Re: [ATM] What is the "dielectric" in dielectric coatings?

["Jerry Reddell" <jerry.p.reddell63@sbcglobal.net>]

If I understand your scenario, it is very simplified  way of looking at it,
but, as with many things, the real answer is in the small details. It is
correct to a point but could lead one to the wrong conclusion, and primarily
has to do with the assumptions 1 and 2. Also, you only used one reflection
from the metal and incorrectly let it all exit through the dielectric to air
boundary. Some of this reflected wave is again reflected from the front of
the dielectric back toward the metal. So there are really an infinite number
of reflections back and forth between the metal and the dielectric surface.
A portion of each gets through the dielectric-air boundary and adds to the
overall systemreflection.  Calculating the reflections and summing an
infinite number of waves becomes a massive set of computations.

The reflection at the boundary of two different materials is dependent on
the electrical (or optical) properties of BOTH materials. In the RF world we
use the RF impedance (a complex value) of the two materials that determines
the reflection. In turn, the RF impedance of a material is dependent on the
relative dielectric constant(a complex value) of the material. Because the
dielectric material has a different dielectric constant than air, the
reflection at the boundary of the metal and the dielectric is different from
the reflection at the metal and air boundary.

A single dielectric layer may have a seemingly minor affect of the
reflection.  The addition of a properly selected different dielectric layers
creates a larger overall affect..

A factor however is the distance from the metal plate to the front surface
of the dielectric. This factor of the problem is where your scenario is
somewhat correct. --The part that says the reflections are in-phase. ---
This distance causes the overall system to behave something like a
transformer, the impedance seen at the the  dielectric - air boundary
surface is such that the reflection coefficient is different than either the
metal-air or the dielectric-air boundaries.

Sorry for the long post


Jerry  Reddell


-----Original Message-----
Of Vladimir Galogaza

But on a second thought.

If reflectivity of metal is (arbitrary) 90% meaning that
without coating 90% of input intensity will be reflected, 10% absorbed.

If there is coating with reflectivity 4%

4% of input intensity will be reflected by coating (arbitrary).
96% will enter coating and fall on the metal
90% of this 96%   will be reflected by metal this amounts to 86.4% of
        input intensity on  the coating
if in phase this 86.4% will be added to 4% which is total 90.4%.

Assumptions
1) There is no additional reflection on the coating-metal boundary
of incoming light
2) same  on the coating-air boundary of the outgoing light.
3) coating has zero absorption (100% transparent).

 0.4% enhancement in reflectivity comes from less light entering metal and
absorbed there .

Multi level coating can increase this gain by reflecting more light
before it reaches metal.
Actually it is possible to avoid metal completely (dielectric mirror),
but the price will be narrower spectral band and greater
dependency on input angle due to interference condition.

Does this sounds reasonable to the public on the list?

Regards
Vladimir.


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