Cecil, W5DXP wrote:
"The reflection from surface "A" is canceled by an equal magnitude and
opposite phase reflection from surface "B:."

Is this not analogous to what happens on a short-circuited 1/4-wave
stub? The hard short reverses the phase. That, combined with travel to
and from the short, produces a total phase rotation of 360-drgrees.

The result is that the open end of the short-circuited stub, the
incident voltage is in-phase and of the same magnitude (no stub loss) so
that no current flows between the incident and reflected sources.

It is as if one connects identical battery cells ib parallel. The
impedance is, in effect, infinite between sources of identical voltage.

Optical experts must have siezed upon the opposite of this somehow.
Their quarter-wave must have ben terminated in the equivalent of an
open-circuit. This 1/4 wave would accept 100% of light presented at its
surface, or would it need to present 377 ohms at its surface?

I am ignorant of optics and find the analogy difficult to understand.

Best regards, Richard Harrison, KB5WZI