Question

If I is the intensity of scattered light of wavelength$\lambda$. Then relation between them is given by:

& A.\quad I\propto \dfrac{1}{{{\lambda }^{4}}} \\\ & B.\quad I\propto {{\lambda }^{4}} \\\ & C.\quad {{I}^{2}}\propto \dfrac{1}{{{\lambda }^{4}}} \\\ & D.\quad I\propto \dfrac{1}{\lambda } \\\ \end{aligned}$$

Answer 1

Hint : Scattering of light refers to the phenomenon of incident light bouncing on surfaces, or small particles or as light passes through a medium. Due to the processes stated above, the light gets absorbed and re-transmitted and reflected, refracted etc. This phenomenon is known as scattering of light.
The intensity of the light scattered depends on the particle size and the wavelength of the light.

Complete step by step solution :
Let’s understand the process of scattering. Let’s consider light emerging from a source. When this light passes from one medium to any other medium or gets reflected, refracted, absorbed by particles of the medium and then re-transmitted into any other direction. This phenomenon is termed as a scattering of light as the light scatters around in all the directions.
The medium from which light gets scattered depends upon the size of the particles in the medium and the wavelength of the light getting scattered.
When the wavelength of the light wave is shorter then, the corresponding frequency of the light wave is high and this wave scatters more as the wave oscillates more. This causes the particles of the medium to have a higher probability of interacting with the light wave. Similarly, for light waves of larger wavelengths having low frequency, the probability of the particles of the scattering medium to interact with the light wave are very less.
Hence, this value of the interaction probability (P) and the wavelength of the light wave is given by: $P\propto \dfrac{1}{{{\lambda }^{4}}}$.
This value of interaction probability translates into intensity as well. Hence, for the incident intensity of the light $\left( {{I}_{0}} \right)$, the scattered intensity (I) becomes: $I\propto \dfrac{{{I}_{0}}}{{{\lambda }^{4}}}$. Hence the dependency of the scattered intensity (I) and the wavelength ($\lambda$) is: $I\propto \dfrac{1}{{{\lambda }^{4}}}$

So, the correct answer is “Option A”.

Note : For molecules of size larger than the wavelength being incident on it, they don’t scatter light using a different phenomenon known as Mie effect. The light being scattered largely appears white. This is also the reason behind the clouds, which are large bags of water droplets appear white, while the sky appears blue.
The sky also appears blue, due to the Rayleigh scattering effect. The wavelength dependence of$\left( \sim {{\lambda }^{-4}} \right)$causes the shorter wavelengths such as (blue) to scatter more strongly than the longer wavelength (red).