# How does a spectrophotometer measure absorbance based on the intensity of detected light?

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 $$I_t/I_0$$ ?

• 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). May 25 '21 at 12:03

• 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 $$\mathrm{Abs} = f(\lambda{})$$.