0
$\begingroup$

I am just reviewing Fresnel's equations where Reflectance $R$ is equal to

$$R=|r|^2$$

where $r$ is the complex reflection coefficient. Is there an equation that relates $R$ to intensity or is there a reason one is related to the other?

$\endgroup$
  • $\begingroup$ Until you get to very high intensities, reflectance is independent of intensity. How high is "very high"? For conventional metal-coated mirrors, "very high" means intensities requiring a pulsed laser in a lab. $\endgroup$ – garyp Mar 18 at 21:18
  • $\begingroup$ Thanks for the response. I'm looking into reflectance spectroscopy and frequently they will plot intensity as a function of wavelength in place of reflectance. I'm just confused on how one just swaps between one and the other in these situations (so situations where wattage is ~5-20mW). $\endgroup$ – MathHelper123 Mar 18 at 21:32
  • $\begingroup$ The reflected intensity $I_R$ is related to the intensity of the incident light $I_0$ by $I_R=RI_0$ so $I_R\propto R$. A graph of $I_R$ against wavelength will look the same as a graph of $R$ against wavelength. Is this what you meant? $\endgroup$ – John Rennie Mar 19 at 6:08
  • $\begingroup$ The reflection r is complex defined through the ratio of the incoming to the The reflected field. So in terms of intensity or irradiance <S>cos(a) is the squared modulus, s pointing vector. $\endgroup$ – user591849 Mar 19 at 9:12
0
$\begingroup$

If the authors plot intensity rather than reflectance, it's probably because the absolute reflectance is not of interest to the study at hand. The interest could me more focused on the location in wavelength of features such as peaks and valleys. The interest is to extract information about the system from the wavelengths.

Absolute reflectance is difficult to measure. Intensity is considerably easier. Reflectance requires a calibrated detector system, or a known reference sample, or both. Even then, the measurement requires quite a bit of care. If the absolute reflectance is of little interest, there is little incentive to measure it. Spectroscopy more often than not is about the location of features.

On the other hand, if the location of features must be known to a high degree of accuracy, absolute measurements might be required, because non-uniform spectral response of a detector can introduce small errors in the position of peaks.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.