0
$\begingroup$

First-time poster here

In Beer's law, the ratio of incident light intensity, $I_0$, to exiting light intensity, $I$, is defined as $\log_{10} (\frac {I_0} {I}) = \epsilon L c$. Then, by rearranging some terms, we see $I = I_010^{-\epsilon L c} $, and so we might say the exiting intensity has a negative exponential relationship with concentration $c$.

My question:

What is the relationship of light intensity exiting at 90 degrees with the concentration (or perhaps turbidity) of a solution? Does it maintain the negative exponential relationship of Beer's law? I'm especially looking for a qualitative/layman's explanation. Thank you.

$\endgroup$
  • $\begingroup$ Such scattering is due to the Tyndall effect. and/or Rayleigh scattering. The relationships depends on particle size and the wavelength of light. $\endgroup$ – MaxW Jul 10 '18 at 14:58
  • $\begingroup$ I'm voting to close this question as off-topic because it belongs on physics.stackexchange $\endgroup$ – A.K. Jul 10 '18 at 15:13
  • $\begingroup$ Lambert-Beer is quite OK for chemists. ,-) I'm not putting this in an answer, but if your dye is not also responsible for the scattering, then the light intensity at every angle shows lambert beer behaviour. However if the dye scatters light, then the relation is of course totally different. $\endgroup$ – Karl Jul 10 '18 at 18:17
  • $\begingroup$ I'll point out that chemists do use nephelometers and turbidimeters. $\endgroup$ – MaxW Jul 10 '18 at 19:00
  • $\begingroup$ Beer's law describes the absorption of light by molecules not light scattering which does not involve absorption. Look up Dynamic light scattering for some theory, and Zimm plots for determining particle properties angular dependence of scattering. $\endgroup$ – porphyrin Jul 11 '18 at 14:46

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Browse other questions tagged or ask your own question.