Okay:
[–]somedave 4341 points 23 hours ago*
Yes, you need to be careful with phrases like “a small amount”.
Mars is around 225 million km away at closest approach average distance. Lets say you have a 1W flashlight and aim it at Mars, the intensity very far away from this flashlight will drop off as the distance squared (also a little extra from absorption and scattering in the atmosphere). Without doing any exact calculations, if we assume scattering is negligible we can say the intensity that hits Mars will be larger than
I > 1W / (2 pi * (225 million km)2) ~ 3 × 10-24 W /m2
Mars has a surface area of 144.8 million km², so the power hitting Mars will be around
I * A/4 ~ 2.3 × 10-10 W
This isn’t a lot of power, but a single photon at optical wavelengths has an energy of around 3 × 10-19 J, so this is still billions of photons a second hitting Mars.
Edit: Lots of people are pointing out the beam divergence and scattering I ignored. Scattering I still don’t think is very significant, about a fraction 10−5 of the light will be scattered for every meter of travel, most of earths atmosphere is within 20 km of the surface so the intensity is reduced by a factor of around
I/I_0 = exp(-20000*10-5) ~ 0.8
which is a 20% loss and thus not significant. If you aimed the beam through more atmosphere or if you had a blue flashlight this gets worse, but never significant.
The beam divergence depends heavily on how wide a flashlight you have to start with, if you had something which is quite compact the divergence is worse than something with a large output. Most of the power is actually in a spherical segment which is, say, 30 degrees in size, where as my calculation assumed this was closer to 90 degrees. To compensate the intensity on Mars would be bigger by a factor of (90/30)2 = 9 ~ 10.