On Thu, 30 Jun 2005 13:43:13 -0500, Cecil Moore
wrote:
An angle of incidence of 30 degrees
is irrelevant to this particular example.
Changing the question to suit the answer, Hmm? As if it mattered!

How much power is reflected from Surface A (first incidence)?
The reflectance is 0.01, so 1%. First incidence 0.01 watts
depending upon the polarization r¬ = 1.583% and r|| = 0.5485%
Splitting the difference (1.066%)
0.0107W
or
0.9893W @ 24° available at the next interface
How much power is reflected from Surface B (first incidence)?
The reflectance is 0.01, so 1%. First incidence is 0.0099 watts
depending upon the polarization r¬ = 1.3381% and r|| = 0.7099%
Splitting the difference (1.024%)
0.0101W
or
0.9792W @ 19° available at the next interface
How much power is transmitted through all interfaces?
One watt net to the "load".
Already provided as 0.9792W

**********************************

Since you couldn't answer the original question, let's explore how
accurate your answer to your own question was:

How much power is reflected from Surface A (first incidence)?
The reflectance is 0.01, so 1%. First incidence 0.01 watts
depending upon the polarization r¬ = 0.9999% and r|| = 0.9999%
Splitting the difference (0.9999%)
0.0100W
or
0.9900W @ 0° available at the next interface
How much power is reflected from Surface B (first incidence)?
The reflectance is 0.01, so 1%. First incidence is 0.0099 watts
depending upon the polarization r¬ = 0.9998% and r|| = 0.9998%
Splitting the difference (0.9998%)
0.0099W
or
0.9801W @ 0° available at the next interface
How much power is transmitted through all interfaces?
One watt net to the "load".
Already provided as 0.9801W

But, hey, what's 2% error in a conservation of energy equation? You
can prove anything (especially your absolute proofs) if you simply
discard precision. Pons and Fleishman proved cold fusion by throwing
away fewer digits than you.

Well, for 1W of light and presuming cancellation (you cannot achieve
full cancellation); this leaves 100µW of light reflected from a
non-reflecting layer - which is quite bright.

So, energy is conserved, and there is no such thing as complete
cancellation.

By the way, the math is available from:
Hecht, Eugene, Optics, 2nd Ed, Addison Wesley, 1987
or perhaps you should invest in:
Hecht, Eugene, Optics, Schaum's Outline Series, McGraw-Hill ,1975