Absorbance of a finite waveband

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Main Question or Discussion Point

For a single wavelength, the absorbance is given by:

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

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.

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I believe what you have is correct. However, I think the letter $A$ is more often used for the absorbance than the letter $\bar{d}$. Using the letter $\bar{d}$ is somewhat misleading, in that I was expecting it to represent the distance $x=\frac{1}{\alpha}$ where $I_t(x)=I_o e^{-\alpha x}$ for the exponential attenuation with distance in the manner of a single wavelength.

Ibix
Agree with Charles. The attenuation coefficient you are using is apparently specific to a particular sample, rather than being for a particular material with the thickness as a parameter. I expected to see $x/\alpha(\lambda)$ (where $x$ is the thickness of your sample and $\alpha(\lambda)$ is the thickness if material needed to attenuate wavelength $\lambda$ by a factor of 10) everywhere you wrote $d$. Otherwise it looks fine.

I believe what you have is correct. However, I think the letter $A$ is more often used for the absorbance than the letter $\bar{d}$. Using the letter $\bar{d}$ is somewhat misleading, in that I was expecting it to represent the distance $x=\frac{1}{\alpha}$ where $I_t(x)=I_o e^{-\alpha x}$ for the exponential attenuation with distance in the manner of a single wavelength.
Thank you for the confirmation.

Yes, I believe $A$ is more commonly used, but $d$ is used too, especially in older texts where absorbance is sometimes called "optical density".

I am using $z$ for distance (or penetration depth). But I think what you are referring to is the quantity called "penetration depth" $\delta=1/\alpha$ (for a purely absorbing/non-scattering medium). The intensity drops to 37% of its initial value at this distance.

So, is there a way to simplify the equation in my last post further?

Agree with Charles. The attenuation coefficient you are using is apparently specific to a particular sample, rather than being for a particular material with the thickness as a parameter. I expected to see $x/\alpha(\lambda)$ (where $x$ is the thickness of your sample and $\alpha(\lambda)$ is the thickness if material needed to attenuate wavelength $\lambda$ by a factor of 10) everywhere you wrote $d$. Otherwise it looks fine.
Yes, I guess it is specific to the sample.

Did you mean:

$$\overline{d}=\log_{10}\left(\frac{\intop_{\lambda_{a}}^{\lambda_{b}}I_{0\lambda}d\lambda}{\intop_{\lambda_{a}}^{\lambda_{b}}I_{0\lambda}10^{-x/\alpha(\lambda)}d\lambda}\right)?$$

Could you please explain how you obtained $d=x/\alpha(\lambda)$?

P. S. My samples do have a considerable amount of scattering, so I am considering an attenuation coefficient $\gamma = \alpha + \beta$, where $\beta$ is the scattering coefficient. In this situation the Beer-Lambert law reads:

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

Ibix
Did you mean:

$$\overline{d}=\log_{10}\left(\frac{\intop_{\lambda_{a}}^{\lambda_{b}}I_{0\lambda}d\lambda}{\intop_{\lambda_{a}}^{\lambda_{b}}I_{0\lambda}10^{-x/\alpha(\lambda)}d\lambda}\right)?$$
Well the left hand side should be $x/\overline{\alpha}$, but yes, basically.
Could you please explain how you obtained $d=x/\alpha(\lambda)$?
It's the Beer-Lambert law you yourself cited, just using a different label for the constant (and base 10 log instead of base e, but that's just a multiplicative factor), and making the "constant" a function of $\lambda$.

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If I can add to the above, since you @roam used a log in base 10, we were expecting to see something of the form $I_t(x)=I_o 10^{-\alpha x }$, where $\alpha=\frac{1}{\bar{d}}$, (which would of course be incorrect). That is not what this $\bar{d}$ is, but by the choice of nomenclature with a $\bar{d}$, that's what we anticipated it was. $\\$ With the spectrum of absorption coefficients that you have, it is impossible to write the absorption in this form, but from initial appearances, with the letter $\bar{d}$, that's what it had indications that it might incorrectly be. As previously stated, $\bar{d}$ is really a poor choice of letter. Capital $D$ would certainly be far better than $\bar{d}$, since the parameter of interest refers to a type of density. $\\$ The $\bar{d}$ immediately has a hint that it is some kind of distance, and obviously, upon further inspection, that is not what it is. It is not the best practice to have nomenclatures that would start the reader on an incorrect path, making it so that they must readjust and see that it actually represents something completely different.

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Noted. I will use $A$ for absorbance to avoid confusion. So, the equation would become:

$$A = \log_{10}\left(\frac{\intop_{\lambda_{a}}^{\lambda_{b}}I_{0\lambda}d\lambda}{\intop_{\lambda_{a}}^{\lambda_{b}}I_{0\lambda}10^{-A_{\lambda}}d\lambda}\right).$$

Do you think this formulation is alright now?

The convention adopted in all textbooks and papers that I have seen is that the Beer-Lambert law is almost always expressed using the natural log, whereas the equation for absorbance/optical density is given using the decadic logarithm. This means that the $\alpha$ in the two equations are different by a factor of $\ln\left(10\right)\approx 2.3$.

I don't really understand the reasoning for transferring to decadic logarithms. But the process seems to be as follows:

$I(x)=I_{0}e^{-\alpha x},$

$T=\frac{I(x)}{I_{0}}=e^{-\alpha x} =e^{-\frac{x}{\delta}},$

$\ln\left(\frac{1}{T}\right)=\alpha x.$

$A = \log_{10}\left(\frac{1}{T}\right)=\log_{10}\left(\frac{I_{0}}{I_{t}}\right)=\frac{\alpha}{\ln\left(2.3\right)}x.$

In my original post, I derived the expression for $I_t$ from the absorbance equation: $I_t = I_0\ 10^{-A}$.