Transcript Slide 1
Optics: Lecture 2
Light in Matter: The response of dielectric materials
Dispersive Prism: Dependence of the index of refraction, n(), on frequency or
wavelength of light.
Sir Isaac Newton used prisms to disperse white light into its
constituent colors over 300 years ago.
When white light
passes through a
prism, the blue
constituent
experiences a larger
index of refraction
than the red
component and
therefore it deviates
at a larger angle, as
we shall see.
The effect of introducing a homogenous, isotropic dielectric changes Maxwell’s
equations to the extent that o and o .
The phase speed in the medium becomes:
v 1 /
The ratio of the speed of an E-M wave in vacuum to that
in matter is defined as the index of refraction n:
c
n
KE KM
v
o o
KE
o
KM
o
Relative Permittivity and Relative Permeability
For most dielectrics of interest that are transparent in the visible, these
are essentially non-magnetic and to a good approximation KM 1.
To a good approximation also known as Maxwell’s Relation:
n KE
KE is presumed to be the static dielectric constant (and works well only for
some simple gases, as shown on next slide).
In reality, KE and n are actually frequency-dependent, n(), known as
dispersion.
Resonant process h E
n
Scattering and Absorption:
h
m n
•Non-resonant scattering:
•Energy is lower than the resonant
frequencies.
•E-M field drives the electron cloud into
oscillation
•The oscillating cloud relative to the
positive nucleus creates an oscillating
dipole that will re-radiate at the same
frequency.
Em
Gas/Solids
Excitation energy can be
transferred via collisions
before a photon is re-emitted.
Works well
Doesn’t work so well
A small displacement x from
equilibrium causes a restoring force F.
F = -kEx
and results in resonant frequency:
o k E / me
E(t)
x-axis
+
-
Light E(t) and produces a
classical forced oscillator.
Amplitude ~ 10-17 m for bright
sun light.
The result can be modeled like a classical forced oscillator with
FE = qeEocos(t) = qeE(t). Using Newton’s 2nd law:
Driving Force – Restoring Force = ma, where Rest. Force = -kEx
2
d
x
2
qe Eo cost meo x me 2
dt
To solve, let x(t) = xocost
qe Eo cost meo xo cost me 2 xo cost
2
qe E o
qe Eo
qe E (t )
xo
; x
cost ; x
2
2
2
2
2
me o
me o
me o 2
*Note that the phase of the displacement x depends on
> o or < o which gives x qE(t).
The electric polarization or density of dipole moments (dipole
moment/vol.) is given by:
qe2 NE / me
P qe xN 2
o 2
Where N = number
electrons per volume.
We learn from the dielectric properties of solids that ( o ) E P.
2
q
P
(
t
)
Therefore
e N / me
o
o 2
E (t )
o 2
Since n2 = KE = /o it follows that we obtain the following
dispersion equation:
*Note that > o n < 1 (above resonance)
2
1 (Displacement is 180 out-of-phase with
q
2
e N
2
n ( ) 1
2 driving force.)
o me o
and < o n >1 (below resonance)
(Displacement is in-phase with driving force.)
Consider classically the average power (Pav) delivered by the
applied electric field:
q E
dx
dx
F (t )
qe E (t ) e 0
2
dt
dt
me
2 02 4 2 2
2
Pav
2
(See, e.g.
Mechanics,
Symon, 3rd Ed.)
02 2
tan
2
Phase angle
This average power is
analogous to absorption
of E-M radiation at the
resonant frequency 0.
For light = ck = 2c/, we can write the dispersion relation as
(n2 1)1 C / 2 C / 2o ; C 4 2c2 o me /( Nqe2 )
Thus, if we plot (n2-1)-1 versus -2 we should arrive at a straight line.
•In reality, there are several transitions in which n > 1 and n < 1 for
increasing , i.e., there are several oi resonant frequencies
corresponding to the complexity of the material.
•Therefore, we generalize the above result for N molecules/vol. with fj
different oscillators having natural frequencies oj, where j = 1, 2, 3..
2
qe N
2
n ( ) 1
o me
fj
j 2 2
oj
A quantum mechanical treatment shows further that
f
j
1
j
where fj are weighting factors known as Q.M. oscillator strengths
and represent the transition probability for each mode j.
The energy oj is the energy of absorption or emission for a
given electronic, atomic, or molecular transition.
•When = oj then n is discontinuous (and blows up). Actual
observations show continuity and finite n.
•The conclusion is that a damping force which is proportional to the
speed medx/dt should generally be included when there are strong
interactions occurring between atoms and molecules, such as in
liquids and solids.
•With damping, (1) energy is lost when oscillators re-radiate and (2)
heat is generated as a result of friction between neighboring atoms
and molecules.
•The corrected dispersion, including damping effects, is as follows:
2
q N
n 2 ( ) 1 e
o me
fj
j 2 2 i
j
oj
This expression often works fine for gases.
•In a dense solid material, the atoms/molecules may experience an additional
field that is induced by the surrounding medium and is given by P(t)/3o.
•With this induced field, the dispersion relation becomes
Nqe2
n2 1
2
n 2 3 o me
fj
j
2
oj
2 i j
•We will see that a complex index of refraction will lead to absorption.
•Presently, we will consider regions of negligible absorption in which n
is real and oj2 2 j.
•Thus
Nqe2
n2 1
2
n 2 3 o me
For various glasses,
j
2
oj
fj
2
oj2 2 , n2 () Const.
Since oj ~ 100 nm in the ultra-violet (UV).
•Note that as oj, n() gradually increases and the behavior is called
“Normal Dispersion.”
•Again, at = oj, n is complex and leads to an absorption band.
•Also, when dn/d < 0, the behavior is called “Anomalous Dispersion.”
When white light
passes through a
glass prism, the
blue constituent
experiences a larger
index of refraction
than the red
component and
therefore it deviates
at a larger angle, as
seen in the first
slide.
•Note the rise of n in the
UV and the fall of n in the
IR, consistent with
“Normal Dispersion.”
•At even lower frequencies
in the radio range, the
materials become again
transparent with n > ~1.
•Transparency occurs
when << o or >> o.
•When ~ o, dissipation,
friction and therefore
absorption occurs, causing
the observed opacity.
Absorption Spectrum of Water: Absorption coefficient:
nI is the imaginary part of the index of refraction.
2nI / c
The water vapor absorption bands are related to molecular vibrations
involving different combinations of the water molecule's three
fundamental vibrational transitions:
V1: symmetric stretch mode
V2: bending mode
V3: asymmetric stretch mode
The absorption feature centered near 970 nm is attributed to a 2V1 + V3
combination, the one near 1200 nm to a V1 + V2 + V3 combination, the
one near 1450 nm to a V1 + V3 combination, and the one near 1950 nm to
a V2 + V3 combination.
In liquid water, rotations tend to be restricted by hydrogen bonds, leading
to vibrations, or rocking motions. Also stretching is shifted to a lower
frequency while the bending frequency increased by hydrogen bonding.
Propagation of Light, Fermat’s Principle (1657)
Involves the principle of least time: The path between two points that is taken
by a beam of light is the one that is transversed in the least amount of time.
t SP
SO OP
t
vi
vt
b 2 (a x) 2
h2 x2
vi
vt
To find the path of least time,
set dt/dx=0.
dt
x
(a x)
dx vi h 2 x 2 vt b 2 (a x) 2
0
sin i sin t
ni Sin i nt Sin t
vi
vt
Since ni = c/vi and nt = c/vt .
Snell’s Law of Refraction
c
1 c o
where o is the
Note that v
n
n
n vacuum wavelength.
In general, for many layers having different n, we can write
t SP
m
sm
s
s1 s2
t ... i
v1 v2
vm i 1 vi
1 m
c
t ni si ; ni
c i 1
vi
Note that if the layers are very thin, we
can write
m
n s
i 1
i i
P
n( s)ds OPL
S
= OPL (Optical Path Length)
We can compute t as simply
OPL
t
.
c
P
m
Note that the spatial path length is
s ds
i 1
i
S
and for a medium possessing a fixed index n1,
OPL
o
s
.
•Fermat’s principle can be re-stated: Light in going from SP traverses the
route having the smallest OPL.
•We will begin next with the E-M approach to light waves incident at an
interface and derive the Fresnel Equations describing transmission and
reflection.