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In Rayleigh scattering by a molecule the intensity of the scattered light is:

$$I \propto I_0({1+\cos^2(\theta))}$$

while the time-averaged Poynting vector of an oscillating dipole is:

$$\langle S \rangle \propto \sin^2(\theta)$$

The magnitude of the Poynting vector is the intensity of EM field.

Since in Rayleigh phenomenon the scattered light is due to an oscillating dipole, why do we get this difference in the angular dependence?

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    $\begingroup$ Study both concepts more in depth from textbooks or internet. What is the meaning of $\theta$ in these formulas? Is it the same? Do both formulas give intensity of total radiation, or does some of them give intensity only due to part of it? $\endgroup$ Jul 23, 2022 at 23:21
  • $\begingroup$ @JánLalinský About second question I think yes: both formulas give intensity of total radiation unless you mean that in Rayleigh scattering only monochromatic wave is scatterd. $\endgroup$
    – Salmon
    Jul 24, 2022 at 18:13
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    $\begingroup$ In the first formula, angle $\theta$ is the scattering angle, i.e. angle between primary and secondary (scattered) wave. In the second formula, it is not so; instead, here $\theta$ is the angle between the direction of electric moment line of oscillation and direction of the secondary wave. Both formulas give only intensity due to scattered wave, they do not give total intensity, which is much higher in the direction of the primary wave. $\endgroup$ Jul 24, 2022 at 23:17

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The formula you give for Rayleigh scattering is for unpolarized light, consisting of equal amounts of two perpendicular polarisation states. These two states excite two perpendicular oscillations in the dipole moments of the scattering objects.

If you Rayleigh scatter linearly polarised light then you will get the $\sin^2\theta$ dependence of a dipole oscillating along one axis.

Rayleigh scattering of unpolarized light involves summing the intensity from two dipoles oscillating in perpendicular directions (both in the plane perpendicular to the direction of the incoming light).

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  • $\begingroup$ Thank you so much so if I study the Poynting vector of two oscillating dipoles in perpendicular directions I will get the $1+cos^2(\theta)$? $\endgroup$
    – Salmon
    Jul 24, 2022 at 18:10

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