I quoted this from Sam's laser FAQ Sam's Laser FAQ - Laser Safety

Standard Sun:


Maximum intensity of sunlight at ground level (directly overhead, no smog, etc.) = 1 kW/m2 or 1 mW/mm2.

Assuming pupil diameter is 2 mm (i.e., radius of 1 mm), the area is approximately 3 mm2. So, the power of the sunlight through the pupil = 3 mW.

Focal length of eye's lens = approximately 22 mm. Angular size of Sun from Earth = 0.5 degree = 9 mR. Thus, diameter of image formed = 22 mm x 9 mR = 0.2 mm and the area of image = 0.03 mm2.

The intensity of the Sun on the retina (Power/Area) = 3 mW/0.03 mm2 = 100 mW/mm2.
Typical 1 mW HeNe laser (or laser pointer):


Power (P) = 1 mW, wavelength (l) = 633 nm, radius of beam (w) = 1 mm, focal length of eye (f) = 22 mm. So, the diameter of spot = (2 x f x l)/(w x pi) = 9 x 10-3 mm and the area of spot = 6 x 10-5 mm2.

The power density of the HeNe laser on the retina is 1 mW/(6 x 10-5 mm2) = 16,667 mW/mm2 = 16.667 watts/mm2.
So the 1 mW laser has the potential to produce an intensity on the retina 167 times that of direct sunlight!

granted, a plant is many times the size of a retina, and I'm pretty sure focal length wouldn't play into it, but i think even after being dispersed over an area say .3M squared a 100mw 780nm laser could produce enough far red to speed up your nights especially if you start adding more lasers since it contains no red, and not any really high infra-red that just makes heat, and looking at a solar chart, of watts energy at ground level per nm wave it looks like the sun is delivering about 1 watt per nm at the far red levels