One interesting phenomenon I encountered at work is that, when I drill holes in acrylic sheets, the bits that comes out has a solid white color, where as the acrylic sheet itself is transparent. So it got me wonder, how come a whole lot of opaque material can form something transparent?
This is similar to the case of graphite and diamond. Both composed of carbon atoms, graphite is black and is found in pencils, diamond is optically transparent and are much rarer. Based on elementary solid state physics knowledge, in the crystal lattice of diamond, carbon atoms are densely arranged in 3D periodic structure. A part of the physics behind transparent material is that the band gap of the material is wide enough so that lights of all wavelengths are allow through. This is the reason why electrical conductors are opaque, because if the band gaps were too large, they wouldn’t be able to conduct electricity.
Back to discussion of acrylic. I looked up the chemical composition of it and here is the Wiki site: https://en.wikipedia.org/wiki/Poly(methyl_methacrylate), it is also known as PMMA. The composition is full of chaotic polymers, so I do not attempt to understand the structure. regardless of its chemical structure, why is it that small pieces that comes out with the drill is opaque while the holistic compound is transparent? This is cannot be analyzed with solid state physics because it does not contain lattice structure. So, set aside all chemistry stuff, what all changed when the small pieces falls off with the drill? it is probably less dense (according to my intuition), the temperature is raised locally, and the chemical bounds are broken by the drill bit. With those three listed changes of environment, I think the most likely dominating factor is the third (again, that’s my intuition speaking). The long polymer chains originally obtained transparency somehow, but by breaking the chains, the property of transparency got lost or broken.
This hypothesis shall remain open to discussion before I learn more about polymer.