# What is the intensity of this light?

I am struggling with a derivation that calculates the cross sections for Mie scattering and since the incident light is considered to be a x-polarized plane wave I thought that we would have $$I_i = \frac{1}{2} \sqrt{\frac{\epsilon}{\mu}} \vert E_0 \vert^2$$, but I do not understand this derivation then, since a factor $2 \pi$ seems to be missing.

It starts with an expression for the scattered field, explains how they got this expression by using some orthogonality properties and then - in my opinion argue - that this $Re(g_n)=1$. But then I do not understand what they take as the incident intensity in order to get the expression $C_{sca}$. Does anybody have an idea?

$$W_s=\frac{\pi|E_0|^2}{k\omega\mu}\sum_{n=1}^\infty(2n+1)\mathcal{Re}\{g_n\}(|a_n|^2+|b_n|^2),$$ where we have used (4.24) and the relation $$\int_0^\pi(\pi_n\pi_m+\tau_n\tau_m)\sin{\theta}\text{ }d\theta=\delta_{n\text{ }m}\frac{2n^2(n+1)^2}{2n+1},$$ which follows from (4.27). The quantity $g_n$, is defined as - $i\xi_n^*\xi_n^{'}$, may be written in form $$g_n=(\chi_n^*\psi_n^{'}-\psi_n^*\chi_n^{'})-i(\psi_n^*\psi_n^{'}+\chi_n^*\chi_n^{'}),$$ where the Riccati-Bessel function $\chi_n$ is - $\rho y_n(\rho)$ and, therefore, $\xi_n=\psi_n-i\chi_n$. The functions $\psi_n$ and $\chi_n$ are real for real argument; therefore, if we use the Wronskian(Antosiewicz, 1964) $$\chi_n\psi_n^{'}-\psi_n\chi_n^{'}=1,\tag{4.60}$$ it follows that the scattering cross section is $$C_{sca}=\frac{W_s}{I_i}=\frac{2\pi}{k^2}\sum_{n=1}^\infty(2n+1)(|a_n|^2+|b_n|^2).\tag{4.61}$$

• Which book is this? Apr 3, 2014 at 10:20

I want to add this as a comment but dont have enough rep, so please dont downvote me lol, but the incident intensity comes from the expansion coefficients $a_n$ and $b_n$. Look further back in the book to see the equation they appear in.