Re: Bandgap of 'white' solids?

JAH wrote:

Hi,

Under 'white visible light' illumination, my understanding is that
colour is percieved as the combined effect of photons that are
reflected instead of absorbed. So an ideal matt black surface absorbes
everything, whereas a matt white surface reflects everything.

That last is a dangerous statement. The holes are only reflective if
they are dielectric with zero absorbance. The blackest surface you've
ever not seen is a lightly bolted stack of de-oiled double-edged razor
blades. The sharp side (all conductive mirrors at an extreme acute
angle) is used as a laser beam dump. The porosity must match the
wavelengths of interest (e.g., etalons).

A really white white is a light-scattering white dielectric surface
paved with *solid* high refractive index glass solid microspheres. It
works for a full moon. A really black black is a black surface (e.g.,
dispersed carbon black or copper chromite) paved with *hollow* glass
microballoons. Cf: Martin Black.

A
transparent surface transmits most of the light, because the band gap
is so large that only high-energy photons (UV range) have sufficient
energy to excite electrons from the valance band to the conduction band
for absorption.

Boundary conditions (refractive indices of the two media plus angle of
incidence) determine reflection losses.

Conversely, the band gap of a 'black surface' is so
small that virtually all photons have sufficient energy for absorption
to occur. Therefore, what type of band gap does a 'white' surface have,
or does it matter? Is it more of a specular surface phenomenon related
to light scattering?

White is caused by non-specular, diffuse scattering. Consider red
into pink. Pink is not on the color chart. Merely diluting red in
water never gets you to pink.

--
Uncle Al
http://www.mazepath.com/uncleal/
(Toxic URL! Unsafe for children and most mammals)
http://www.mazepath.com/uncleal/qz3.pdf
.

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