# How big should a lens of a telescope be so we can see the American flag on the Moon from the Earth's surface? [duplicate]

How big in diameter should the lens of the imaginary telescope to be so we can see the American flag from e.g. some observatory in Hawaii?

• We wouldn't only need a huge telescope; to compensate for the Earths atmosphere we'd need adaptive optics that are equally beyond anything that's been built so far. – Dan Is Fiddling By Firelight Sep 26 '13 at 14:07
• @DanNeely do you know about some theories or ideas that could somehow "fake" that and make us see that without having so big lense? Are there some ideas or hyphoteses like that ;) ? – Derfder Sep 26 '13 at 18:35
• Interferometry can get equivalent resolution to a single huge telescope with several smaller ones spaced as far apart as the single huge one. But if I understand correctly, what is done is to take an image of the interference pattern and use a computer to figure out what input would create the pattern; it doesn't give an image directly. – Dan Is Fiddling By Firelight Sep 26 '13 at 18:40
• For radio interferometers they can record the signal with precise timing info and perform the interferometry with a computer later, but for optical interferometry it is not possible to record the light signal with sufficient precision because the frequencies are just too high. Optical interferometry require the actual light rays to be brought together to construct the image. – Philip Gibbs - inactive Sep 26 '13 at 21:52
• possible duplicate of Observing lunar lander and footprints on the moon? – user10851 Sep 27 '13 at 1:06

The angular resolution of a telescope is approximated by the formula:

$$\sin \theta \approx\theta\approx1.220\frac{\lambda}{D}$$

So, if we know the angle, we can calculate the diameter $D$.

The rest really depends on how big the flag is and where on moon it has been planted. Assuming 50$\text {cm}$ as the diameter of the flag, and assuming the flag is planted somewhere perpendicular to earth's view line, and also doing it when moon is in its perigee($d=363,295\text{km}$); the diameter would be(for visible blue light):

$$D\approx\frac{1.22 \lambda}{\theta}\approx\frac{1.22 \times450\text{nm}}{\frac{50\text{cm}}{363295\text{km}}}\approx400\text{m}$$

If you don't trust my numbers, you may look at this link.

• I guess you'd need a bigger lens to clearly distinguish between a flag and a boulder. Much bigger to distinguish US, Russian and Chinese flags. – RedGrittyBrick Sep 26 '13 at 11:58
• Fair point @RedGrittyBrick . – Ali Sep 26 '13 at 12:32
• Or if you required, say, a $100 \times 100$ pixel image to "see" the flag... – DJohnM Sep 26 '13 at 19:02

We are fortunate that NASA released some photos from the Lunar Reconnaissance Observatory (LRO) last year! Even from a mere 31 miles (50 km) up, it is impossible to see the actual flag, though we can see its shadow.

Assuming the resolution is limited only by diffraction, e.g. not by atmospheric turbulence, then the angular resolution is given by:

$$\sin \theta = 1.22 \frac{\lambda}{D}$$

where $D$ is the telescope aperture. If the distance is $d_m$ and the size of the flag (or the feature on the flag you want to resolve) is $d_f$, then $\sin \theta = d_f/d_m$ so:

$$d_f/d_m = 1.22 \frac{\lambda}{D}$$

or

$$D = 1.22 \lambda \frac{d_m}{d_f}$$

If you put the size of the flag as 1 metre the corresponding aperture size is about 240m.

You dont need a single large mirror to resolve small objects. You can use two mirrors separated by a distance as an interferometer. With adaptive optics it is possible to overcome atmospheric distortion.

The Keck pair of telescopes use this technique to get an effective mirror size of 85m, equal to the distance between the two mirrors. The resolution achieved is 5 milliarcseconds (mas) at 2.2 micron wavelength. (The VLT array of four telescopes resolves 3.5 mas)

The flag on the moon subtends an angle of 0.5 mas, so if the resolution and adaptive optics could be scaled up it should be possible to see the flag with two mirrors separated by something like 850m. This is more than the other answers but I think it is more realistic taking into account the adaptive optics.