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Physics help (again) please - energy given up at reflection - Page 4

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Posted by Morris Dovey on June 26, 2009, 3:25 pm
Jean Marc wrote:

Bonjour et merci! I did find several articles. It appears that for
direct sunlight, [c] random (unpolarized) is the practical answer - and
now I'm trying to decide if that plays a significant role in this project.

The problem with learning new things is that every answer creates at
least three new questions. :)

Morris Dovey
DeSoto Solar
DeSoto, Iowa USA

Posted by daestrom on July 2, 2009, 8:27 pm
Morris Dovey wrote:

AIUI, when light is reflected at low angles, certain 'polarities' are
reflected better than others.  This is why Polaroid Sunglasses work well
against such glare.  THe light reflected from road surface, etc.. is
only the components that were of a particular polarization.  By having
the lenses in the glasses oriented 90 degrees to that, almost all the
directly reflected sunlight is blocked, reducing the glare.

So if you're using shallow angles of reflection, this might somehow be


Posted by Morris Dovey on July 2, 2009, 9:04 pm
 daestrom wrote:

I think for the moment I don't care - which is probably a Good Thing :)

More later, after the filing...

Morris Dovey
DeSoto Solar
DeSoto, Iowa USA

Posted by Frogwatch on July 14, 2009, 2:09 am
In general, when light is reflected, Reflectivity is a function of
wavelength, angle of incidence and material properties.  For a
material that partially transmits light, the proper relationships are
given by the Fresnel equations (google em).  This probably does not
apply if your reflector is good or for most metal surfaces.  In these
cases, you might find tables of reflectivity in the Handbook of Chem
and Physics or can look up reflectivity of the material surface.
These tables generally assume nearly perpendicular rays to the
surface.  As the angle of insidence goes up, the reflectivity goes up
just as predicted by the Fresnel equations.  Generally, the wavelength
does NOT change on reflection.
Light that is not reflected is either absorbed, or transmitted into
the material OR is scattered from the surface although scatter is a
special case of reflection.  Light that is absorbed has its energy go
into the motion of the elctrons of the material.
Consider a beam of light being reflected back and forth betwen two
parallel plates.  Assume the reflectivity is 95% which is good for
most materials.  After 20 reflections, the fraction of energy left in
the beam is then .95 raised to the 20th power or just under 36%
Most light absorbers are rough surfaces so as to scatter light into
large angles where it has another chance to be absorbed by the rough
surface.  Consider a metal surface, a highly polished surface refelcts
well but a rough one scatters the light and absorbs some too.  A very
mooth surface of carbon is highly reflective whereas a rough carbon
surface is highly absorbing.
The scattered amount of a beam of light is given by:  i=Iexp(-4*pi
(sigma*sin(theta)/lambda)^2)  where sigma is the roughness of the
surface in the sane units as your wavelength lambda.  Theta is the
angle WITH RESPECT TO THE SURFACE, not the angle of incidence.

Posted by Morris Dovey on July 14, 2009, 10:00 am
 Frogwatch wrote:

Thank you! One of the things I was after was identification of the
variables, and you've filled that in rather nicely. If "material
properties" includes more than smoothness/roughness, then I still have
digging to do, but this all is a good start.

Ok - will do. Thank you again because I wouldn't have known to look
without your help.

I've been looking at metal and ceramic surfaces, but will still do the
reading on the Fresnel equations.

The info I've found so far is for perpendicular rays, and I'm going to
need to find some way to derive at least a close approximation of
reflectivity for other angles.

Ok - I'm taking your word for the first, and I've observed the second.
Now I'm beginning to crave knowledge of /why/ - and trying to not let
myself become distracted... :)

Makes sense to me.

Right - my little software model says 35.8485922% - and this is
essentially the game I play with my flat panel with "stacks" of formed
reflecting black aluminum ribbons. The surfaces aren't parallel (because
they're curved), but the geometry guarantees a minimum number of
reflections over a useful range of sun azimuths.

Hmm - I think I need to learn more about how to quantify and model this
behavior (and about how to economically produce surfaces to match
calculated roughness/smoothness values).

I suspect there may be a optimum trade-off between reflectivity (and
number of reflections) and absorptivity (is that a word?) of the surface...

I understand the relevance (and don't have any difficulty accepting the
correctness) but I think I have a bit of studying to do before I'll be
ready to apply this usefully with any degree of confidence.

Again, thanks for your time and effort to help me understand more about
this stuff. It's very helpful!

Morris Dovey
DeSoto Solar
DeSoto, Iowa USA

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