# Effective attenuation coefficient for polychromatic light

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.

• What's to explain? You invented this value, now find a use for it. – Ivan Neretin Oct 11 '18 at 11:19
• I don't see what is unclear about this question. How do I make two attenuation coefficients into one? – A.K. Oct 11 '18 at 16:14
• @IvanNeretin This could be very useful in the area of material processing by laser beams. Some lasers can have two wavelengths present simultaneously (for example, when you frequency-double a laser). – Merin Oct 12 '18 at 5:21

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.