= Beer-Lambert calibration model
{c}
Choose an ideal absorbing <chemical species> with $\varepsilon=8000\,\mathrm{L\,mol^{-1}\,cm^{-1}}$ at a fixed <wavelength>, and a <cuvette> with geometric <length> $l=1\,\mathrm{cm}$. These are stipulated model parameters. A micromolar <molar concentration> is $1\,\mu\mathrm M=10^{-6}\,\mathrm{mol\,L^{-1}}$, so each additional $25\,\mu\mathrm M$ gives $\Delta A=8000\times1\times25\times10^{-6}=0.2$. Five chosen standards therefore predict
$$
\begin{array}{c|ccccc}
c\;(\mu\mathrm M)&0&25&50&75&100\\
A&0&0.2&0.4&0.6&0.8\\
T&1&0.63096&0.39811&0.25119&0.15849
\end{array}
$$
The <absorbance> <graph> is a <straight line> of <slope> $0.008\,(\mu\mathrm M)^{-1}$, while the <transmittance> <graph> shows <exponential decay>. Equal increments of <molar concentration> multiply <transmittance> by $10^{-0.2}\approx0.63096$; they do not subtract a fixed transmitted fraction.
\Image[/beer-lambert-absorbance-transmission-calibration.png]
{title=Linear <absorbance> and exponential <transmittance> describe the same calibration model}
{id=image-beer-lambert-absorbance-transmission-calibration}
{description=Original exact model curves for $\varepsilon=8000\,\mathrm{L\,mol^{-1}\,cm^{-1}}$ and $l=1\,\mathrm{cm}$; circles mark the five stipulated standards, not measured material data. The highlighted $T=0.40$ reading corresponds to $A=\log_{10}(2.5)$ and $c\approx49.7425\,\mu\mathrm M$. The common horizontal <molar concentration> scale lets the two panels show the same inversion.}
Suppose a measurement returns the stipulated blank-normalized reading $T=0.40$ under those same conditions. The inferred <absorbance> and <molar concentration> are
$$
A=-\log_{10}(0.40)=\log_{10}(2.5)\approx0.3979400,
\qquad
c=\frac{\log_{10}(2.5)}{8000\times1}\,\mathrm{mol\,L^{-1}}
\approx49.7425\,\mu\mathrm M.
$$
This lies slightly below the $50\,\mu\mathrm M$ standard because $0.40$ is slightly above $10^{-0.4}\approx0.39811$. A calibration determined experimentally by <linear regression> may instead have $A=mc+b$. Then use $c=(A-b)/m$, keeping the <slope> $m$ and <intercept> $b$ in their measured units. A nonzero <intercept> calls for checking the <reference blank> and sample preparation; it does not uniquely identify the cause.
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