We know that $\displaystyle A = -\log\frac{I_t}{I_0}$, where $I_0$ is the intensity of light without a sample absorbing, and $I_t$ with the sample. For determining the absorption $A$, we need both $I_t$ and $I_0$, but how does a spectrophometer detector measure $A$ knowing only $I_t$ ? (The detector is placed after the sample in the optical path, so it does not measure $I_0$ in front of it.)

Does the detector measure $I_t$ or the ratio $I_t/I_0$ ?

  • 3
    $\begingroup$ I'm not sure I understand the question because you haven't defined any variables and it's not clear what the difference between "It" and "If" is, but I guess you are missing the fact that the sample solution and the reference (aka "blank") are measured separately (or there is a scheme allowing for automatic subtraction of a "blank" signal). $\endgroup$
    – andselisk
    May 25, 2021 at 12:03

1 Answer 1


There are two possibilities:

  • If there is only one optical pathway, you record i) the blank sample for all wavelengths of interest and store the detector's intensity information in function of $\lambda$ in a working memory. Light passing the blank is then assumed to equate $I_0$. Then you record ii) the wavelength dependent intensities $I_\mathrm{abs}$ with your sample of interest, and report the subsequently corrected data as spectrum $\mathrm{Abs} = f(\lambda{})$.


  • Your instrument permits the simultaneous recording of sample and blank to record both intensity data and reporting the already corrected absorption in one run. Be sure, though, to assert which of the two paths (= two sample holders) is about your sample, and the blank (which may depend on spectrometer and software).

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