I'm not sure I understand your question. There should not be any correlation between the color of reflection and temperature (emission from a heated black body would be different story, where temperature and color are well correlated, the hotter the shorter the wavelength). "That something else" in my previous post referred to anything that happens to appear brighter than the dark green vegetation without the filter, but does not get "brightened up" (relatively speaking) when seen through the IR filter. Chances are that it is reflecting more light and has a lower temperature (hence cooler in temperature, not necessarily in color).
Here you are talking about the filter and that's fine, but since
the human eye cannot see the IR light..then what is making the
foliage lighter?
Infrared filter is not a binary device like our mathematic
abstraction would make it out to be: 100% pass-through for photons
with wave length at 721nm or longer, 0% pass-through for photons at
720nm or shorter.
In reality, a typical infrared filter more likely has a transfer
curve of, say, 50% of pass-through at 720nm, 85% at 750nm, 96% at
800nm or longer (surface multicoating decides whether the max
pass-thru gets beyond 96% or so), 25% at 700nm, 10% at 650nm, 1% or
less for wavelength shorter than 600nm (green color). 1% of a the
brightest broad daylight is still noticeabe amount of light, about
8 stops below, 1/2000 shutter vs. 1/20 shutter speed. In other
words, no worse than typical evening indoor lighting.
Jim
ps.
My wife says I have a talent for explaining things in the most
complex way possible ("scientificly precise" is what I say). Since
she is a science teacher and I'm an electrical engineer by training
who sees infrared filter as a low-pass filter, I will give you her
version: an infrared filter is just like a tinted sunglass; the
filter drasticly reduces the amount of visible light that can pass
through in the visible spectrum, but does less to infrared.