Here's an answer I wrote for a question on Space.SE, but which applies equally well here. Let's talk about the Hubble space telescope, which would be much better at observing these comets than any ground telescope:
Using the equation: (d / D) × c = φ
where d is the diameter of the Oort Cloud comet (some estimates put this number at an upper limit of 300 km for the diameter of a cometary nucleus), $D$ is the distance from the Oort Cloud to Hubble (0.3 light years, or 3×1015 metres – distance at which it is theorized there is the highest density of Oort Cloud objects), c is a constant (c = 206265) and φ is the telescope resolution.
So what resolution do we need to image an Oort Cloud object, 300 km in diameter, from 0.3 light years away? If we plug in the numbers we get:
φ = 2.06×10-5 = 0.00002 arc-seconds
The resolving power of Hubble is 0.1 arc-seconds, and is therefore useless at detecting anything below this angular size; Oort Cloud comets (although pretty big at an upper limit of 300 km) simply cannot be observed by the world’s most advanced space-based optical observatory.
How big would a telescope have to be? Well, from the same article:
But how big would an Oort Cloud observing telescope have to be to resolve a cometary nucleus 300 km wide at a distance of 0.3 light years away? Using the simple relationship R = 11.6 / w, where R is the resolving power (R = 0.00002/2; the reason for halving our resolving power is given by Phil), and w is the width of the telescope mirror, we rearrange to get:
w = 11.6 / 0.00001 = 1.16×106 cm = 11.6 km
As you can see, such a telescope would be huge.