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Question: Answered & Verified by Expert
Let $S_1$ be the amount of Rayleigh scattered light of wavelength $\lambda_1$ and $S_2$ that of light of wavelength $\lambda_2$ from a particle of size $a$. Which of the following statement is true?
PhysicsRay OpticsTS EAMCETTS EAMCET 2018 (04 May Shift 2)
Options:
  • A $\frac{S_1}{S_2}=\left(\frac{\lambda_2}{\lambda_1}\right)^4$, if $\lambda_1, \lambda_2>a$
  • B $\frac{S_1}{S_2}=\left(\frac{\lambda_1}{\lambda_2}\right)^4$, if $\lambda_1, \lambda_2 \gg a$
  • C $\frac{S_1}{S_2}=\left(\frac{\lambda_2}{\lambda_1}\right)^4$, if $\lambda_1, \lambda_2 \ll a$
  • D $\frac{S_1}{S_2}=\left(\frac{\lambda_1}{\lambda_2}\right)^4$, if $\lambda_1, \lambda_2 \ll a$
Solution:
2162 Upvotes Verified Answer
The correct answer is: $\frac{S_1}{S_2}=\left(\frac{\lambda_2}{\lambda_1}\right)^4$, if $\lambda_1, \lambda_2>a$
Condition for scattering is
$$
\lambda>>a
$$
and according to Rayleigh, intensity (or amount) of scattered wavelength is inversely proportional to fourth power of the wavelength of incident wave, i.e.
$$
S \propto \frac{1}{\lambda^4} \Rightarrow \frac{S_1}{S_2}=\left(\frac{\lambda_2}{\lambda_1}\right)^4 \text { and } \lambda>>a
$$

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