# Why absorption/transmission spectroscopic data is normalized?

A typical data processing step for acquired absorption/transmission data is normalizing, i.e. stretching the curve such that it is bounded between 0 and 100%.

However, the absorption/transmission curve changes with the density and thickness of the sample. Moreover, absorption is a non linear phenomena (for example Beer–Lambert law), so I would expect that the relative height of the various peak would not remain constant across several samples of the same material.

Why, therefore, is it justified to normalize acquired spectroscopic data? Why not present data with original abs./trans. values, while providing details about the sample itself (for example thickness)?

In addition, how can spectroscopy be quantitative if the relative height of the peaks changes with illumination/thickness/density/etc.?

• The Beer Lambert Law is explicitly a linear relation. I'm some confused about this question. At constant thickness, the Law is linear with chemical concentration. At constant concentration, the Law is linear with pathlength (thickness) – Lighthart Dec 18 '15 at 20:54
• @Lighthart how is BL linear? it is exponentially decaying with penetration depth. – Sparkler Dec 18 '15 at 20:57
• Absorbance equals molar absorptivity times concentration times path length? life.nthu.edu.tw/~labcjw/BioPhyChem/Spectroscopy/beerslaw.htm – Lighthart Dec 18 '15 at 22:06
• @Lighthart yeah, but intensity is A = -log T = - log (I / Io). Aren't all measurements intensity based? – Sparkler Dec 18 '15 at 22:29

There is confusion here about absorption at one wavelength and the absorption spectrum. At each wavelength the Beer-Lamber law applies $$I_{trans_{\lambda}}=I_{0_{\lambda}}\exp(-\epsilon_{\lambda} c l)$$ for a compound with extinction coefficient $\epsilon$ at wavelength $\lambda$ and concentration c and path length l.
When an absorption spectrum is produced it is normal to show it as optical density vs wavelength, i.e. $OD_{\lambda}=\log(I/I_0)_{\lambda}=-\epsilon_{\lambda} c l$ and in this case the ratio of peaks for the same compound is constant no matter what the concentration. (Naturally assuming that it does not absorb so much that the instrument fails).