| Author | Message | ||
     DJDahm |
I've abstracted the following from the comments written by Howard Mark. In them he mentions the Eastern Analytical Symposium session at which we will both be talking. I certainly hope to see many of you there. "The penetration depth in a remission measurement depends on the sample. In fact, it can even change within a sample as you measure different wavelengths, since penetration depth depends on the sample absorbance at each wavelength. It turns out that depth of penetration is roughly inversely related to the product of the absorption and scatter." The mentioning of the approximation was motivated by my saying that "the Absorbance function in remission { log(1/R) } from a sample is roughly proportional to { a/s } [absorption divided by scatter]. As we go to longer wavelengths, scatter goes down; hence, { log(1/R)} goes up." [Keep in mind that the Absorbance function in remission is not a measure of reflection, but a measure of the lack of reflection.] These approximations are very valuable as a way to carry around understanding in bite size form. However, keep in mind that they are best when we are dealing with low levels of absorption and scattering. Where do these approximations come from? (If I screw this up, I'm sure Howard will correct me.) Most of you are probably more used to thinking in terms of absorption and scattering coefficients than fractions, but I find fractions easier to think about. Hence I referred to fractions designated by 'a' and 's'. For thin slices, the fraction absorbed (or scattered) is just the coefficient times the thickness of the slice. For thicker slices of a sample, the intensity of the directly transmitted light as a function of thickness {Tt} as the beam experiences loss due to absorption is given by { Tt = exp(-kt) }, where k is the absorption coefficient, and t is the thickness of the slice. Similarly, the attenuation of the directly transmitted light due to scatter as a function is given by { Tt = exp(-zt) }, where z is the scattering coefficient. When there is both absorption and scatter, the directly transmitted intensity is given by { Tt = exp(-(k + z)t . Since we can write this as { Tt = exp(-kt) exp(-zt) }, we have a measure of transmission given by the product of quantities related to a loss due to absorption {a} and scatter {s}. Howard's reasoning goes that when the transmission loss gets too big, very little light from the back of the slice returns to the surface, and we are at a limit of penetration. The penetration depth is then roughly inversely related to the product of absorption and scatter. The line of reasoning that I was using goes like this. The Absorbance function in remission is very roughly proportional to the Kubelka-Munk function. The Kubleka-Munk function is roughly proportional to { a/s (2-a-s) }, which for small values of a and s can be approximated as { a/s }. The final part of my paragraph ("As we go to longer wavelengths, scatter goes down; hence, log(1/R) goes up." ) was motivated by the observation by Gabi Levin that at high wavelengths "the signal is quite weak" in remission due to scatter. I was saying that at longer wavelengths, where the scatter is low, the absorbance can get very high, so that it is hard to distinguish the signal from the background caused by reflection from the front surface. | ||
| hlmark |
Good point about the wavelength dependence, Don, but the question was about depth of penetration, not amount reflected. Turns out that depth of penetration is VERY VERY ROUGHLY (meaning don't pick on me because you have a more exact expression for it) inversely proportional to the product of the absorbance and scatter, rather than their ratio (which determines the reflectance) \o/ /_\ | ||
| djdahm |
I think Gabi mis-spoke, though I'm not quarrelling with his advice. He said "longer wavelengths � are scattered more than the shorter ones". Actually, it the other way round. For example, Blue light (shorter wavelength) is scattered off of air molecules more than red light (longer wavelengths), hence the sky is blue. However, the Absorbance function in remission [log(1/R)] from a sample is roughly proportional to "a/s" [absorption divided by scatter]. As we go to longer wavelengths, scatter goes down; hence, log(1/R) goes up. | ||
| Gabi Levin |
Sorry for being late - Howard is right, but there is another factor affecting the depth in the longer wavelengths, they are scattered more than the shorter ones. This is why in transmission through seeds we usually end up using the range 900 to 1400, because beyond that the signal is quite weak. Simple coarse test: Place a 3mm sheet of polyethylene in a petri dish, and place some thickness of powder of your compound, detect the polyethylene peak among the spectrum of your compound. Keep adding powder until the polyethylene "disappears". This is a rough, but could give you an idea. Thanks, Gabi Levin | ||
| hlmark |
Eric - that's a REALLY tricky question. Your estimate for the reflection measurements is not too bad, at least as far as the area examined - for one set of special conditions: the original "historical" instruments that NIR started off with. The penetration depth, however, depends more on the sample than on the instrument. In fact, it can even change within a sample as you measure different wavelengths, since penetration depth depends on the sample absorbance at each wavelength. I will be giving a talk on just this subject at EAS (at Don Dahm's award symposium), so it you can come to EAS, you probably should take in this talk. More recent instruments, however, have a much wider range of measurement areas, especially those incorporating microscopes, and those that produce images. In both those cases, the area measured by a single pixel is literally microscpic, and depends, as with any microscope, on the magnification used. \o/ /_\ | ||
| Eric LALOUM |
Dear All, I would like to have an idea about the volume of sample that can be analyzed in NIR (either transmission or diffuse reflectance). I don't need theoretical estimation but rather an order of magnitude that is currently observed with commercial equipment in order to compare NIR with other analytical techniques. My guess is that in transmission it is roughly 0.5 mm (pathlenght) x 10 mm2 and in diffuse reflection 0.1 mm (penetration) x 100 mm2. Is it correct ? Thank you |