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,


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.

  • 2
    $\begingroup$ What's to explain? You invented this value, now find a use for it. $\endgroup$ Oct 11, 2018 at 11:19
  • $\begingroup$ I don't see what is unclear about this question. How do I make two attenuation coefficients into one? $\endgroup$
    – A.K.
    Oct 11, 2018 at 16:14
  • 1
    $\begingroup$ @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). $\endgroup$
    – Merin
    Oct 12, 2018 at 5:21

1 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.


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