At what direction does a ray reflect at a 90 degrees corner? Say we pointed a laser that is projected into a $90^{\circ}$corner of a mirror made by perpedicular walls.

I know that you cant use tangent since a derivative of boundary of mirror would not exist at a corner. But this didn't convince me that a reflection could not exist.
So I imagined projecting a incident ray of the same slope as close as possible to the corner.

I thought that the ray of incidence would be in the same as the ray of incidence except the reflected ray moves in the opposite direction.
 
However, I am unsure if this is correct as I do not have a corner mirror. I'm also not sure if a reflection can exist in a corner.
So am I correct or do laws of physics state otherwise? And if I am not correct where would the relfected ray project?
 A: I think you are perhaps confusing reflection at a corner with a "corner reflector" (https://en.wikipedia.org/wiki/Corner_reflector).  The latter is a type of mirror which reflects light back to where it came from, as shown in your diagrams. The reflection comes from the 3 perpendicular mirrors, rather than at the join between them.
There is no "specular" (mirror-like) reflection exactly in the corner formed by 3 mirrors, or along an edge formed by 2 mirrors. You will not see an image in such a corner or edge no matter what direction you look at it.  Light which strikes at or very close to the corner or edge will be scattered or diffracted in a number of directions, because of the wave nature of light.  Scattering depends on the details of how the corner is made, what imperfections it has, and the wavelength of the light. This is "diffuse" reflection, which occurs from microscopically irregular surfaces.  If the join is very well made you may not notice by eye the actual corner or edge in the reflected image (unless you magnify it).  Otherwise, if you do notice anything it will just be a thin dark line or spot, or (in magnification) a blurring of the image.
I think that your guess is wrong : a ray pointed into a corner will not be reflected directly back upon itself, even if the join is perfect on sub-wavelength scales.  If that were the case you would not be able to see the edge or corner at all.
