Inna Vishik over at Quora:
Theoretically but not practically.
Three examples of theoretical perfect mirrors are a perfect metal (electrons never scatter) below its plasma frequency, a dielectric mirror, and total internal reflection. In practical implementations, these deviate from perfection to varying degrees.
A perfect metal
Within the for AC conductivity, if the scattering time of electrons goes to infinity (i.e. they don’t scatter), the metal will reflect 100% of light if the light has low enough frequency. Introducing a reasonable scattering time degrades reflectivity a bit, and inter-band absorption in real metals can make it even worse. Many metals are not as bad as the example below, with 99% reflectance being achievable for certain wavelength ranges.
Reflectivity for an undamped metal (dotted line), a metal with some damping (dashed line), and aluminum. image source:
A employs a stack of dielectrics with differing indices of refraction. Thicknesses of each layer are chosen such that there is constructive interference for certain wavelengths. Because light is either transmitted or reflected at every interface, you don’t get the same losses as you do in a metal. Reflectivities better than 99.999% can be achieved for a limited wavelength range and angle of incidence. Deviations from perfect reflection occur because the materials have defects which absorb light and also some portion of light will make it through the entire stack without getting reflected.
When light hits the boundary between two materials of different indices of refraction (n) with a shallow enough angle, it will be perfectly reflected at the interface inside the material with the higher n. This is the operating principle behind optical fibers. In practice, the perfect reflectivity is disrupted by roughness at the interface, and for the case of optical fibers, the light will degrade over some (long) distance because of absorption.