Yes, it is possible.
Consider a mirror in the shape of a closed hemisphere - that is, half of a concave spherical mirror combined with a flat circular mirror passing through its center.
If you are inside this hemispherical mirror, you will see a real image of yourself rotated 180 degrees around the axis of the hemisphere. Here's one way to show this is true: The flat mirror creates a virtual image of you which is your reflection in the plane (duh), and the spherical mirror takes that as an object and creates a real image which is the inversion of that through the center of the sphere. The composition of the reflection and the inversion is a rotation.
Now, for practical reasons you don't want to be entirely enclosed by the mirror (it would be hard to get in and out, and you'd have to have your own light source...), so you could build just a part of the hemisphere. But the part you built would have to contain both some of the spherical surface and some of the flat surface for it to work.
If you have trouble understanding the orientation of everything, here is a concrete description: The flat mirror is a circle in the $z=0$ (horizontal) plane, $x^2+y^2\le1$. The spherical mirror is a hemisphere described by $z>0$, $x^2+y^2+z^2=1$. You're standing inside so your hand is at a point near the axis, for example, (0.01, 0, 0.2). What the mirrors do is this: For an object at a point $(x,y,z)$, the flat mirror creates an image at $(x,y,-z)$ (this image has opposite handedness because one coordinate is flipped). Then the spherical mirror takes that image at $(x,y,-z)$ and creates another image at $(-x,-y,z)$. This image has the same handedness as the original object. It is also "right side up" because the z coordinate is not flipped. The image of your right hand at (0.01, 0, 0.2) will appear as another right hand at (-0.01, 0, 0.2) which is in the appropriate orientation for you to shake hands.
If you have the budget for special purpose stuff, I would go for an anti-reflective coating. Since the angle of incidence is fixed and well defined and since you are filtering the IR illumination ideally what would happen is you use a very narrow pass IR filter and an anti-reflective coating that is specifically 'tuned' for the frequency and angle of illumination. You should be able to eliminate most of the reflection.
Best Answer
Actually this is a bit of a tricky problem that only was solved relatively recently. The primary rainbow is generated by a beam that refracts once inside the drop, reflects off the back, then refracts again upon exit.
You have to use snell's law and then some calculus to find the minimum scattering angle for the angle that produces the brightest light as illustrated below from ray tracing:
For a fully worked example try here. And this site has some nice charts of the primary and secondary bows.