Jim, I think you often don't word things carefully enough. The resulting discussions get pointless in a big hurry.
The optics used for focusing are NOT mirrors, they are (hopefully) transparent at the frequency under use. A mirror on the other hand is required to be OPAQUE with respect to transmission, we want full, 100%, reflectivity. That means that every photon that hits that mirror interacts, loses some energy, and gets re-emitted. ^^^^^^^^^^^^^^^^^
Are you implying that the wavelength for incident photons changes upon interaction with the mirror?
The energy loss at the mirror is lost photons not altered wavelengths. The lost photons have varying fates.
You stated that every photon interacts, loses energy and is re-emitted.
I think the reflected beam has the same wavelength as the incident beam. Your blurb about absorption and cascades is only true for some fraction of the lost photons that constitute the inefficiency of the mirror. Others have a different fate. Maybe that's what you meant but you did say "every photon." And here's an exchange with Tim :
At 6:30 PM -0500 7/24/01, Jim Choate wrote:
And these are reasonably low power lasers...
http://www.ee.surrey.ac.uk/SSC/IJSSE/issue1/unwin/unwin.html
The simple fact is that the thermodynamic impact of a laser beam that is several feet across and emitting more photons than the surface of the sun will not be easy to reflect unless immense cooling is taken. Cost/weight factors alone argue it in the negative.
"More photons than the surface of the sun" for HOW LONG?
A minute? A second? A millisecond? A microsecond?
You confuse fluence with flux, a classic mistake.
(A pulse "brighter than the sun" but lasting only milliseconds will have far less heating effect than other flux level pulses lasting longer. Calculations matter. And, yes, I used to do these calculations when I was refuting Kosta Tsipis' calculations of the late 70s. Fluence matters.)
--Tim May
The sun produces shitloads ( check your CRC Handbook for conversions between the shitload and more familiar units ) of power : http://seds.lpl.arizona.edu/nineplanets/nineplanets/sol.html says 386 billion billion megawatts If we know the spectral characteristics of the sun ( the black body spectrum perhaps? ) we could come up with a photon count. I'm not sure whether you mean to talk about photon counts and adjust the power and wavelength variables or you really mean to discuss something that operates somewhere between IR and UV. Let's assume the latter. It is after all a LASER. You did say "surface of the sun". To me that means integrate over 4 pi. 3.86E26 W regardless of the radius. I doubt if anyone has made a laser that operates at that power level even for one fs. Let's try the other approach... The power output from the sun is about 1350 W/m^2 as measured here. Maybe that was what you meant as a reference power level. Let's see, 1350 W/m^2 -> 1.35E-3 W/mm^2 so a 1 mW laser with a beam area of .74mm^2 is "as bright as the sun" at least in terms of gross energy density. That disregards spectral effects. Not too tough to be "brighter than the sun". I don't think you could even light a bucket of gasoline 1 m away with it no less knock down a rocket. It's also pretty easy to handle with a basic mirror. I'd say that's a pretty wussy power level for something that needs to melt a rocket in flight. Focussed to a spot that is 1/1000 the area of the parent beam it starts to get interesting but let's see you hold that spot steady from a 747 in turbulence long enough to burn a hole in a nice shiny casing going 8000kph 200km away. So if we're going to discuss physics let's do it with a bit of care. Maybe it will be more interesting. I'm no expert but I'm willing to try. Yawn, Mike