- #1

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$$d=\log_{10}\left(\frac{I_{0}}{I_{t}}\right), \tag{1}$$

where ##I_0## is the light intensity incident on the material and ##I_t## is the transmitted intensity (so that ##I_t/I_0## represents the fraction transmitted).

Many lasers don't have a single wavelength, but a rather a finite waveband ##\lambda_a - \lambda_b## (for instance, here is the relatively broad spectrum of a laser I have used recently).

In this case, I tried to express the effective absorbance ##\bar{d}## due to all those wavelengths as follows:

$$\text{Effective Transmitance}=\frac{\intop_{\lambda_{a}}^{\lambda_{b}}I_{0\lambda}10^{-d_{\lambda}}d\lambda}{\intop_{\lambda_{a}}^{\lambda_{b}}I_{0\lambda}d\lambda}, \tag{2}$$

so using Eqn. (1) the effective absorbance will be:

$$\overline{d}=\log_{10}\left(\frac{\intop_{\lambda_{a}}^{\lambda_{b}}I_{0\lambda}d\lambda}{\intop_{\lambda_{a}}^{\lambda_{b}}I_{0\lambda}10^{-d_{\lambda}}d\lambda}\right). \tag{3}$$

Is my expression mathematically and physically correct?

P. S. I have seen a very similar equation to Eqn. 2 used in a slightly different context. They used this in the area of spectrophotometry to account for the fact that a monochromator always selects a finite waveband, rather than a single wavelength.