$\let\l=\lambda$
I'm sorry you received no answers till now. Yet your question is far
from trivial. I hope what follows may remedy, at lest partly.
It might help to begin with a "big" mirror and then think of reducing
its size. Until mirror is much larger than wavelength it will behave
as geometrical optics says. Assume you send on the mirror a wide beam
of parallel rays. You'll see a reflected beam the same size as mirror
is and beyond the mirror a shadow, also sized and shaped like the
mirror.
If mirror is still larger than wavelength, but not very much,
diffraction effects will begin to be felt, both in reflected beam and
in the shadow. You know that diffraction will involve a region of
angular size $a/\l$, where $a$ is mirror's size and $\l$ is
wavelength. Diffraction will cause an alternation of light and dark
along the borders of reflected beam and of geometric shadow regions. Note that if you put a screen at a distance $D$ of the order $a^2/\l$ the diffraction zone will cover the full reflection and shadow regions - diffraction is dominant if $D \ge a^2/\l$.
If mirror's size is comparable with $\l$ then $a^2/\l\simeq a$ and
geometric optics is no longer applicable, even as a rough
approximation. Actually the diffraction region is so large that there
is no sharp distinction between backward reflection and forward
shadow: you may pass from one to the other in a continuous way.
This pattern is still more apparent if the mirror is smaller than
wavelength. In that case physicists speak of "scattering" rather than
of diffraction.
Could all this be calculated? In principle it could, but it's a
formidable task. As far as I can understand the only treatable case is
the one of normal incidence on a mirror of simple form, e.g. circular.
I've never seen such calculation but I'd bet it was undertaken around end of 19th century, when wave propagation was one of the main topics of theoretical research. I have some idea of how the calculation could be done - not easy but feasible. Unfortunately I have no time to engage in it. But if you need some further clarification don't hesitate to ask.
