Dear Rootosaurus, when you're looking at an image of a chair behind you in a flat mirror, then you're observing the so-called virtual image of the chair. If the mirror's surface is located in the $x=0$ plane and the coordinate of the real chair is at $(x,y,z)$, then the virtual image of the chair is at $(-x,y,z)$.
However, the light rays coming from the real chair that are reflected by the mirror and that ultimately end up in your eyes have exactly the same directions as the light rays of a hypothetical chair that would be actually located at the point $(-x,y,z)$ behind the mirror. So geometrically, you can't really distinguish a reflection of a chair that seems to be located at $(-x,y,z)$ from a real chair that is located at $(-x,y,z)$ and that you're observing through a window without a mirror. The geometry of the light rays is identical. That's why the concept of images is so useful.
In particular, myopia means that one has some trouble to observe distant objects. Distant objects - imagine a distant point - have the property that they emit light rays that are nearly parallel. The further an object is, the more parallel its light rays look when they arrive to your eye. However, the lenses in your eyes have to convert these parallel mirrors into converging mirrors - so that all the light rays coming from the distant star end up at one point of the retina.
Myopic eyes are good in converting "steeply divergent" light rays from nearby objects to converging ones, but they're doing too much of a good thing. When you get too parallel light rays, myopic eyes make them converge too much, too early - the intersection will be inside the liquid in your eye. Hyperopia is the opposite disorder in which eyes make the light converge less than is needed. But what's relevant for your question is that the virtual image of the chair at $(-x,y,z)$ "emits" the same excessively parallel rays as a real chair at the same point, so a myopic eye will have the same trouble seeing it. After all, it shouldn't be paradoxical: the total distance that the light rays have traveled includes the distance of you from the mirror as well as the distance of the mirror from the real chair - because the colorful photons ultimately came from the chair and they were just reflected, not created, at the plane of the mirror.
This is easily explained if you use the method of images. The man sees a mirror image of himself which has an apparent location reflected about the plane of the mirror, i.e. equal distance from the mirror but in the opposite direction. Hence the apparent distance from man to his image is always twice the distance from the man to the mirror, hence the 50% scale.
Best Answer
Edit: Reading other people's answers, I forgot to mention I assumed a flat mirror.
Excellent question, but the answer is no. The reason is because the object (in the strict optics meaning) in the case of the photograph is actually on the paper whereas in the case of the mirror it is still at the same place, far behind: the rays of light coming from it are reflected on the mirror but still require tuning from the eye muscles to get the focal point right (the eyes are in fact sort of a tunable lens with a detector similarly to cameras).
I would recommend you draw a diagram of the light rays coming from the edges of the objects into the eye for both cases to get a better idea.
Example
As an example, consider the following: if the mirror has dust on it, try looking first at the dust and then at an object that the mirror displays behind you. You will have to strain your eyes to focus on the dust, which proves that what the mirror displays is not actually at its surface (the image, again in the optics strict sense, is behind the surface of the mirror).