The US lunar lander has a plaque that says "WE CAME IN PEACE FOR ALL MANKIND", in a successful effort to avoid sparking a global thermonuclear war. It’s 384 megameters away on average, although sometimes it gets as close as 356.4 megameters. Can you read it from Earth?
An interferometer, which is kind of a generalization of a telescope, distinguishes things by the phase of light reflected from them. A lens or telescope mirror is a device to adjust the phase such that light waves emanating from the same point will be focused onto the same point, and more importantly for the question here, light waves emanating from different points will be focused onto different points. It uses the difference in phase across the width (“baseline”) of the interferometer to identify each point.
To be clearly distinguishable by interferometer, the phase difference of two points ought to differ by a significant fraction of a wavelength. We can take half a wavelength as a convenient number. And let’s figure we can’t conveniently use radiation any harder than, say, 300 nm wavelength, for example because it’s hard to focus or because the Sun doesn’t emit it or it doesn’t penetrate our atmosphere.
I don’t know how big the letters on the plaque are, but let’s suppose that centimeter resolution is good enough.
So how big of an interferometer do you need? Does it fit on Earth?
The Euclidean sum of 384 megameters and 1 centimeter is a bit tricky to calculate. But you can easily calculate that 1 centimeter over 384 megameters is about 30 picoradians, and 150 nm is 30 picoradians at a radius (or baseline!) of 5 km. So even if the letters are a bit smaller than I imagine, you can read them from Earth with optical interferometry.
You can do better, though! You can laser-print a new plaque by selectively melting the lunar soil from Earth with phased-array lasers, thus changing its color! If you require a 50km baseline, you can get down to one-millimeter resolution. You need high enough power to melt it, not just slightly warm it; but exactly how much power you need goes down as the resolution improves, but it depends on the thermal conductivity of the soil, its density, its heat of fusion, its melting point, the depth of penetration of light into it, and its specific heat. Also, to estimate how fast this happens, you need to know its albedo and the spectral selectivity of its emission spectrum.
I found some information about lunar soil physical properties online. Lunar soil particles are very fluffy, with half a square meter of surface area per gram. Specific gravity of the solid mass of regolith particles is about 3.1 g/cc; porosity is typically 50%; consequently lunar soil is about 1.6 g/cc, although varying from 0.8 to 2.3, and perhaps even lower in the top surface layer. Thermal conductivity in the top centimeters of lunar soil was measured at 1.5 × 10⁻⁵ W/cm/° in the Apollo 15 and 17 missions. Its albedo is about 0.07.
I don’t know the spectral selectivity, heat capacity, heat of fusion, or light penetration depth of the lunar soil, but let’s estimate. The soil includes some components that would melt at lower temperatures, including many glasses that won’t have a heat of fusion at all, and it might be adequate to melt just those. But I'm going to use a somewhat more pessimistic analysis to make sure my conclusions are ironclad. Let’s suppose it has no particular spectral selectivity, melts at about 1000°; has a heat capacity similar to quartz’s — 65 J/mol/° / (28 + 2·16 = 60) g/mol ≈ 1 J/g/°; and has a heat of fusion similar to quartz’s — 9 kJ/mol ≈ 150 J/g.
The lunar surface can get as hot as 220 K during the day even before we start shooting lasers at it.
So we need, hypothetically, to heat the surface of the regolith to 1000°, sucking up about 780 J/g, and then melt it with 150 more J per gram. At 1000° a black body soil would be emitting 57kW/m², so that’s probably about what we need to illuminate it with in order to keep heating it up once it’s already at that temperature. But how much actual power is that?
It depends on the area you can illuminate, of course — the tighter you can focus the beam, the easier it gets. Above we posited a square-millimeter focus, which is 10⁻⁶m², at which point you’d need 57mW — an eminently feasible number! As long as you can focus 57mW onto a one-millimeter square in a vacuum, it will eventually heat up to 1000°. How fast?
1.5 × 10⁻⁵ W/cm/° at this temperature is about 1.5 × 10⁻² W/cm, which is to say, 1.5 mW mm/mm² — we can expect on the order of a milliwatt to escape to a depth of on the order of a millimeter through an area on the order of a square millimeter. None of this is very precise because of course the planar focus spot gradually morphs into hemispherical shells of heat spreading into the regolith, so really at the depth of a millimeter you need to be considering about two square millimeters of almost-hemispherical shell, but it’s close enough to show that the vast majority of heat will be lost radiatively rather than conductively. You might need 58mW or 59mW or 60mW to reach equilibrium, but really you want to be far from equilibrium, like at least 2× and ideally over 10×.
So supposing we’re focusing 600mW and up on this square millimeter on the lunar surface, losing 7% of it to reflection, two or three milliwatts to conduction, and 60mW to thermal reradiation, leaving 500mW, focused on a millimeter-sized spot that’s about a millimeter deep, with a density of about 1.6 g/cc (thus about 1.6 milligrams) and a heat capacity of 1 J/g/°. This gives us a temperature rise of about 300°/s, so it takes three or four seconds to melt the spot, so we can write on the moon at about 0.3 mm²/s with 600 milliwatts.
This should scale linearly with applied power. We should be able to do 3 mm²/s at 6 watts, or 30 mm²/s at 60 watts. Even 6 kW should be doable without any extra difficulty (remember, we're talking about a phase-locked phased-array ultraviolet laser system spread over a thousand square kilometers or so) and should allow you to print on the moon at 3000 mm²/s.
Lunar soil is accumulating at about 1½mm per million years, so whatever you write there might be readable for tens of millennia if daily dust redistribution doesn’t cover it up.
The total information capacity of the moon’s surface would be close to 1 bit per mm² if encoded in this way. The moon's radius is 1738 km, so the part facing us has space for on the order of 5 × 10¹⁸ bits encoded in this way, which is several hundred petabytes.
Calculating the bandwidth of this data storage channel is trickier.
Relevant to this is NASA N78-13420, “Analysis and Design of a High Power Laser Adaptive Phased Array Transmitter”, from 1978. They propose to use photovoltaic-powered terrestrial lasers, phase-locked to produce a phased array, to deliver 5 megawatts of infrared laser power to satellites, from a 6-meter transmitter to a 2-meter collecting aperture, for example to power orbital transfer maneuvers. They predicted overall power transfer efficiency of 53%, which seems inconceivably large to me given the generally low efficiencies of lasers, especially in the 1970s — and indeed, it turns out they’re talking about 53% of the light coming out of the laser making it to the satellite, and they suggest using an isotopically pure laser that wasn’t yet developed (and maybe still hasn’t been), plus six other areas of “advanced technology development required”, which I think all do exist now.
In the high-spatial-resolution phased-array field, there's a 1993 proposal for an OVLA (NASA N93-13583), or “optical very large array”, like the radio-telescopy Very Large Array, using optical heterodyning to get the equivalent of a very large aperture.
The recent DE-STAR proposal from Cal-Poly is the closest in spirit: it proposes a high-power phased-array laser to divert or evaporate asteroids and comets in danger of hitting Earth.