The article mentions 13.5nm. Is 19nm the wavelength used to blast the tin that re-emits as the the 13.5nm EUV source? Is that where the power is going to result in a few watts at the printing surface?
Second question, the loss of power is due to two factors: 1) Conversion efficiency of CO2 laser pulse to EUV light pulse. This is in the single digit range as a percentage. 2) Transmission loss of EUV from light source to wafer. EUV mirrors reflect in the ballpark of 50% of the light, and there are many of them inside the scanner, so you have .5 * .5 * .5...