330/880))4=((3/8))4=(81/4096) I2=(4096/81) A=(50.57)A

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

$10 A$

$20 A$

$40 A$

$50.5 A$

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

Q. 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

10A

20A

40A

50.5A