ENG2000 Chapter 10 Optical Properties of Materials ENG2000: R. Hornsey Optic: 1 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 ENG2000: R. Hornsey Optic: 2 Nature of light • Light is an electromagnetic wave: § with a velocity given by c = 1/√(ε0µ0) = 3 x 108 m/s • In view of this, it is not surprising that the electric field component of the wave should interact with electrons electrostatically ENG2000: R. Hornsey http://www.com/light/emanim.gif Optic: 3 • Many of the electronic properties of materials, information on the bonding, material composition etc.
was discovered using spectroscopy, the study of absorbed or emitted radiation § evidence for energy levels in atoms § evidence for energy bands and band-gaps § photoelectric effect ENG2000: R. Hornsey Optic: 4 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 http://www.uk/tekano/translucent. Hornsey Optic: 5 • 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 = - .I § where α is the absorption coefficient (dimensions are m-1) • In the limit of x 0, we get dI =− I dx • The solution of which is I = I0 exp(– x) • Taking “ln” of both sides, we have: I x = − ln I0 § which is known as Lambert’s Law (he also has a unit of light intensity named for him) ENG2000: R.
Hornsey Optic: 6 • Thus, if we can plot -ln(I) against x, we should find from the gradient • Depending on the material and the wavelength, light can be absorbed by § nuclei – all materials § electrons – metals and small band-gap materials ENG2000: R. Hornsey Optic: 7 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 ENG2000: R. Hornsey Optic: 8 • If we think of our atom-on-springs model, there is a single resonance peak: absorption f f0 • But things are more complex when the atoms are connected – phonons § recall transverse and longitudinal optical phonons ENG2000: R. Hornsey Optic: 9 Electronic absorption • Absorption or emission due to excitation or relaxation of the electrons in the atoms ENG2000: R.
Hornsey http://www.edu/~kieran/reuhome/vizqm/figs/hydrogen.gif Optic: 10 Molecular materials • Materials such as organic (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 • e. water molecule: ENG2000: R. Hornsey http://www.uk/water/images/molecul5.jpg Optic: 11 • The spectrum of liquid water http://www.uk/water/images/watopt.jpg Optic: 12 ENG2000: R. Hornsey • 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 ENG2000: R.
Hornsey Optic: 13 Optical properties of metals • Recall that the energy diagram of a metal looks like: empty T = 0K levels EF full levels § EF is the energy below which, at 0K, all electron states are full and above which they are empty § this is the Fermi Energy • For T > 0, EF is the energy at which half of the available energy states are occupied • Semiconductors also have a Fermi level § for an intrinsic material EF is in the middle of the bandgap § nearer Ec for n-type; nearer Ev for p-type ENG2000: R. Hornsey Optic: 14 • This structure for metals means that almost any frequency of light can be absorbed • Since there is a very high concentration of electrons, practically all the light is absorbed within about 0.1µm of the surface • Metal films thinner than this will transmit light § e. gold coatings on space suit helmets • Penetration depths (I/I0 = 1/e) for some materials are: § water: 32 cm § glass: 29 cm § graphite: 0. Hornsey Optic: 15 • So what happens to the excited atoms in the surface layers of metal atoms? § they relax again, emitting a photon • The energy lost by the descending electron is the same as the one originally incident • So the metal reflects the light very well – about 95% for most metals § metals are both opaque and reflective § the remaining energy is usually lost as heat • In terms of electrostatics, the field of the radiation causes the free electrons to move and a moving charge emits electromagnetic radiation § hence the wave is re-emitted = reflected ENG2000: R.
Hornsey Optic: 16 • The metal appears “silvery” since it acts as a perfect mirror • OK then, why are gold and copper not silvery? § because the band structure of a real metal is not always as simple as we have assumed § there can be some empty levels below EF and the energy re- emitted from these absorptions is not in the visible spectrum • Metals are more transparent to very high energy radiation (x- & - rays) when the inertia of the electrons themselves is the limiting factor ENG2000: R. Hornsey Optic: 17 • Reflection spectra for gold and aluminum are: aluminum spectrum is relatively flat gold reflects lots of red wavelengths blue red ENG2000: R. Hornsey http://www.com/eThermo/CMA/Images/Various/109Image_12275.gif Optic: 18 Electronic absorption in non-metals • Dielectrics and semiconductors behave essentially the same way, the only difference being in the size of the bandgap • We know that photons with energies greater than Eg will be absorbed by giving their energy to electron-hole pairs EC EG EV hole § which may or may not re-emit light when they relax ENG2000: R. Hornsey Optic: 19 • Hence, the absorption coefficients of various semiconductors look like: Photon energy (eV) 5 4 3 2 1 0.47As Si 106 (m-1) GaAs InP 105 a-Si:H 104 103 0.19: Absorption coefficient ( ) vs.
wavelength ( ) for various semiconductors (Data selectively collected and combined ENG2000: R. Hornsey from various sources.) Optic: 20 From Principles of Electronic Materials and Devices, Second Edition, S. Kasap (© McGraw-Hill, 2002) http://Materials.Ca • Semiconductors can appear “metallic” if visible photons are all reflected (like Ge) but those with smaller Eg, such as CdS look coloured § yellow for CdS which absorbs 540nm and above • The above picture is good for pure materials but impurities can add extra absorption features EC hf1 phonon hf2 EV ENG2000: R. Hornsey Optic: 21 • Impurity levels divide up the bandgap to allow transitions with energies less than Eg • Recombination can be either radiative (photon) or non-radiative (phonon) depending on the transition probabilities • Practical p-n diodes usually contain a small amount of impurity to help recombination because Si has a relatively low recombination “efficiency” § for the same reason that Si is inefficient at generating light ENG2000: R.
Hornsey Optic: 22 Refraction in non-metals • One of the most important optical properties of non-metallic materials is refraction • This refers to the bending of a light beam as it passes from one material into another § e. from air to glass • We define the index of refraction to be n = c/v § where c is the speed of light in a vacuum and v is the speed of light in the material (which is in general wavelength- dependent) • A familiar example is the prism where the different amounts of bending separates out the wavelengths ENG2000: R. Hornsey Optic: 23 • Refraction is also vital for other applications, such as: § optical fibres – keeps the light in § semiconductor laser – keeps the light in the amplifying cavity of the laser • Given that 1 1 v= and c = 0 0 § where µ and µ0 (= µrµ0) are the permeability of the material and free space, respectively (a magnetic property) § and ε and ε0 (= εrε0) are the permittivity of the material and free space, respectively (an electrostatic property) • We find that n = (µr r) ( r for many materials) ENG2000: R. Hornsey Optic: 24 • Since light is an electromagnetic wave, the connection with both the dielectric permittivity ( ) and the magnetic permeability (µ) is not surprising • The index of refraction is therefore a consequence of electrical polarization, especially electronic polarization – + • Hence, the radiation loses energy to the electrons ENG2000: R.
Hornsey Optic: 25 • Since E = hv/ , and doesn’t change, the velocity must be smaller in the material than in free space § since we lose E to the atoms, v must also decrease • Electronic polarization tends to be easier for larger atoms so n is higher in those materials § e.1 (which makes glasses and chandeliers more sparkly!) • n can be anisotropic for crystals which have non- cubic lattices ENG2000: R. Hornsey Optic: 26 Reflection in non-metals • Reflection occurs at the interface between two materials and is therefore related to index of refraction • Reflectivity, R = IR/I0, where the I’s are intensities • Assuming the light is normally incident to the interface: n2 − n1 2 R = n2 + n1 n1 n2 § where n1 and n2 are the indices for the two materials • Optical lenses are frequently coated with antireflection layers such as MgF2 which work by reducing the overall reflectivity § some lenses have multiple coatings for different wavelengths ENG2000: R. Hornsey Optic: 27 Spectra • So we have seen that reflection and absorption are dependent on wavelength § and transmission is what’s left over! • Thus the three components for a green glass are: ENG2000: R. Hornsey Callister Fig.8 Optic: 28 Colours • Small differences in composition can lead to large differences in appearance • For example, high-purity single-crystal Al2O3 is colourless § sapphire • If we add only 0.0% of Cr2O3 we find that the material looks red § ruby • The Cr substitutes for the Al and introduces impurity levels in the bandgap of the sapphire • These levels give strong absorptions at: § 400nm (green) and 600nm (blue) § leaving only red to be transmitted ENG2000: R.
Hornsey Optic: 29 • The spectra for ruby and sapphire look like: • A similar technique is used to colour glasses or pottery glaze by adding impurities into the molten state: § Cu2+: blue-green, Cr3+: green § Co2+: blue-violet, Mn2+: yellow ENG2000: R. Hornsey http://www.com/images/sapp.jpg Optic: 30 http://home.net/~jtalbot/glossary/photopumping.