Re: IR & Water: a robust debate continues...

[Steve Hodges <shodges@xxxxxxxxxxxxxx>]

"Editor - P.O.V. Image Service" wrote:
> 
> OK time for the math kiddies...
> 
> YAY!!!
> 
> Guess what?  If water attenuates Near IR by 10% each centimeter, how many
> centimeters of water does the light need to pass through until 50% of the near IR
> is attenuated..?

Lets see T = t^n

where T is the total transmission,
      t is the transmission per unit
      n is the number of units.

and A = 1 - T

where A = total absorption

so A = 1 - t^n
   0.5 = 1 - 0.9^n
   0.5 = 0.9^n
   ln 0.5 = n ln 0.9
   n = ln 0.5 / ln 0.9
     = -0.6931 / -0.1054
     = 6.58 cm

> 
> <Final Jeopardy Music Playing>
> 
> Approximately 6.2 cm, according to YOUR book..

Poor maths in that book...

> Last time I checked, that means at about 15.75 inches, 50% of the available IR
> light will be absorbed...

Ooooh!  Last time _I_ checked, 6.58 cm was approximately 2.6 inches.  So
after a round trip of 2.6 inches (1.3 inches of water - the light has to
go both ways) you lose a stop.

> Thanks for the data..
> 
> So, if the water is totally pure, you will have a 50% attenuation of Near IR  by
> the water at 15 3/4 inches... Real world water is unlikely to be of the pure
> de-ionized variety... Therefore, the attenuation will be increased...

How about, you lose a stop for each 1.3 inches of water depth (by these
figures).

Steve

p.s.  Note that I am just correcting sloppy maths, not making any
judgment on the accuracy of the quoted figures for absorbency.  But I do
tend to believe what I see in the graphs.

Also I note that my swimming pool is almost totally black when viewed
through a filter that permits only the very near IR (just visible)
light.
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