Difference between revisions of "Beer"
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| − | \ln\frac{I(0)}{I(x)}=\ | + | \ln\frac{I(0)}{I(x)}=\alpha c x |
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| + | <span style="font-family:Georgia;">whence, on choosing <math>x</math> to be the exit plane for the radiation, and with <math>\varepsilon=\alpha ln(10)=2.303\alpha</math>: | ||
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| + | <math> | ||
| + | \log_{10}\frac{I(0)}{I(L)}=\varepsilon c L | ||
</math> | </math> | ||
Latest revision as of 22:35, 3 February 2017
1. The classical law and its history
When electromagnetic radiation, such as ultraviolet or visible light, passes through a transparent medium that contains an absorber of that illumination, the radiation’s intensity diminishes steadily with passage through the medium. Commonly the radiation is in the form of a collimated beam that impinges perpendicularly on a slab of width [math]L[/math] of the medium, as suggested diagrammatically in Figure 1.One may conjecture that, at any illuminated plane [math]x[/math] within the medium, the decrease in the intensity [math]I[/math] of the radiation with distance would be proportional to the uniform concentration [math]c[/math] of the absorber and to the local intensity of the light at that point; that is
[math] \frac{\mathrm d}{\mathrm d x} ( I(x) )=-\alpha cI(x) (1) [/math]
where [math]\alpha[/math] is a proportionality constant. Integration of this equation leads to
[math] \ln\frac{I(0)}{I(x)}=\alpha c x [/math]
whence, on choosing [math]x[/math] to be the exit plane for the radiation, and with [math]\varepsilon=\alpha ln(10)=2.303\alpha[/math]:
[math] \log_{10}\frac{I(0)}{I(L)}=\varepsilon c L [/math]
