From my (exceedingly limited) knowledge on lithography, the exposure is energy-dependent, not power dependent. Yes? Even with a 50% attenuation, doubling power or doubling exposure time negates this, yes?
I too don't have much knowledge with lithography (hence why I'm asking question in a laser/optic forum) but, yes, exposure is energy dependent. The datasheet for the photoresist specifically tells the optimal energy exposure to be in the range of 50-200 mJ/cm2.
I could increase the laser power to output more light energy to compensate the lens attenuation, common sense tells me that this should work in theory... but then again who knows.
Here is the link for the UV sensitive Dupont photoresist: