For a smooth mirror like surface, the angle of incidence equals the angle of reflection which is one of the rules of the classical optics. There are several models to explain the mechanism of this. Some of them are more or less described in the answers posted to this question: "Why is angle of incidence equal to angle of reflection?"
Wave theory explanation is that is due to constructive and destructive interference cancelling each other out in all other regions on the mirror except in a narrow region called 'specular region' where constructive interference dominates such that reflection is only observed as coming from this region per each incidence or reflection angle. This is also backed up by 'Feynmann path integrals' in Quantum Mechanics where timing differences in paths cause all other probability amplitude arrows in non specular regions to cancel each other out and only the arrows representing paths that contain the specular region remain.
My questions are
Does this interference happen in the matter at surface level (the way electrons oscillate and how they influence each other etc.) or en-route to the observer after reflecting off the surface.
What is the length of this specular region in terms of wavelength of light for a perfectly smooth mirror like surface and how this length changes with the distance of the path of the ray for the same incidence or reflection angle (for example how it changes with the distance of the observer from the surface for the same incidence or reflection angle)?
(If you are curious what kind of technical knowledge the asker of this question possesses, yes, I am a layman.)
