Modern light scattering uses a polarized laser, but since much of the older literature used unpolarized light sources, it is useful to understand them. Unpolarized light can be represented as a combination of a vertically and horizontally polarized waves-the first one with electric field oscillating along the vertical $x$ axis and the second one-along the horizontal $y$ axis. The intensities of these two parts of the incident light are $I_x=I_y=I_i / 2$. The intensity per unit scattering volume of the vertically potarized scattered wave is $\bar{I}_{\text {polar }} / 2$.
(i) Show that the intensity at scattering angle $\theta$, per unit scattering volume of the horizontally polarized scattered wave, is $\left(\bar{I}_{\mathrm{polar}} / 2\right) \cos ^2 \theta$.
(ii) Show that the intensity of the scattered light per unit scattering volume using an unpolarized light source valid for any radial position of the detector with scattering angle $\theta$ is
$$
I_{\text {unpolar }}=\frac{2 \pi^2 n^2}{\lambda^4 r^2}\left(1+\cos ^2 \theta\right)\left(\frac{\mathrm{d} n}{\mathrm{~d} c}\right)^2 \frac{c M}{\mathcal{N}_{\mathrm{Av}}} I_{\mathrm{i}}
$$
(iii) Demonstrate that the Rayleigh ratio from an unpolarized light source [Eq. (1.87)] is equal to
$$
\begin{aligned}
R_\theta^{\text {unpolar }} & \equiv \frac{\bar{I}_{\text {unpolar }} r^2}{I_i}=\frac{2 \pi^2 n^2}{\lambda^4}\left(\frac{\mathrm{~d} n}{\mathrm{~d} c}\right)^2 \frac{c M}{\mathcal{N}_{\mathrm{Av}}}\left(1+\cos ^2 \theta\right) \\
& =K c M \frac{1+\cos ^2 \theta}{2}
\end{aligned}
$$
where $K$ is the optical ratio defined in Eq. (1.89).