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Optical Properties of Materials
Atomic and Molecular Origins of Color – 2.1, 2.6, 2.7
Colors originate from the interaction of light waves with
matter. Specifically, with electrons in matter.
The human eye detects light in the
350350-750nm wavelength range,
or ~3.5~3.5-1.5eV (1240=1eV*λ
(1240=1eV*λ)
The longer the wavelength the
lower the energy.
E = hv = hc / λ
Optical Properties of Materials
Sources of color are:
1. Energy absorption – requires a source of light
(i.e. electronic transitions, molecular vibrations)
2. Energy emission – can be seen in the dark
(i.e. phosphorescence, fluorescence)
Transmission or Reflection –
The perceived color of the
object will be the complementary
color of the light absorbed.
Ch. 2, #1. If six paints are mixed in equal portions (red, orange,
orange, yellow, green, blue
violet) the paint is black. However if six lights of the same color
color shine at the same
spot the spot is white. Why?
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Optical Properties of Materials
Color centers (F(F-centers) – Example of free electrons in
solids or liquids. For materials not usually colored, their
band gaps are >3 eV,
eV, i.e. insulators
*Arises because of a site defect in the lattice
For example, occurrence
in NaCl gives a yellow
color, and in KCl a
violet color.
Optical Properties of Materials
The trapped electron is a classic example of a particle in a box.
box. The
spacing of the energy levels and the color observed depends on the
the
host crystal and not the source of electrons.
N-doped
Examples of colored
diamonds.
Model: Radiation dislodges a
carbon atom from the
diamond structure and an
electron takes its place.
B-doped
* When the electron escapes the trap (through heating or solar
radiation), the electrons and atoms assume more regular lattice
positions and the color is lost.
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In-Class Problem Set
(1) Rubies are red because of the absorption of light by Cr3+ ions. The
color of emeralds also has the same origin in absorption of light
light by
Cr3+ ions. Explain whether the crystal field experienced by Cr3+ is
less in emerald or in ruby.
(2) How do the host lattices influence the energy level spacings (and
color) of the free electron? For example, why do the color centers
centers in
NaCl appear yellow and the color centers in KCl appear violet?
(3) Ch. 2, #13. Most clothing stores have fluorescent lighting. In this
light, two items can appear to be color matched, but when viewed in
sunlight, they are not well matched. Explain.
Colors in Metals and Semiconductors
Metals – free electron gas; Delocalized electrons span a near
continuum of closelyclosely-spaced energy levels at Ef.
Æ Metals can absorb all wavelengths (energy) of light
T = 0K
P(E)
T > 0K
P(E) =
1
kT +
(E-Ef)/
Ef)/kT
e(E-
1
Ef
E (eV
(eV))
* At 0K, states above Ef are empty (dashed line)
* At temperatures >0K, states above Ef are partially filled
(red), and states below Ef are partially empty (green).
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Colors in Metals and Semiconductors
Semiconductors – The electrons must cross an energy
gap, Eg, in order to reach the conduction band.
Æ Can absorb all light with E > band gap
ZnS,ZnO
ZnS,ZnO
CdS
HgS
GaAs
Kurt Nassau “The Causes of Color”
Colors in Metals and Semiconductors
Doped Semiconductors – Semiconductors that have
internal atomic impurities.
Æ Impurities can lower the energy to cross the band gap
n-type
p-type
LED – pn
junction
N-doped
B-doped
Kurt Nassau “The Causes of Color”
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Optical Properties of Materials
Color From Interactions of Light Waves with Bulk Matter
Sec. 4.1, 4.2, 4.5
Refraction – Change in speed of light from one
material to another; n = v / v’
v’ = index of refraction
n1sinθ
sinθ1 = n2sinθ
sinθ2
n1
n2
n2 > n1
θ1
θ2
Light bends down
Fermat’
Fermat’s Principle of
Least Time
How could this effect lead to color?
Optical Properties of Materials
Light refraction as a source of color
Why are the colors of light separated
upon passing through the prism?
Refraction angle depends on: 1. Wavelength (shorter is refracted
refracted more)
2. Larger n refracted more
(depends on identity of material)
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Optical Properties of Materials
Light refraction as a source of color
Total internal reflection (n1>n2):
Substance
Air
Ice
Water
Diamond
n
1.0003
1.305
1.33
2.42
Critical Angle(o)*
88.6
50.02
* sin-1(n2/n1); n2 ~ 1 (air)
48.7
24.4
Where/How could this lead to color?
Optical Properties of Materials
Example: Total Internal Reflection of a Rain Droplet
(See also: Diamonds, M.A. White)
Perceived position of colors
Below 42o, sky is bright – light reflection
Above 42o, sky is dark – no reflection
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Optical Properties of Materials
Diffraction of Light Waves as a Source of Color
Materials having stacked layers with distances (d) in the visible
visible
range of light (nm); nλ = 2dsinθ
2dsinθ
Colloidal crystals – Opals
nλ = 2dsinθ
2dsinθ
270nm spheres –
SiO2, polystyrene
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