Question

Given that the amplitude of the scattered light is
(i) directly proportional to amplitude of incident light
(ii) directly proportional to the volume of the scattering dust particle
(iii) inversely proportional to its distance from the scattering particle and
(iv) dependent upon the wavelength $\lambda$ of the light.
Then, the relation of intensity of scattered light with the wavelength is

• A. $\frac{1}{\lambda^{2}}$
• B. $\frac{1}{\lambda^{4}}$
• C. $\frac{1}{\lambda^{6}}$
• D. $\frac{1}{\lambda^{7}}$

Answer 1

Answer: (B)

According to the question, the expression for the scattered amplitude of light $\left(A_{s}\right)$ in terms of amplitude of incident light $\left(A_{i}\right)$, volume $(V)$, distance from scattering particle $(x)$ and wavelength $(\lambda)$ can be given as
$\therefore A_{s}=k A_{i}^{1} V^{1} x^{-1} \lambda^{d}$
where, $k$ is the constant of proportionality.
Writing the dimensions on both sides of the above equation, we get
$[ L ]=[ L ]\left[ L ^{3}\right]\left[ L ^{-1}\right]\left[ L ^{d}\right]=\left[ L ^{3+d}\right]$
Comparing the powers of $L$ on both sides, we get
or $ 1=3+d$
or $ d=-2$
i.e. $ A_{s} \propto \frac{1}{\lambda^{2}}$
But intensity $\left(I_{s}\right) \propto\left[\text { amplitude }\left(A_{s}\right)\right]^{2}$
$\therefore I_{s} \propto \frac{1}{\lambda^{4}}$