- 16 Jul '09 12:38 / 1 editWhy is the boundary between the “near infrared” range and the “intermediate infrared” range defined as 1500nm (wavelength) and the boundary between the “intermediate infrared” range and the “far infrared" range defined as 20000nm (wavelength) ?

-what I mean is are these values just totally arbitrary values that someone just haphazardly and mindlessly conjured up in his head or is there some either practical or theoretical or scientific significance to these approximate boundary values and, if so, what is it? -just always wondered.

Does anyone here know? - 16 Jul '09 13:14

All sorts of practical reasons, apparently:*Originally posted by Andrew Hamilton***Why is the boundary between the “near infrared” range and the “intermediate infrared” range defined as 1500nm (wavelength) and the boundary between the “intermediate infrared” range and the “far infrared" range defined as 20000nm (wavelength) ?**

-what I mean is are these values just totally arbitrary values that someone just haphazardly and mindless ...[text shortened]... oximate boundary values and, if so, what is it? -just always wondered.

Does anyone here know?

http://en.wikipedia.org/wiki/Infrared#Different_regions_in_the_infrared - 16 Jul '09 13:36

Strangely, none of the boundary values there are the ones I was taught at collage.*Originally posted by PBE6***All sorts of practical reasons, apparently:**

http://en.wikipedia.org/wiki/Infrared#Different_regions_in_the_infrared

I assume when it talks about “water absorption” it is talking about liquid water and not water vapour? - 16 Jul '09 14:01 / 1 editI am trying to understand the exact mathematical meaning of the term “Absorption coefficients” for water at the graph at the bottom of:

http://www.lsbu.ac.uk/water/vibrat.html

-but just can’t quite understand what it says.

I want to understand it in terms of what proportion of the radiation is absorbed per, say, one cm distance of liquid water the radiation travels through.

For example, according to the graph, the “Absorption coefficients” for 2000nm wavelength of infrared is “100 per cm”; so if a beam of infrared all of 2000nm wavelength is shone through one cm distance of liquid water, what percentage of that beam's energy would be absorbed by that water? - 16 Jul '09 14:46

See footnote "d":*Originally posted by Andrew Hamilton***I am trying to understand the exact mathematical meaning of the term “Absorption coefficients” for water at the graph at the bottom of:**

http://www.lsbu.ac.uk/water/vibrat.html

-but just can’t quite understand what it says.

I want to understand it in terms of what proportion of the radiation is absorbed per, say, one cm distance of liquid water ...[text shortened]... istance of liquid water, what percentage of that beam's energy would be absorbed by that water?

http://www.lsbu.ac.uk/water/vibrat.html#d - 17 Jul '09 17:49

It means the fraction of transmitted light is exp(-100).*Originally posted by Andrew Hamilton***I am trying to understand the exact mathematical meaning of the term “Absorption coefficients” for water at the graph at the bottom of:**

http://www.lsbu.ac.uk/water/vibrat.html

-but just can’t quite understand what it says.

I want to understand it in terms of what proportion of the radiation is absorbed per, say, one cm distance of liquid water ...[text shortened]... istance of liquid water, what percentage of that beam's energy would be absorbed by that water? - 17 Jul '09 19:07

Oh, I see now! Thanks for that*Originally posted by KazetNagorra***It means the fraction of transmitted light is exp(-100).**

So, for example, if we want to know how much light of 2000nm wavelength is absorbed if it passes through, say, 0.01 of a cm ( =0.1mm ) of liquid water, then, according to the graph, the ‘absorption coefficient’ is 100 for that wavelength so the proportion that gets through that 1/100th of a cm is:

e^( -100 * 0.01) = 0.3678 (approximately)

So only about 37% of that 2000nm wavelength of infrared light would get through that 0.1mm layer of liquid water (the rest being absorbed -and also scattered? -or is it only absorption we are talking about here?) - 18 Jul '09 10:28

I haven't really read the article so I can't tell how they defined it. I guess it's only absorption, though there will be scattering as well.*Originally posted by Andrew Hamilton***Oh, I see now! Thanks for that**

So, for example, if we want to know how much light of 2000nm wavelength is absorbed if it passes through, say, 0.01 of a cm ( =0.1mm ) of liquid water, then, according to the graph, the ‘absorption coefficient’ is 100 for that wavelength so the proportion that gets through that 1/100th of a cm is:

e^( -100 * 0 ...[text shortened]... e rest being absorbed -and also scattered? -or is it only absorption we are talking about here?) - 20 Jul '09 13:37

If it was pure water, IE DI water, it would not have much scattering due to the fact there are no particles of any size to scatter off, just H2O molecules and it's associated structures.*Originally posted by KazetNagorra***I haven't really read the article so I can't tell how they defined it. I guess it's only absorption, though there will be scattering as well.**