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How water absorbtion influences off-surface reflectance

Pret

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Greeting all
It is known that water absorbs IR and UV (at various levels depending on actual wavelength).
Would it be appropriate to assume that the off-surface reflection will also be weakened in terms of reflected energy ?
Or, alternatively, if using wavelength that is less absorbed (penetrates to greater depth) by water, such as blue or green, would it be appropriate to assume that since larger percentage of energy gets penetrated - less energy will be reflected than in the case of highly absorbed wavelength ?

In other words, if one would like to throw a beam into water with the goal to obtain minimum reflected energy off the water surface (lets assume calm water for simplicity and incident angle that does not generate total reflection), would it be most apporpriate the wavelength that is as much absorbed by water as possible (weak water penetration) or it is more appropriate to use the opposite case - the wavelength that penetrates well into water (weak absorbtion) ?
Would be interested to hear your opinion

Thanks
 



Pret

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Perhaps I did not asked clearly enough, trying to rephrase..
I meant the power of reflected energy at different wavelengths when directed at the same angle
The deep IP and UV are absorbed heavily by water, however the blue and green penetrates quite well (which means they are much less absorbed by water)
So the question is, given the same incident angle, how the amount of reflected energy dependent on the water absorption properties of wavelengths?

If the particular wavelength gets absorbed heavily by water (deep IRfor instance), does that mean the reflected energy will be less ?
Or the opposite is more likely - the less absorbed wavelength penetrates deep into water to less energy left to get reflected off the surface ?
 

steve001

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Perhaps I did not asked clearly enough, trying to rephrase..
I meant the power of reflected energy at different wavelengths when directed at the same angle
The deep IP and UV are absorbed heavily by water, however the blue and green penetrates quite well (which means they are much less absorbed by water)
So the question is, given the same incident angle, how the amount of reflected energy dependent on the water absorption properties of wavelengths?

If the particular wavelength gets absorbed heavily by water (deep IRfor instance), does that mean the reflected energy will be less ?
Or the opposite is more likely - the less absorbed wavelength penetrates deep into water to less energy left to get reflected off the surface ?
In physics all surfaces are mirrors. The angle of incidence matters.
 

RedCowboy

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In physics all surfaces are mirrors. The angle of incidence matters.
More than the angle the material matters.

The black surface of a laser power meter head is not a mirror, also any surface in Vantablack will absorb 99.9 % of light so not what anyone could call a mirror.

 
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steve001

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More than the angle the material matters.

The black surface of a laser power meter head is not a mirror, also any surface in Vantablack will absorb 99.9 % of light so not what anyone could call a mirror.

Anything that reflects light is technically in discipline of physics a mirror though it may not be a mirror that maintains specular reflection. For another technical example consider cold vs heat. In common usage things are cold or hot. In scientific usage there are only things with lesser degrees of heat. Another is conductors vs insulators. Given enough electrical input an insulator can conduct electricity. Everything is a conductor or an insulator of one degree or another. You see the assigned definitions are necessary and yet arbitrary.

P.S. Seeing around corners.
 
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RedCowboy

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I'm not confused at all, I disagree with your terminology because it's wrong, the correct way to say it would be " all surfaces are reflective to some degree " not all surfaces are mirrors because they are not.

Where in physics does it say that all surfaces are mirrors verbatim ? It doesn't, all surfaces are reflective by some amount but that does not a mirror make.

Of course I understand energy and boiling water is colder than high pressure steam but boiling water is not cold, it contains less heat energy per unit and of course I understand that we couldn't see something that can't reflect some light, such as water or glass is transparent but still has some level/amount of reflectivity and that changes with the angle of incidence, so it's correct to say that all surfaces are reflective to some degree, but all surfaces are not mirrors, the bottom of my shoe is not a mirror, actually a mirror is defined as :
 
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steve001

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I'm not confused at all, I disagree with your terminology because it's wrong, the correct way to say it would be " all surfaces are reflective to some degree " not all surfaces are mirrors because they are not.

Where in physics does it say that all surfaces are mirrors verbatim ? It doesn't, all surfaces are reflective by some amount but that does not a mirror make.

Of course I understand energy and boiling water is colder than high pressure steam but boiling water is not cold, it contains less heat energy per unit and of course I understand that we couldn't see something that can't reflect some light, such as water or glass is transparent but still has some level/amount of reflectivity and that changes with the angle of incidence, so it's correct to say that all surfaces are reflective to some degree, but all surfaces are not mirrors, the bottom of my shoe is not a mirror, actually a mirror is defined as :
Oy!
 
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Encap

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Perhaps I did not asked clearly enough, trying to rephrase..
I meant the power of reflected energy at different wavelengths when directed at the same angle
The deep IP and UV are absorbed heavily by water, however the blue and green penetrates quite well (which means they are much less absorbed by water)
So the question is, given the same incident angle, how the amount of reflected energy dependent on the water absorption properties of wavelengths?

If the particular wavelength gets absorbed heavily by water (deep IRfor instance), does that mean the reflected energy will be less ?
Or the opposite is more likely - the less absorbed wavelength penetrates deep into water to less energy left to get reflected off the surface ?
Please make a Welcome post in the Welcome sub-forum, especially if you want member to respond to a very complicated question that has no real stated purpose.

Some almost meaningless questions. Obviously if light is absorbed or transmitted by water or anything else it is not reflected by definition of the words/concepts/terms.

How and to what extent that happens? Is very complicated thing, not a simple matter--too complicated to answer in a LFP post--see article below.
See: http://www1.lsbu.ac.uk/water/water_vibrational_spectrum.html#d

PS "Pure water is colorless, whether in a vapor, liquid, or solid phase. That is, the molecules of water cannot absorb visible light and so cannot yield colors like dye or pigment molecules. "
See: https://www.uu.edu/dept/physics/scienceguys/2001Feb.cfm
+ There is more going on than absorption and reflectance---there is also transmission of light.
See: https://light-measurement.com/reflection-absorption/
 
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Pret

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Thanks a lot
Looks like indeed quite complicated matter
Recently have been envolved in certain project that combines LED/laser beaming into water and came across the graph of wavelengths absorption by water per 1/m, so got curious how the absorption influence the amount of energy reflected off the surface (per given incidence angle)
 







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