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2. optical properties

(1) introductionluminescencelaseroptical fiber

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(2) light and electromagnetic spectrum

visible light – electromagnetic radiation with wavelength 400 to 750 nm

ultraviolet – 10 – 400 nminfrared – 750 – 106 nmlight is in form of waves and consist of particles called photons

ΔE = hυ = hc/λΔE : energy λ : wavelength υ : frequencyc : speed of light 3 × 108 m/sh : plank’s constant 6.62 × 10-34 J · sex. a photon in ZnS drops from an impurity at

1.38 eV below its conduction band to valence band, what is the wavelength ? (ZnS band gap = 3.54 eV)

3.54 – 1.38 = 2.16 eV 1 eV = 1.6 × 10-19 J λ = hc/ΔE = (3 × 108)(6.62 × 10-34) / (2.16)(1.6 × 10-19 )

= 5.747 × 10-7 m = 574.7 nm2

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(3) refraction of lightwhen photons are transmitted through a transparent material, they lose some energy and speed and the direction changes

index of refraction c

n = ─vc : velocity of light in vacuumv : velocity of light in a medium

the values range from about 1.4 to 2.6

the refractive indices of materials are a function of wavelength and frequency

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Snell’s law of light refractionif light passes through one media to another

n sin φ’─ = ───n’ sin φ

n: refractive index of first median’: refractive index of second mediaφ : angle of incidenceφ ’: angle of refraction

if light passes from media of high refractive index to a media of low refractive index, φ ’ = 90o at φ = φc critical angle of incidencetotal internal reflection if angle φ > φc

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ex. What is the critical angle for light to be totally reflected when entering the air from soda-lime-silica glass (n = 1.51)? n sin φ’ 1.51 sin 90o

─ = ─── ─── = ────n’ sin φ 1 sin φc

sin φc = 1/ 1.51 = 0.662φc = 41.5o

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(4) absorption, transmission, reflection of lightfor a particular wavelength λ(reflected fraction)λ + (absorbed fraction)λ + (transmitted fraction)λ = 1(a) metals• metals strongly reflect and/or absorb light• amount of energy absorbed depends on

electronic structure• incident beam easily elevates electrons to

higher levels• Au absorbs shorter wavelength (blue &

green) and reflects longer wavelengths (yellow and red)Ag and Al reflect all parts of visible light

(b) silicate glasses• amount reflection of light from a glass

surface mainly depends on refractive indexn and angle of incidence

n – 1 2fraction of light reflected R = ───

n + 1R is called reflectivity (φi = 900)n : refractive index

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ex. calculate the reflectivity of incident light from the polished flat surface of silicate glass (n = 1.46)

n – 1 2 1.46 – 1 2R = ─── = ──── = (0.46/2.46)2 = 0.035

n + 1 1.46 + 1 3.5% light is reflected

• absorption of light by glass plate the light intensity decreases as light path increases

I── = e –αt

IoI : fraction of light exitingIo : fraction of light enteringα : linear absorption coefficientt : thickness

ex. incident light strikes a polished glass plate 0.50 cm thick (n = 1.50), what fraction of light is absorbed by the glass as the light pass between the surface of glass (α = 0.03 cm-1)

I── = e –αt = e –(0.03)(0.5) = 0.985Io1 – 0.985 = 0.015 1.5% absorbed

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• amount transmitted depends upon amountof light reflected from upper and lowersurface and light absorbed

fraction on incident light reaching lower surface = (1 - R)(I0e-αt)fraction of incident light reflected fromlower surface = R(1 - R)(I0e-αt)fraction of light transmitted

I = [(1 - R)(I0e-αt) ] - [R(1 - R)(I0e-αt) ]= (1 - R)2(I0e-αt)

90% incident light is transmitted by silica glass if λ > 300 nm

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(c) plastics• many plastics have excellent transparency• if crystalline regions having high refractive

index are larger than wavelength of light, the light will be scattered

(d) semiconductors• in pure semiconductors, photons may cause

electrons to jump across energy band gapthe energy of photons should be greater than Eg

• impure semiconductors (n-type or p-type) absorb photons of lower energy

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• semiconductors are opaque to high- and intermediate-energy light photons and transparent to low-energy, very longwavelength photons

ex. calculate the minimum wavelength for photons to be absorbed by intrinsic Si at room temperature (Eg =1.10 eV)

hc (6.62 × 10-34)(3 × 108)λ = —— = ————————— = 1.13 × 10-6 m

Eg (1.10 eV)(1.6 × 10-19)minimum wavelength is 1130 μm

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(5) luminescenceluminescence – the process by which substance absorbs energy and spontaneously emits visible or near-visible radiation

• electrons are excited by input energy anddrop to lower energy level

• fluorescence – emissions occur within 10-8

seconds after excitation. • phosphorescence – emissions occur longer

than10-8 seconds after excitation• produced by material called phosphors• emission spectra can be controlled by

adding activators

1 electron-hole pairs are created2 electrons can be thermally excited from one trap

to another or into conduction band3 electrons can drop to upper activator levels and

then to lower acceptor levels, emitting visible light

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(a) photoluminescence• ultraviolet radiation from a mercury arc is

converted into visible light by using halophosphate phosphor

• in fluorescent lights, calcium halophosphate(Ca10F2P6O24) with 20% F- replaced by Cl-

is used• Sb3+ produce blue emission and

Mn2+ provide orange-red emission band

(b) cathodoluminescence• produced by energized

cathode that generates a beam of high energy bombarding electronsex. electron microscope,

CRO, TV screen

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• in TV screen, the signal is rapidly scanned over the screen deposited with blue, greenand red emitting phosphors to produce images

• commonly used phosphors for the colors:blue – ZnS with Ag+ acceptor and Cl- donorgreen – (Zn, Cd)S with Cu+ acceptor and

Al3+ donorred – Y2O2S with 3% Eu

(c) intensity of luminescenceI t

ln ─ = ─I0 τI0 : initial intensityτ : relaxation time constantI : fraction of luminescence after time t

ex. color TV phosphor has a relaxation time of 3.9 × 10-3 s, how long will it take for the intensity of this phosphor material to decrease to 10% of its original intensity?

I t 1 tln ─ = ─ ln ── = ────I0 τ 10 3.9 × 10-3

t = 9.0 × 10-3 s

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(6) stimulated emission of radiation and lasersconventional light – photons with similar wavelength, independent, random, incoherentand out of phaseLASER : Light Amplification by Stimulated Emission of Radiation.laser – photons that are in phase, coherent, parallel, directional and monochromaticactive photons in laser stimulate manyothers of same frequency

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(a) ruby laser• single crystal of Al2O3 contains 0.05% Cr3+

ions in substitutional lattice spaces, these ions drop to lower energy levels causing photon emission at specific wave lengths

• Cr3+ ions are excited by xenon flash lightpartial transmitting mirror at front end allows beam to pass through

• electrons are pumped to E2 and E3 energy levels, the return to E1 level emittingphotons that stimulate more electrons to jump from E2 to E1

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• large number of photons are hence producedthat are in phase

• a wavelength of 694.3 nm laser beam• pulsed type laser

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(b) neodymium-YAG (Nd: YAG) lasers• made by combining 1 part per hundred of

Nd in YAG (yttrium-aluminum-garnet) crystal

• near infrared wavelength (10.6 μm) withcontinuous power

• used in drilling, scribing and cutting

(c) carbon dioxide lasers• most powerful lasers• few milliwats continuous power to large

pulses of 10000 J• operates by electron collision exciting

nitrogen molecules that transfer energy to CO2 molecules

• used in cutting, welding and heat treatment

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(d) semiconductor lasers• the smallest lasers produced• consists of pn junction of semiconducting

compound• population inversion is achieved by strong

forward bias of heavily doped pn junction• great number of electron-hole pairs are

generated that recombine to emit photons • used in compact disks

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(7) optical fibershair thin (≈ 1.25 μm) optical fibers made ofSiO2 are used for optical-fiber communicationcentral core glass has higher refractive index than outer clad

(a) light loss in optical fibers10 I

-loss (dB/km) = ─── log ─l (km) Iol : length I, I0 : initial and final intensities

impurities (Fe2+ ) should be very low for low light loss (attenuation) ex. an optical fiber has a 0.2 dB/km light

attenuation (a) what is the fraction of light remaining after pass through 1 km? (b) whatis the fraction of light remaining after 40 km transmission?

(a) -0.2 = (10/1)log(I/Io) log(I/Io) = -0.02 I/Io = 0.95(b) -0.2 = (10/40)log(I/Io) log(I/Io) = -0.80 I/Io = 0.16

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single mode – one acceptable guided ray pathmultiple mode has a graded refractive index –many wave modes pass simultaneously.

(b) fabricated through modified chemicalvapor deposition ( MCVD) process

• vapor of SiCl4, GeCl4, and fluorinated hydrocarbons are passed through rotating pure silica tube with pure oxygen

• oxyhydrogen torch allows contents to react to form doped silica glass

• composition of vapors is adjusted to vary refractive index

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• silica tube is then heated and the tubecollapses uniformly into a solid role calledpreform

• the fiber is then drawn and polymer is coated for protection

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(c) modern optical-fiber communication systems

• modern systems use single mode fiber withan InGaAsP double heterojunction laserdiode transmitter and InGaAs/InP PIN photodiode is usually used for the detector

• erbium-doped optical-fiber amplifier(EDFA) is a length of optical fiber doped with rare earth element Erit boosts all signal passing through it

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(8) superconducting materialselectrical resistvity of a normal metal (e.g. Cu) decreases as temperature decreasing andreaches a low residual value at 0 Kelectrical resistvity of Hg drops to zero at 4.2 K

(a) superconductivity – the resistivity of a metal drops suddenly to a immeasurable valueupon cooling to critical temperature Tc• for a material to be superconducting, its

critical temperature Tc, magnetic field Band current density J must not be exceeded

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(b) magnetic properties of superconductorsif a critical magnetic field Hc is applied to a superconductor below Tc, superconductor returns to normal state.

T 2Hc = H0 1- —

TcHc = critical magnetic field at temperature TH0 = critical field at 0 K

sufficiently high (critical) current density Jcwill also destroy superconductivity.ex. calculate approximate value of critical field necessary to cause the superconductivity of pure Nb disappear at 6 K

T 2 6 2Hc = H0 1- — = 0.196 [1 – (——) ] = 0.112 T

Tc 9.15

(i) type I superconductor Pb, Snbelow Tc and Hc, applied magnetic field will be expelled from specimen except for a very thin penetration layer (~ 10-5 cm) at the surface – Meissner effect

(ii) type II superconductors NiTi, Ni3Sbmagnetic flux is excluded from material up to a lower critical field Hc1from Hc1 to Hc2, field starts to penetrate and after upper critical field Hc2, the behavior is normal

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(c) current flow and magnetic fields in superconductors

• type I superconductors are poor carriers of electrical current as current flowsonly in surface layer

• in type II superconductor, between Hc1 and Hc2 the current can be carried inside the superconductor by filament

• the field penetrates the superconductor in the form of flux bundles (fluxoids)

• with increasing magnetic field strength, more fluxoids enter the superconductor andform a periodic array

• at Hc2 the vortex structure collapses and material returns to normal conducting state26

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(d) high current high field superconductors• type II superconductors have small current

carrying capacity below Tc as fluxoids are weakly tied to crystal lattice

• mobility of fluxoids are impeded by dislocations, grain boundaries and fine precipitates

• heat treatment of Nb-45 wt% Ti alloy isused to precipitate a hexagonal α phase inBCC matrix to help pin down fluxoids.

• Nb-Ti filaments are embedded in copper matrix to improve formingex. Nb-46.5 wt% Ti – Cu composite

Cu : NbTi vol. ratio of 1.57250 filaments of 6-μmdiameters Jc = 2990 A/mm2 at 5 KJc = 1256 A/mm2 at 8 K

• applications for NbTi and Nb3Sn :nuclear magnetic imaging system, magnetic levitation of vehicles (high-speed train)

(e) high critical temperature superconducting oxides• YBa2Cu3Oy has Tc of 90 K and has defective

perovskite structure, y = 9 for ideal structure, however, y = 6.65 ~ 6.90 for this material

• at y = 6.90, Tc = 90 K and material is normal at y = 6.65

• oxygen vacancies play an important role in the superconducting behavior

• to have high Tc, oxygen atoms on (001) planes are ordered so that vacancies are in ‘a’ direction.

• superconductivity is confined to CuO2planes, oxygen vacancies provide electron coupling between CuO2 planes.

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oxygen content vs. unit cell parameters

oxygen content vs. unit cell parameters

HR-TEM for YBa2Cu3Oy