【正文】 0) are the permeability of the material and free space, respectively (a magic property) ? and ? and ?0 (= ?r?0) are the permittivity of the material and free space, respectively (an electrostatic property) ? We find that n = √(181。Optic: 1 第四章 固體材料的宏觀光學性質 Optic: 2 Overview ? The study of the optical properties of materials is a huge field and we will only be able to touch on some of the most basic parts ? So we will consider the essential properties such as absorption/reflection/transmission and refraction ? Then we will look at other phenomena like luminescence and fluorescence ? Finally we will mention applications, in particular optical fibres and lasers Optic: 3 Nature of light ? Light is an electromagic wave: ? with a velocity given by c = 1/?(?0?0) = 3 x 108 m/s ? In view of this, it is not surprising that the electric field ponent of the wave should interact with electrons electrostatically Optic: 4 ? Many of the electronic properties of materials, information on the bonding, material position etc. was discovered using spectroscopy, the study of absorbed or emitted radiation ? evidence for energy levels in atoms ? evidence for energy bands and bandgaps ? photoelectric effect Optic: 5 General description of absorption ? Because of conservation of energy, we can say that I0 = IT + IA + IR ? Io is the intensity (W/m2) of incident light and subscripts refer to transmitted, absorbed or reflected ? Alternatively T + A + R = 1 where T, A, and R are fractions of the amount of incident light ? T = IT/I0, etc. ? So materials are broadly classed as ? transparent:relatively little absorption and reflection ? translucent:light scattered within the material (see right) ? opaque:relatively little transmission Optic: 6 ? If the material is not perfectly transparent, the intensity decreases exponentially with distance ? Consider a small thickness of material, ?x ? The fall of intensity in ?x is ?I so ?I = a.? ? where a is the absorption coefficient (dimensions are m1) ? In the limit of ?x ? 0, we get ? The solution of which is I = I0 exp(–ax) ? Taking “l(fā)n” of both sides, we have: ? which is known as Lambert’s Law (he also has a unit of light intensity named for him) ??dIdx ? ?aI??a x ? ? ln II0????????????Optic: 7 ? Thus, if we can plot ln(I) against x, we should find a from the gradient ? Depending on the material and the wavelength, light can be absorbed by ? nuclei – all materials ? electrons – metals and small bandgap materials Optic: 8 ATOMIC ABSORPTION ? How the solid absorbs the radiation depends on what it is! ? Solids which bond ionically, show high absorption because ions of opposite charge move in opposite directions ? in the same electric field ? hence we get effectively twice the interaction between the light and the atoms ? Generally, we would expect absorption mainly in the infrared ? because these frequencies match the thermal vibrations of the atoms Optic: 9 ? If we think of our atomonsprings model, there is a single resonance peak: ? But things are more plex when the atoms are connected – phonons ? recall transverse and longitudinal optical phonons f0 f absorption Optic: 10 Electronic absorption ? Absorption or emission due to excitation or relaxation of the electrons in the atoms Optic: 11 Molecular materials ? Materials such as anic (carbon containing) solids or water consist of molecules which are relatively weakly connected to other molecules ? Hence, the absorption spectrum is dominated by absorptions due to the molecules themselves ? . water molecule: Optic: 12 ? The spectrum of liquid water Optic: 13 ? Since the bonds have different “spring constants”, the frequencies of the modes are different ? when the incident illumination is of a wavelength that excites one of these modes, the illumination is preferentially absorbed ? This technique allows us to measure concentrations of different gas species in, for example, the atmosphere ? by fitting spectra of known gases to the measured atmospheric spectra, we can figure out the quantities of each of the gases Optic: 14 Optical properties of metals