Is there a way to photograph the very edge of black hole? According to Professor Stephen Hawking, black holes emit radiations known as Hawking radiations. The bigger the black hole, the fainter the radiations. That means, I personally believe, if we built large enough telescopes, we could see the very edge of black holes.
Does this mean, in the near future we can take a photo of a not-so-black hole? If so, how large a telescope to be to do this job?
 A: Hawking radiation is such a miniscule effect we can be sure we'll never detect it for a real astrophysical black hole.
The Wikipedia article gives some numbers: For a black hole of the mass of the Sun, the power emitted in Hawking radiation amounts to $9\times10^{-29}\ \mathrm{W}$. Even if all this energy were converted to visible-light photons ($4\times10^{-19}\ \mathrm{J}$ each), that still amounts to about a single photon emitted once per century. And larger black holes are worse, since the total power emitted scales with the mass as $M^{-2}$.
You could build a telescope extending from here to Alpha Centauri using most of the matter in this sector of the galaxy and still not detect a single photon's worth of Hawking radiation from the nearest black hole in your lifetime.
However, if you're willing to accept other mechanisms for energy production, the situation is much better. Material falling into a black hole will tend to emit all kinds of energy classically, without needing to call upon esoteric quantum-field-theory-in-curved-spacetime concepts.
The black hole at the center of our galaxy must have a radius of something like $12\times10^6\ \mathrm{km}$. Currently, radio telescopes have resolved features of its emission down to scales of $44\times10^6\ \mathrm{km}$, or about the radius of Mercury's orbit. The Event Horizon Telescope is one concerted effort to get even better angular resolution in the coming decade or so.
A: The light due to Hawking radiation will only ever be detected from very tiny black holes. The Hawking radiation scales as the inverse square of the black hole mass, but the radiation causes the black hole mass to decrease. This causes accelerated emission, such that all tiny black holes will go through a phase of emitting all their rest mass  as $10^{22}$ J of radiation in their final 1 second before evaporation. It is possible that such flashes could be detectable.
However, stellar mass black holes do not evaporate and the Hawking radiation remains undetectable.
Another problem is that radiation emitted just above the event horizon would be red shifted to such an extent that a telescope could not see it.
Thus a black hole will be black, but if it is an accreting black hole it will be surrounded by a disc of accreting matter that is likely hot and emitting copious radiation. If so, it may be possible to "see" the black hole based on its dramatic distortion of the light from the disc. Or, if it were not accreting it would produce a dramatic distortion of the background starlight. Just google a picture of the black hole from the movie "Interstellar" to see an approximate example. This distortion is at a scale just a bit bigger than its event horizon.
To arrive at such a picture would need a telescope capable of resolving the event horizon, of radius $3 M/M_{\odot}$ km.
The nearest black hole could be only of order 10 parsecs away (if we could find it) assuming that most 20-plus solar-mass stars end their lives in this way. The angular size of the event horizon (assuming a 10 solar mass black hole diameter would be $2\times 10^{-13}$ radians. A telescope's resolving capability is given approximately by $\lambda/D$, where $\lambda$ is the wavelength it works at and $D$ is the telescope diameter.
Thus to resolve an (as yet undiscovered) nearby 10 solar mass black hole at 10 pc requires an optical ($\lambda = 500$ nm) telescope of diameter 2500 km.
A better bet would be the black hole at the centre of our Galaxy, which should have an event horizon of angular size about $10^{-10}$ radians, requiring an optical telescope diameter of a mere 5 km!
