This is an example of fluorescence. Usually, when you shine a laser beam on a material and the photon is absorbed by an atom or molecule, a photon of the same frequency will be re-emitted. Sometimes, though, when a photon is absorbed, exciting an electron to a higher state, the electron doesn't decay to the ground state right away, but in several steps. In this case, when you shine light of one frequency on the material, it may emit light of a lower frequency. This is called fluorescence. Apparently whatever dye is using to color the orange plastic fluoresces orange when you shine green light of the right frequency on it.
There are lots of materials which fluoresce in the visible spectrum when UV light is shined on them. You don't see fluorescence with red lasers, because red is the lowest frequency of visible light, and photons emitted by fluorescence are almost always lower in frequency than the ones absorbed (there are apparently a few exceptions; see the wikipedia article). But it should be possible to see red, orange, or maybe yellow fluorescence with green lasers.
And here is a discussion of this phenomenon (with videos) on a blog.
This is one of the places where wave particle dualism gets some people in trouble. Many are taught that it means that light can be a wave and a particle, and that phrasing can lead to some confusion. I find it more intuitive to just rip the bandaid off quickly and say light is neither a wave nor a particle. It is something which, in some situations, can be well modeled as a wave, and in some situations can be well modeled as a particle, but it is its own thing (which can be well modeled in all known cases using a more complicated concept, a "wavefunction").
You can think of photons getting randomly reflected or transmitted on the boundary, but the truth is that the billiard-ball photon model really isn't very effective at describing what happens at this boundary. This is one of the regions where wave mechanics models the effects very well, while particle models don't do so well. If you use wave mechanics, the idea of a wave getting partially reflected and partially transmitted isn't difficult to believe at all. In fact, it's pretty easy to prove.
Thinking in wave terms at these boundaries also gives correct answers in peculiar situations where the particle model simply falls on its face. Consider the interesting case of an "evanescent wave."
In this setup, the laser and prism are set up at the correct angles to cause "total internal reflection." This means that, by the simple models, 100% of the light should bounce off the side of the prism and into the detector. Indeed, if the prism is in the open air, we do see 100% reflection (well, within the error bars of absorption). However, bring an object close to the prisim (but not touching) and things change. You end up seeing effects from the object, even though 100% of the light was supposed to be reflected!
If you think of light like photons, this is hard to explain. If you look at it as a wave governed by Maxwell's equations, you see that you would violate the law of conservation of energy if there was a "pure" reflection. Instead it creates a reflection and an "evanescent wave" which is outside the prism, and its strength falls off exponentially, which is really hard to explain with particles!
Of course, these too are all simplifications. The real answer to your question is that the wavefunction of the light interacts with the electromagnetic fields of the atoms in the prisim, and the result of that interaction leads to reflection, refraction, diffusion, absorption, and eveansecent waves. However, naturally those equations are a bit harder to understand, so we use the older, simpler models from before quantum mechanics. We just have to be sure to use the one which is most applicable in any given situation, because none of them are quite right.
Best Answer
I'll just go ahead and write this down even though it's already been covered in the other thread.. but I didn't post there so :)
First, if you think of light as a (scalar) wave (which is really a semi-classical way of thinking but might be enough to answer your question) you can invoke the Huygen-Fresnel principle which in this case boils down to considering every point on the reflecting surface as an origin of a re-emitted spherical wave with a starting phase directly related to the phase the point got from the incident wavefront.
The superposition of these wavefronts, after you let them destructively interfere with each other, will amount to a new combined wavefront which propagates according to Snells law (angle of incidence = angle of reflectance). See this image for the corresponding illustration of refraction (which is very similar, I couldn't quickly find a good image of reflection):
Now, light really doesn't behave "sometimes as a particle, sometimes as a wave". It is always detected as quanta (particles) but the probability amplitudes (phases) propagate in a wavelike fashion. One way of expressing the propagation is to say that a photon is sort of split and takes every possible path between A and B (or, in the case of a reflector, from A to any point on the reflector and then from there to point B by any means). Every path gets a phase contribution, and all the indistinguishable paths are summed. Most paths are simply cancelling each other but some constructively interfere, creating a large contribution (in case you don't know QM, the probability amplitude squared is the probability of the described event so a large contribution means this result will most probably occur). There is a VERY good image and description of this process in Feynman - QED The Strange Theory of Light and Matter (as I wrote in the comment above).
In the case of the reflector, the large contribution occurs at the classical angle of reflectance (Snell's law again). Notice the similarity between this formulation (called the path integral approach) and the semi-classical principle outlined above; this is not a coincidence of course.
Also to briefly digress on your implied question on the per-atom non-zero "reflectance" time - saying that an electron-orbit absorbs the photon energy for a while the re-emits it a non-zero time later is of course also a slight simplification. In reality the electron interacts with the a photon, changes its momentum a bit, the re-emits (interacts) with the new photon and changes its momentum again. This scattering process occurs at all allowed momenta and intermediate times, which are then all superpositioned like above and hence I'm not sure it's meaningful to talk about any appreciable time of reflectance. Notice that this scattering is in practice very different from scattering which can excite the electron to another orbit.