Effective attenuation coefficient for polychromatic light

Question

The Beer-Lambert law gives the optical intensity of collimated light as a function of depth $z$ as:

$$I(z)=I_{0}\mathrm e^{-\gamma z},$$

where $\gamma = \alpha + \beta$ is the wavelength-dependent attenuation coefficient, with $\alpha$ and $\beta$ being the absorption and scattering coefficients respectively.

Suppose two completely different wavelengths $\lambda_1$ and $\lambda_2$ are present in the light beam. Is it possible to define the "effective attenuation coefficient" of the material $\overline{\gamma}$ for two or more wavelengths as an addition of the different attenuation coefficients?

That is,

$$\overline{\gamma}=p_{\lambda_1}\gamma_{\lambda_1}+p_{\lambda_2}\gamma_{\lambda_2},$$

where $p$ is the fraction of the beam which is of a given wavelength (indicated by the subscript). The mathematics seems to work out, but I have never seen this used in literature before. Any explanation is greatly appreciated.

Answer 1

The intensity transmitted at wavelength $a$ where the extinction coefficient is $\epsilon_a$ is $I_a=I_{0a}\exp(-\epsilon_a L C)$ with path length $L$ and concentration $C$, and similarly for a second wavelength $b$. If, for clarity, we let $x=\epsilon l C $ then then $I_a=I_{0a}\exp(-x_a)$ and $I_b=I_{0a}\exp(-x_b)$. The total transmitted light is $I_t=I_a+I_b$ and the total initial amount $I_{0t}=I_{0a}+I_{0b}$. Thus the total transmittance ratio is

$$ \frac{I_t}{I_{0t}}=\frac{I_{0a}e^{-x_a}+I_{0b}e^{-x_b}}{I_{0a}+I_{0b}}$$

You want to make this equal to $I_{0ab}e^{-x_{ab}}$ but there seems no systematic way to do this and this confirms the fact that the simple Beer-Lambert law only applies for monochromatic light.