Download CHAPTER 1 Wave Nature of Light

CHAPTER 1
Wave Nature of Light
1
1.1
LIGHT WAVES IN
HOMOGENEOUS MEDIUM
2
A.
Plane Electromagnetic Wave
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Nature of Light
• Photons
– Photoelectric effect
– Compton effect
• Electromagnetic waves
– Interference
– Diffraction
“Physicists use the wave theory on Mondays, Wednesdays and Fridays
and the particle theory on Tuesdays, Thursdays and Saturdays.”
Sir William Henry Bragg (July, 2, 1862-March, 10, 1942)
The Nobel Prize in Physics 1915
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Electromagnetic Wave (EM wave)
Electric field
Magnetic field
• An electromagnetic wave is a traveling wave which
has time varying electric and magnetic fields which
are perpendicular to each other and the direction of
propagation, z.
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The simplest traveling wave along z
• A sinusoidal wave
Ex ( z, t )  E0 cos(t  kz  0 )
(1)
– Ex : the electric field at position z at time t
– k: the propagation constant or wave number
• k = 2/, : the wavelength
– : the angular frequency
•  = 2/T = 2, T : period, : frequency
– E0 : the amplitude
– 0 : a phase constant (0 accounts for the fact that at t =
0 and z = 0, Ex may or may not necessarily be zero
depending on the choice of origin)
–  = (t – kz + 0) : the phase of the wave
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Monochromatic plane wave
• Ex ( z, t )  E0 cos(t  kz  0 ) (1)
describes a monochromatic plane wave of infinite
extent traveling along in the positive z direction.
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Monochromatic plane wave
• In any plane perpendicular to the direction of
propagation (along z), the phase of the wave
is constant, which means that the field is this
plane is also constant.
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Wavefront
• Wavefront : a surface over which the
phase of a wave is constant.
• A wavefront of a plane wave is a plane
perpendicular to the direction of
propagation.
E and B have constant phase
in this xy plane; a wavefront
z
E
E
B
Ex
k
Propagation
Ex = Eo sin( t–kz)
z
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Electromagnetic Induction
• Faraday’s law:
– Time varying magnetic fields result in time varying
electric fields.
• Maxwell’s suggestion:
– Time varying electric fields result in time varying
magnetic fields.
• A traveling electric field
Ex ( z, t )  E0 cos(t  kz  0 )
(1)
always be accompanied by a traveling magnetic field
By ( z, t )  B0 cos( t  kz  0 )
– same wave frequency  and propagation constant k, but
EB
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Optical Field
• Electric field displaces the electrons in
molecules of ions in the crystal and thereby
gives rise to the polarization of matter.
• Electric field is dominant.
• We generally describe the interaction of light
wave with matter through electric filed rather
than magnetic field.
• The optical field refers to the electric field.
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Exponential notation of traveling wave
exp( j )  cos   j sin  ,
j  1
 cos   Re[exp( j )]
E x ( z , t )  E0 cos( t  kz  0 )
(1)
 E x ( z , t )  Re[ E0 exp( j0 ) exp j (t  kz)]
or E x ( z , t )  Re[ Ec exp j (t  kz)]
where Ec  E0 exp( j0 )
(2)
Complex amplitude including the constant
phase information 0
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Wave vector k
• We indicate the direction of propagation with a vector k
2 ˆ
k
k  kkˆ

– k : the propagation constant
– k̂ : unit vector along the direction of propagation
• k  the constant phase planes
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A traveling wave along a direction k
• When the EM wave is propagating along some
arbitrary direction k, then the electric field E(r,t) at a
point r on a plane perpendicular to k is
E (r , t )  E0 cos(t  k  r  0 ) (3)
 E0 cos(t  (k x x  k y y  k z z )  0 )
 kz
• k  r  kr        
if propagation along z
 (3)  E ( z, t )  E0 cos( t  kz  0 ) (1)
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Phase velocity
•
Ex ( z, t )  E0 cos(t  kz  0 )
(1)
• The relationship between time and space for a given
phase, , is described by
  t  kz  0  constant
 z constant phase  (t  0  constant)/ k
• During a time interval t, this constant phase is moved a
distance z. Thus the phase velocity is
z dz
 2
v 
 
t dt constant phase k 2 / 
 v  
(4)
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Phase difference
(1)
• Ex ( z, t )  E0 cos(t  kz  0 )
• The phase difference  at a given time
between two points on a wave that are
separated by z is
  kz 
2z

– in phase
 = 2n , n = 0, 1, 2, …
– out of phase  = (n+1/2) , n = 0, 1, 2, …
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B.
Maxwell’s Wave Equation and
Diverging Waves
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Plane wave
• The propagation vectors
everywhere are all parallel
and the plane wave
propagates without the wave
diverging.
The plane wave has no
divergence.
• Amplitude E0 is the same at
all point on a given plane
perpendicular to k.
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Plane wave is an idealization
• Planes extend to 
 energy  
• We need an infinite large EM
source to generate a perfect plane
wave!
• In reality, the electric field in a plane
at right angles to k does not extend
to infinity since the light beam would
a finite cross sectional area and
finite power.
 A plane is an idealization that is
useful in analyzing many
phenomena.
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Maxwell’s EM wave equation in an
isotopic and linear medium
• Isotropic medium
– relative permittivity r is the same in all directions
• Linear medium
– relative permittivity r is independent of the electric field
• In an isotropic and linear dielectric medium (we
assume conductivity  = 0) , E obeys Maxwell’s EM
wave equation
2E 2E 2E
2E
 2  2   0 r 0 2
2
x
y
z
t
– 0: absolute permittivity
– r: relative permittivity
– 0: absolute permeability
(5)
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Possible waves that satisfy
Maxwell’s EM wave equation
2E 2E 2E
2E
 2  2   0 r 0 2
2
x
y
z
t
(5)
• To find the time and space dependence of the field,
we must solve Eq. (5) in conjunction with the
initial and boundary conditions.
• There are many possible waves that satisfy Eq. (5):
– Plane wave
– Spherical wave
– Cylindrical wave
–
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Spherical wave
• A spherical wave is described
by a traveling field that
emerges from a point EM
source:
A
E  cos(t  kr)
r
A : a constant.
(6)
 The wavefronts are spheres
centered at the point source O.
A / r r
 0

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Spherical wave
A
• E  cos(t  kr)
r
 Amplitude decays with
distance r from the source:
A / r r
 0

• k wavevectors diverge
out and, as the wave
propagates, the constant
surfaces becomes larger.
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Optical divergence
• Optical divergence
– refers to the angular separation
of wavevectors on a given
wavefront.
• Spherical wave
– 360° divergence
(fully diverging wavevectors)
• Plane wave
– 0° divergence
(perfectly parallel wave vectors)
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Waves from ideal EM sources
• Infinitely large source • Point source
produces plane wave
produces spherical wave
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Light ray of geometric optics
• Light ray of geometric optics are drawn to be
normal to constant phase surfaces
(wavefronts).
• Light rays follow the wavevector direction.
Wave fronts
(constant phase surfaces)
Wave fronts
k

P

P
k

Wave fronts
E
r
O
z
A perfect plane wave
(a)
A perfect spherical wave
(b)
A divergent beam
(c)
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Wave from a practical EM source
• In reality, an EM source would have a
finite size and finite power.
 The light beam exhibits some
inevitable divergence while
propagating.
 The wavefronts are slowly bent
away thereby spreading the wave.
 Light rays slowly diverge away
from each other.
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The reason for favoring plane waves
• At a distance far away from a source, over a
small spatial region,
 the wavefronts will appear to be plane even if
they are actually spherical.
Wave fronts
(constant phase surfaces)
Wave fronts
k

P

P
k

Wave fronts
E
r
O
z
A perfect plane wave
(a)
A perfect spherical wave
(b)
A divergent beam
(c)
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Gaussian beams
•
Many light beams can be described by assuming that they are Gaussian
beams.
– Ex. The output from a laser
•
•
A result of radiation from a source of finite extent
Properties:
–
–
–
–
Still exp j(t – kz) dependence
Amplitude varies spatially away from the axis and also along the axis.
Slow diverges
y
Intensity distribution across the beam cross-section is Gaussian.
Wave fronts
(b)
x
2wo O
z Beam axis
Intensity

Gaussia n
(c)
r
(a)
2w
(a) Wavefronts of a Gaussian light beam. (b) Light intensity across beam cross
section. (c) Light irradiance (intensity) vs. radial distance r from beam axis (z).
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Gaussian distribution
• I = Io exp(-2r2/w2)
• Beam diameter 2w
– It is defined in such
way that the cross
sectional area w2
contains 85% of the
beam power.
– It increases as the
beam traveling along z.
I0
I (r )  I 0 e
2 r 2 / w2
0.135I0
r=w
I (r  w)  I 0 e
2 w 2 / w 2
 I 0 e  2  0.135I 0
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Power contained in the area of w2
• Total power
Ptotal  

0

2
0

I (r )rdrd  2  I 0 e
 2 r 2 / w2
0
1 2
rdr  w I 0
2
• Power contained in the area of w2
Parea  
w
0

2
0
w
I (r )rdrd  2  I 0 e
0
 2 r 2 / w2
1 2
rdr  w I 0 (1  e 2 )
2
Parea

 (1  e  2 )  0.865
Ptotal
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Gaussian beam
• It starts from O with a finite width 2w0 where the
wavefronts are parallel and then the beam slowly
diverges as the wavefronts curve out during
propagation along z.
– waist: 2w0
(where the wavefronts are parallel)
– waist radius: w0
– spot size: 2w0
– beam divergence: 2
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Beam divergence
• The increase in beam diameter 2w with z makes an
angles 2 at O which is called the beam divergence.
4
2 
 (2 w0 )
1
  
w0
• The greater the waist, the narrower the divergence.
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The minimum spot size to which a
Gaussian beam can be focused
• Suppose that we reflect the Gaussian beam back on
itself so that the beam is traveling in the –z direction
and converging towards O.
• The beam would still have the same diameter 2w0
(waist) at O.
• From then on, the beam again diverges out just as it
did traveling in +z direction.
The minimum spot size to which
a Gaussian beam can be focused
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EXAMPLE 1.1.1
A diverging laser beam
• HeNe laser
–  = 633 nm = 633  10-9 m
• Spot size 2w0 = 10 mm = 10  10-3 m
• Assuming a Gaussian beam, what is the
divergence of the beam?
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EXAMPLE 1.1.1
• Solution
divergence
4
4  (633 10 m)
2 

3
 (2w0 )   (10 10 m)
9
5
 8.06 10 rad  0.0046

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1.2
REFRACTIVE INDEX
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Propagation of polarization
• When an EM wave is traveling in a
dielectric medium, the oscillating electric
field polarizes the molecules at the
frequency of the wave.
EM wave propagation can be
considered to be the propagation of
this polarization in the medium.
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Polarization-induced delay of wave
propagation

 Induced molecular dipoles
• Field interactio
n
• Net effect:
– The polarization mechanism delays the
propagation of the EM wave.
– It slows down the EM wave.
– The stronger the interaction, the slower the
propagation of the wave.
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Relative permittivity r
• The relative permittivity (dielectric constant) r
measures the ease with which the medium
becomes polarized.
• It indicates the extent of interaction between
field and induced dipoles.
•  ~ optical region
– r will due to electronic polarization
•  ~ infrared or below
– r will due to both electronic and ionic polarizations.
 r (LF) > r (optical)
40
Phase velocity in a medium
• The phase velocity of EM wave in a medium with r is
given by
1
v
•
 r  0 0
(1)
r (LF) > r (optical) V(LF) < V(optical)
• In free space, r = 1
v vacuum  c 
1
 0 0
 3 108 m/s
This is the velocity of light in vacuum
41
Refractive index n
• n  the ratio of the speed of light in free space to
its speed in a medium
1
v 
 r  0 0
, v vacuum  c 
 r  0 o
n 
 r
v
 0 o
c
1
 0 0
(2)
• Light propagates more slowly in a denser medium
that has a higher refractive index.
42
Wave vector and wavelength in
medium
• In free space
k  2 / 
– k : wave vector in free space
– : wavelength in free space
• In the medium
c


n 

v medium medium
 medium 
 k medium 

n
2
medium
2
2

n
 nk
/n

43
Refractive index of noncrystalline
materials
• Noncrystalline materials
– Glasses
– Liquids
• The material structure is the same in all
direction and index n does not depend
on the direction.
• The refractive index is isotropic.
44
Refractive index of crystals
• Crystals
– the atomic arrangements and interatomic bonding are
different along different direction
• r is different along different crystal directions.
• n depends on r along the direction of the
oscillating electric field (the direction of
polarization).
• The refractive index is anisotropic.
45
Phase velocity depends on the
direction of polarization
• A wave is traveling along z in a particular
crystal with polarization along x direction.
• Relative permittivity along x is rx , then
nx   rx
electric field
• The wave propagates
with phase velocity
v  c / nx  c /  rx
46
Phase velocity depends on the
direction of polarization
• A wave is traveling along z in a particular
crystal with polarization along y direction.
• Relative permittivity along y is ry , then
n y   ry
• The wave propagates
with phase velocity
v  c / n y  c /  ry
electric field
47
Optically isotropic materials
• Noncrystalline solids
– Ex. glasses
• Liquids
• Cubic crystals
– Ex. diamond
• Only one refractive
index for all directions
48
Ex. 1.2.1 Relative permittivity and
refractive index
• r depends on the frequency of the EM wave.
• r can be vastly different at high and low
frequencies because of different
polarization mechanisms.
• Polarization mechanisms
– Electronic polarization
– Ionic polarization
• n = r1/2 must be applied at the same
frequency for both n and r .
49
Electronic Polarization
• Electronic polarization involves the
displacement of light electrons with
respect to heavy positive ions of the
crystal.
• This process can readily respond to the
field oscillations up to optical or even UV
frequencies.
50
• At low frequency (IR or below)
– All polarization mechanisms present can
contribute to r
– r = r(LF)
• At optical frequencies
– Only electronic polarization can respond
to the oscillating field
– r = r(optical)
51
Table 1
52
Si and Diamond
• Both are covalent solids.
• Electronic polarization is the only
polarization mechanisms at low and high
frequencies.
• There is an excellent agreement between r(LF)
and n.
 r (LF)  n(optical )
53
GaAs and SiO2
• The bonding is not totally covalent and there is a
degree of ionic bonding that contributes to
polarization at frequencies below far-infrared
wavelengths.
• At low frequencies both of these solids possess a
degree of ionic polarization.
 r (LF)  n(optical )
54
Water
• r(LF) is dominated by orientational or
dipolar polarization, which is far too
sluggish to respond to high frequency
oscillations of the field at optical
frequencies.
• r(LF) is large.
 r (LF)  n(optical )
55
A approximate expression for
relative permittivity
p  E and P  Np  NE
 P   0  e E  NE
  e  N /  0
  r  1   e  1  N /  0
•
•
•
•
•
•
•
p: the induced dipole moment per molecule
: the polarizability per molecule
E: the electric field
P: the dipole moment per unit volume
N: the number of molecules per unit volume
e: the electric susceptibility of the medium
r: the relative permittivity of the medium
56
Factors that affect n
•
 r  1  N /  0
n   r  1  N /  0
– N: the number of molecules per unit volume
– : the polarizability per molecule
• If N or , then r and n 
• Both density and polarizability increases n.
• Ex., glasses of given type but with greater
density tend to have higher n.
57
1.3
GROUP VELOCITY AND
GROUP INDEX
58
A group of wave differing slightly in
wavelength
• Since there are no perfect monochromatic waves
in practice, we have to consider the way in which
a group of wave differing slightly in wavelength
will travel along the z-direction.
59
Two harmonic waves interfere
• wave 1
 + 
– frequency  +
– wavevector k + k
 
–
• wave 2
– frequency  
– wavevectors k  k
• wave 1 + wave 2
 wave packet
–
–
–
–
Emax
Emax
Wave packet
k


Two slightly different wavelength waves travelling in the sam
direction result in a wave packet that has an amplitude variati
which travels at the group velocity.
mean frequency 
© 1999 S.O. Kasap, Optoelectronics (Prentice Hall)
amplitude modulated by slowly varying field of frequency 
the maximum amplitude moves with a wavevector k
group velocity
d
vg 
dk
(1)
60
Group velocity
• The speed of the envelope of the amplitude
variation.
• The speed with which energy or information is
propagates.
d

• v 
g
dk
• Emax advances with a velocity vg whereas the
phase variation are propagating at the phase
velocity v.
61
Group velocity in vacuum
• In vacuum
  ck
d
 v g (vacuum) 
 c  phase velocity
dk
 group velocity  phase velocity
62
Group velocity in a medium
• In a medium, index n depends on the
wavelength, n = n ().
c
v 
also depends on 
n( )
 c   2  2c
 vk  




 n( )     n
• Group velocity
d d d
d  1  d
Vg (medium) 

 2c   
dk d dk
dk  n  dk
63
Group velocity in a medium
d  1  d
Vg (medium)  2c   
d  n  dk
2
dk  2
d  2
k 

 2 


d

dk
2
dn 

n  

2

 c   dn 

d



Vg (medium)  2c 

    1 
2
n 
 2  n  n d 
c

1
  dn 
n1 

 n d 
64
Group velocity and group index of
the medium
Vg (medium) 
c
  dn 
n1 

 n d 
1

c
  dn 
n1 

 n d 

c
dn
n
d
This can be written as
c
Vg (medium) 
Ng
(4)
dn
Ng  n  
d
(5)
in which
group refractive index of the medium
65
Dispersive medium
• In general, for many materials the
refractive index n and the group index
Ng depend on the wavelength of light by
virtue of r being frequency dependent.
Both the phase velocity v and the group
velocity vg depend on the wavelength.
• The medium is called dispersive medium.
66
Pure SiO2 (silica) glass
• Refractive index n and
the group index Ng of
pure SiO2 (silica) glass
are important parameters
in optical fiber design in
optical communications.
• ~1300 nm Ng is minimum Ng is wavelength
independent Light waves with ~1300 nm travel with
the same group velocity and do not experience
dispersion.
Significant in the propagation of light in optical fibers
67
EXAMPLE 1.3.1 Group velocity
Consider two sinusoidal waves that are close in frequency
E x1 ( z, t )  E0 cos  t  k  k z 
E x 2 ( z, t )  E0 cos  t  k  k z 
 E x ( z, t )  E x1 ( z, t )  E x 2 ( z , t )
 E0 cos  t  k  k z   E0 cos  t  k  k z 
1

1

By using cos A  cos B  2 cos  ( A  B) cos  ( A  B)
2

2

 E x ( z, t )  2 E0 cost  k z cos(t  kz)
 + 
 
–
Emax
Emax
Wave packet
k


68
EXAMPLE 1.3.1 Group velocity
modulation
Ex ( z, t )  2E0 cost  k z  cos(t  kz)
amplitude
The maximum in the field occurs when
()t  (k ) z   2m  constant, m is an integer
which trav els with a velocity
dz 
d

or v g 
dt k
dk
Emax
Emax
Wave packet
 + 
 
–
k


69
EXAMPLE 1.3.2 Group and phase velocities
• A light wave in a pure SiO2 (silica) glass
medium:
–  = 1m
– n (= 1m) = 1.450
• phase velocity v = ?
• group index Ng = ?
• group velocity Vg = ?
70
EXAMPLE 1.3.2 Group and phase velocities
• Solution
– phase velocity:
c 3 108 m/s
v 
 2.069 108 m/s
n
1.45
– group index:
Fig. 1.7Ng = 1.460 @  = 1m
– group velocity:
c
3 108 m/s
vg 

 2.055 108 m/s
Ng
1.460
vg < v, vg is about ~ 0.7 % slower than phase velocity
71
1.4
MAGNETIC FIELD, IRRADIANCE
AND POYNTING VECTOR
72
Magnetic field
• The magnetic filed (magnetic induction) component By
always accompanies Ex in an EM wave propagation.
• For an EM wave in an isotropic dielectric medium
c
E x  vB y  B y
(1)
n
1
where v 
phase velocity
 0 r 0
n   r refractive index
73
Energy density
• Energy density (energy per unit volume) in the Ex field
1
u E   0 r E x2
2
• Energy density in the By field
uM 
•
1
20
B y2
The energy densities are the same
1
1
1
1
1 1
 0 r Ex2   0 r (vBy ) 2   0 r (
By ) 2 
By 2
2
2
2
2 0
 0 r 0
• Total energy density in the wave
u EM   0 r E x2
74
Energy flow
• As the EM wave propagates in the direction of k, there
is an energy flow in this direction.
• If S is the EM power flow per unit area, then
S  Energy flow per unit time per unit area
( Avt )( 0 r E )
S
 v 0 r E x2  v 2 0 r E x B y
At
2
x
Ex = vBy
(3)
75
Poynting vector and irrandiance
• EB is in a direction of wave propagation.
• The EM wave power flow per unit area:
S  v  0 r E  B
2
(4)
• S, called the Poyntin vector, represents the
energy flow per unit time per unit area in a
direction determined by EB (direction of
propagation)
• S = |v20rEB| = v20rExBy = v0rEx2 = the power
flow per unit area, is called (instantaneous)
irradiance.
76
Average irradiance (intensity)
E x  E0 sin(  t )
If
1 2
E
0 E0 sin(  t ) dt  2 E0
1
2
2
S  v 0 r E x  I  S average  v 0 r E0
(5)
2
• Saverage is called the average irradiance
(intensity).
2
x
1

T
T
2
I  Saverage  E
2
0
77
Average irradiance (intensity)
v  c/n
r = n2
1
1
2
2
I  S average  v 0 r E0  c 0 nE0
2
2
8
12
2
2
c  3 10 m/s,  0  8.85419 10 C /N  m ,
1
 c 0  1.328 10 3 C 2 /J  s,
2
 I  S average  (1.33 10 3 )nE02
Watt/m 2 (6)
in N/C (or V/m)
78
Instantaneous irradiance and
average irradiance
S  v 0 r Ex2  v 0 r E02 cos 2 (t ) (instantan eous irradiance )
1
I  Saverage  v 0 r E02 (average irradiance )
2
• S can be measured only if the power meter can
respond more quickly than the oscillations of
electric field.
• Optical frequency ~ 1014 Hz
 All detectors have a response rate much slower
than the frequency of the wave.
 All practical measurements invariably yield Saverage.
79
EXAMPLE 1.4.1 Electric and
magnetic fields in light
• Laser beam from a He-Ne laser
Intensity (average irradiance) I = 1 mW/cm2
• E0 = ? , B0 = ? in air
• E0 = ? , B0 = ? in a glass (n = 1.45)
80
EXAMPLE 1.4.1 Electric and
magnetic fields in light
• Solution
The intensity (average irradiance)
1
2I
2
I  Saverage  c 0 nE0  E0 
2
c 0 n
• In air, n = 1
E0 
2  (1 10 3  10 4 Wm -2 )
(3  108 m/s)(8.85  10 12 Fm -1 )(1)
 87 Vm -1 or 0.87 Vcm -1
B0  E0 / c  (87 Vm -1 ) /(3  108 m/s)  0.29 μT
(recall E0  vB0 )
81
• In a glass medium, n = 1.45
E0 (medium ) 
2  (110 3 10 4 Wm -2 )
(3 108 m/s)(8.85 -2 10 12 Fm-1 )(1.45)
 72 Vm -1
B0 (medium)  E 0 /v  nE0 /c  (1.45)(72 Vm -1 ) /(3 108 m/s)
 0.35 μT
82
1.5
SNELL’S LAW AND TOTAL
INTERNAL REFLECTION (TIR)
83
A light wave suffers reflection and
refraction at the boundary
• Consider a traveling plane EM
wave in medium (1) (index n1) is
propagating towards a medium
(2) (index n2)
• Incident light in medium (1)
– wave vector ki
– incident angle i
• Refracted light (transmitted
wave) in medium (2)
– wave vector kt
– refracted angle t
• Reflected light in medium (1)
– wave vector kr
– reflected angle r
84
Reflected light
• kr = ki = 2 / 
since both the incident and
reflected waves are in the same
medium
• The two waves along Ai, Bi are in
phase.
 the reflected waves along Ar, Br
must still be in phase, otherwise
they will interfere destructively
and destroy each other.
i
r
Ai A  AA  AAr  Bi B  BB  BBr
 AA  BB
since AA  AB sin  r , BB  AB sin  i
  r  i
law of reflection
85
Refracted light
• The wavefront AB in medium 1
becomes the front A'B' in medium 2.
 Time for the phase A on wave Ai to
reach A' = Time for the phase B on
wave Bi to reach B'
BB '  v 1t  ct / n1 , AA'  v 2t  ct / n2
• From geometrical considerations
t
i
r
AB'  BB ' / sin i , AB'  AA' / sin t
• So that
v 1t
v t
 2
sin  i sin  t
sin  i v 1 n2



sin  t v 2 n1
AB' 
Snell’s law (law of refraction)
86
Critical angle
• Snell’s law:
sin  i n2

sin  t n1
• If n1 > n2  t > i
• When t = 90
 the incident i is called the critical angle c
n2
sin  c 
n1
87
Total Internal Reflection (TIR)
• When i > c
 there is no transmitted wave
but only a reflected wave
• This phenomena is called
total internal reflection.
• TIR phenomenon leads to
the propagation of waves in
a dielectric medium
surrounded by a medium of
smaller refractive index.
sin  c 
n2
n1
– Ex., optical fibers
88
Evanescent wave
• When i > c
n1
n1
sin  t  sin  i  sin  c  1
n2
n2
 sin  t  1
  t : imaginary angle of refraction
• There is a wave that
propagates along the boundary
called the evanescent wave.
sin  c 
n2
n1
89
1.6
FRESNEL’S EQUATIONS
90
Augustin Jean Fresnel
•
Augustin Jean Fresnel (pronounced
/frei’nl/ fray-nell) (1788 - 1827) was a
French physicist, and a civil engineer for
the French government, who was one
of the principal proponents of the
wave theory of light. He made a
number of distinct contributions to optics
including the well-known Fresnel lens
that was used in light houses in the 19th
century. He fell out with Napoleon in
1815 and was subsequently put into
house-arrest until the end of Napoleon's
reign. During his enforced leisure time he
formulated his wave ideas of light into a
mathematical theory.
•
“IF you cannot saw with a file or file with a
saw, then you will be no good as an
experimentalist”
91
1.6 FRESNEL’S EQUATIONS
A. Amplitude Reflection and Transmission
Coefficient (r and t)
92
Incidence, reflection, and refraction of light
at a boundary
• Plane of incidence: the plane that contains the incident and the
reflected rays.
• We can resolve the field E into two components:
 Ei , // : parallel to the plane of incidence
Incident field Ei  
 Ei , : prependicu lar to the plance of incidence
 Er , //
 Et , //
Reflected field Er  
Transmitte d wave Et  
 Er , 
 Et ,
93
TE and TM waves
• TE waves (transverse electric field waves)
– Waves with Ei, , Er,, and Et,, have only their electric
field components perpendicular to the plane of incidence.
• TM waves (transverse Magnetic field waves)
– Waves with Ei,// , Er,//, and Et,,// have only their magnetic
field components perpendicular to the plane of incidence.
94
Exponential representation of
traveling wave
Ei  Ei 0 exp j (t  k i  r )
Er  Er 0 exp j (t  k r  r )
Et  Et 0 exp j (t  k t  r )
• Er0, Et0 : complex amplitudes (including the
phase changes r , t respect to the phase
of incident wave)
95
Boundary conditions
•
•
•
•
E tangential(1) = Etangential(2)
H tangential(1) = Htangential(2)
B = 0 r H B tangential(1)r1 = Btangential(2)r2
For nonmagnetic media
– Magnetic susceptibility |m| << 1 (M  m H)
– Relative permeability r1 = 1+ m1  r2 =1+ m1  1
B tangential(1) = Btangential(2)
96
Fresnel’s equations for E
• Reflection coefficient for E
r 
Er 0 , 
Ei 0,
cos  i  [n 2  sin 2  i ]1/ 2

cos  i  [n 2  sin 2  i ]1/ 2
(1a )
• Transmission coefficient for E
t 
Et 0,
Ei 0,
2 cos  i

cos  i  [n 2  sin 2  i ]1/ 2
(1b)
– where n  n2/n1 is the relative refractive index
• r + 1 = t
97
Fresnel’s equations for E//
• Reflection coefficient for E//
Er 0, // [n 2  sin 2  i ]1/ 2  n 2 cos  i
r// 
 2
Ei 0, // [n  sin 2  i ]1/ 2  n 2 cos  i
(2a)
• Transmission coefficient for E//
t // 
Et 0, //
Ei 0, //
2n cos  i
 2
2
2
1/ 2
n cos i  [n  sin  i ]
(2b)
– where n  n2/n1 is the relative refractive index
• r// + nt// = 1
98
Phase changes
• We can take Ei0 to be a real number so that
the phase angles of r and t correspond to
the phase changes measured with respect
to the incident wave.
• Ex., r  r exp( j )
r : relative amplitude,  : relative phase
• When r is real
– If r > 0,  r =| r | and  =0 (no phase shift)
– If r < 0,  r =  | r | and  =180 (phase shift = )
99
r 
Er 0 , 
r// 
Er 0, //
Ei 0,
Ei 0, //
cos  i  [n 2  sin 2  i ]1/ 2

,
2
2
1/ 2
cos  i  [n  sin  i ]
t 
Et 0,
Ei 0,

2 cos  i
cos  i  [n 2  sin 2  i ]1/ 2
Et 0, //
[n 2  sin 2  i ]1/ 2  n 2 cos  i
2n cos  i
 2
,
t


//
[n  sin 2  i ]1/ 2  n 2 cos  i
Ei 0, // n 2 cos  i  [n 2  sin 2  i ]1/ 2
• If complex coefficients  [n2  sin2i ]1/2 is imaginary
 n2  sin2i = (n2 /n1)2  sin2i < 0  (n2 /n1)2 < sin2i  1
n2 / n1  1

sin  c  sin i
(sin  c  n2 / n1 )
n1  n2
    (critical angle)
c
 i
 Conditions for total internal reflection
 the phase changes other than 0 or 180 occur only when
there is total internal reflection
100
c =
Critical angle
sin-1(n
2/n1)
= 43.98
r  r exp( j )
44
Brewster’s angle
101
Normal incidence (i = 0)
• From Fresnel’s equation
n1  n2
r//  r 
 0 (for n1  n2 )
n1  n2
//    0  no phase change
102
Brewster’s angle (polarization angle)
[n 2  sin 2  i ]1/ 2  n 2 cos  i
 2
[n  sin 2  i ]1/ 2  n 2 cos  i
r// 
Er 0, //
If
r//  0 for  i   p
Ei 0, //
( 2a )
 [n 2  sin 2  p ]1/ 2  n 2 cos  p  0
 n 2  sin 2  p  n 4 cos 2  p  n 4 [1  sin 2  p ]  n 4  n 4 sin 2  p
 (n 4  1) sin 2  p  n 4  n 2  n 2 (n 2  1)
n 2 (n 2  1)
n 2 (n 2  1)
n2
 sin  p 
 2
 2
4
2
n 1
(n  1)( n  1) (n  1)
n
 sin  p 
 tan  p  n
2
n 1
n
This special angle is called the polarization
 tan  p  2
n1
angle or Brewster’s angle.
2
103
David Brewster
• David Brewster (1781-1868), a British physicist,
formulated the polarization law in 1815.
104
r  r exp( j )
44
Brewster’s angle
p = tan-1 (n2/n1) = 34.8
105
Linearly polarized wave
• For i = p (p = tan-1 (n2/n1))
The field in the reflected wave is always
perpendicular to the plane of incidence.
The reflected wave is a linearly polarized
wave.
Electric field oscillations are contained
with a well defined plane.
106
From Fresnel equation
r// 
Er 0, //
Ei 0, //
[n 2  sin 2  i ]1/ 2  n 2 cos  i
 2
[n  sin 2  i ]1/ 2  n 2 cos  i
( 2a )
For  p   i   c  r//  0
 a phase shift  180
107
r 
Er 0 , 
Ei 0,
cos  i  [n 2  sin 2  i ]1/ 2

cos  i  [n 2  sin 2  i ]1/ 2
(1a)
For  i   c  r  r exp( j )
 r  1  Total Internal Reflection

2
2 1/ 2
  1  [sin  i  n ]
 tan    
 0    180
cos  i
 2 
r  1 exp( j )  a complex number

108
r// 
Er 0, //
Ei 0, //
[n 2  sin 2  i ]1/ 2  n 2 cos  i
 2
[n  sin 2  i ]1/ 2  n 2 cos  i
( 2a )
For  i   c  r//  r// exp( j// )
 r//  1  Total Internal Reflection

  [sin 2  i  n 2 ]1/ 2 1
1
 1
 tan  //   

tan(
 )  180  //  0
2
2
2
n cos  i
n
2
 2
r//  1 exp( j// )  a complex number

109
• The fact that // has an additional  shift
that makes // negative for i > c is due to
the choice for the direction of the reflected
optical field Er,// in figure 1.11. This  shift
can be ignored if we simply invert Er,// .
110
What happens to the transmitted
wave when i > c ?
• There is a wave traveling near the surface of
boundary along the z direction
 an evanescent wave
Et , ( y, z, t )  e  2 y exp j (t  kiz z )
where kiz  ki sin  i
the wave vector of the incident
wave along the z-direction
 2 : the attenuatio n coefficien t
2n2 n1 2 2
2 
[( ) sin  i  1]1/ 2 ,  : free space wavelengt h

n2
e
 2 y
: amplitude decays exponentia lly along y
111
Penetration depth
Et , ( y, z , t )  e  2 y exp j (t  kiz z )
2n2
n1 2 2
1/ 2
2 
[( ) sin  i  1]

n2
If y 
1
2
   e  2  e 1
  : penetratio n depth
1



 2 2n [( n1 ) 2 sin 2   1]1/ 2
2
i
n2
112
Internal and external reflection
• If n1 > n2  internal reflection (e.g., glass  air)
– normal incidence no phase change
• If n1 < n2  external reflection (e.g., air  glass)
– for normal incidence:
• r < 0, r// < 0  phase shift 180
– for i = p (Brewster angle tanp = n2/n1):
• r// = 0  reflected wave is polarized in the E component
113
Transmitted light
t 
Et 0,
t // 
Et 0, //
Ei 0,
Ei 0, //
2 cos  i

2
2
1/ 2
cos  i  [n  sin  i ]
(1b)
2 cos  i
 2
n cos  i  [n 2  sin 2  i ]1/ 2
(2b)
• For both n1 > n2 (internal reflection when i
< c) and n1 < n2 (external reflection)
 t, t// > 0
 phase shift = 0
114
EXAMPLE 1.6.1 Evanescent wave
• Total internal reflection (TIR) of a light from
a boundary (n1 > n2 ) is accompanied by a
evanescent wave propagation in medium
2 near the boundary.
• Find the functional form of the evanescent
wave.
115
EXAMPLE 1.6.1 Evanescent wave
• Solution
The transmitted wave have general form
y
Et ,  t  Ei 0, exp j (t  k t  r )
ktsint
ktcost t
the transmission coefficient
 k t  kt cos  t yˆ  kt sin  t zˆ
 k t  r  ykt cos  t  zkt sin  t
n2
y

x
z
i r
kt
z
n1
 Et ,  t  Ei 0, exp j (t  ykt cos  t  zkt sin  t )
116
From Snell' s law
n1
n1 sin  i  n2 sin  t  sin  t  sin  i
n2
if n1  n2 and  i   c
n1
(recall sin  c  1)
n2
n1
 sin  t  sin  i  1
n2
 cos  t  1  sin  t   jA2
2
pure imaginary number
117
Et ,  t  Ei 0, exp j (t  ykt cos  t  zk t sin  t )
cos  t  1  sin 2  t   jA2
Taking cos  t   jA2
jA2 must be ignored because it implies
a wave with growing amplitude
Et ,  t  Ei 0, exp j (t  ykt ( jA2 )  zk t sin  t )
 t  Ei 0, exp(  2 y ) exp j (t  zk t sin  t )
where  2  kt A2 

2n2

2n2

sin 2  t  1
(n1 / n2 ) 2 sin 2  i  1
118
Et ,  t  Ei 0, exp j ( 2 y ) exp j (t  zkt sin  t )
• The traveling wave part
exp j (t  zk t sin  t )
 exp j (t  zk i sin  i )
Snell’s law kt sint  ki sini
 exp j (t  kiz  z )
The evanescent wave
propagates along z at the same
speed as the incident and
reflected wave along z.
y
n2
kiz
y

x
z
i r
z
ki
kr
n1
kisini  kiz
119
Et ,  t  Ei 0, exp j ( 2 y ) exp j (t  kiz z )
• The transmission coefficient is
t 
Et 0,
Ei 0,
2 cos  i

cos  i  [n 2  sin  i ]1/ 2
2 cos  i

cos  i  [( n2 / n1 ) 2  sin  i ]1/ 2
 t 0 exp( j  )
For TIR , sini > n2 / n1
 [(n2/n1)2  sin2i ] < 0
a complex number
t0 : a real number,  : a phase change
• Note that t does not change the general behavior of
propagation along z and the penetration along y.
120
Evanescent wave
Et ,  t  Ei 0, exp j ( 2 y ) exp j (t  kiz z )
2 cos  i
where t  
1/ 2
2
cos  i  [( n2 / n1 )  sin  i ]
2 
2n2

kiz  ki sin  i
(n1 / n2 ) sin  i  1
2
2
y

x
z
121
1.6 FRESNEL’S EQUATIONS
B. Intensity, Reflectance, and Transmittance
122
Intensity (Irradiance) I
• I  the energy flow per unit time per unit area
• In medium with a velocity v, relative
permittivity r, the light intensity (or irradiance)
1
1
2
2
I  v 0 r E0  c 0 nE0
2
2
1
 0 nE02 : the energy in the field per unit volum e
2
123
Reflectance R
• R measures the intensity of the reflected light
with respect to that of the incident light
I r cos  r I r | Er |2
1
2
R
 
(

I

c

nE
0
0)
2
I i cos  i I i | Ei |
2
R 
| Er 0 ,  |2
R // 
| Er 0, // |2
| Ei 0, |2
| Ei 0, // |2
| r |2
| r// |2
r, r//: complex numbers
R, R//: real numbers
124
Normal incidence
 n1  n2 
• R  R   R //  

n n 
 1 2
2
• Ex., glass n2 = 1.5, air n1  1
– Reflectance from air-glass surface
 1  1.5 
R 
  0.04  4 %
 1  1.5 
2
125
Transmittance T
• T measures the intensity of the transmitted
wave to that of the incident wave.
I t cos  t
T 
I i cos  i
n2 cos  t
T 
n1 cos  i
 E0 t

 E0 i



2
1
( I  c 0 nE02 )
2
126
Normal incidence (i = t = 0)
n2 | Et 0, |2
n2
2
T 
 | t |
2
n1 | Ei 0, |
n1
n2 | Et 0, // |
2
n2
2
T // 

|
t
|
//
2
n1 | Ei 0, // |
n1
and | t  | | t // | 
2
2
4
(1  n2 / n1 ) 2
4n1n2
T T  T // 
(n1  n2 ) 2
 R T  1
127
EXAMPLE 1.6.2 Reflection of light from
a less dense medium (internal reflection)
• light  = 1m (in free space)
• glass 1 (n1 = 1.450)  glass 2 (n2 = 1.430)
a. What should the minimum incident angle for
TIR be ?
b. What is the phase change in the reflected
wave when i = 85and when i = 90 ?
c. What is the penetration depth of the
evanescent wave into medium 2 when i = 85
and i = 90 ?
128
EXAMPLE 1.6.2 Reflection of light from
a less dense medium (internal reflection)
• Solution
a. The minimum incident angle for TIR is
the critical angle c
n2 1.43
sin  c 

n1 1.45
1.43

  c  sin (
)  80.47
1.45
1
129
EXAMPLE 1.6.2 Reflection of light from
a less dense medium (internal reflection)
•
b.


Solution
For i = 85 > c = 80.47  TIR
There is phase shift in the reflected wave
The phase change in Er, is given by 
2
2 1/ 2
 1  [sin  i  n ]
tan    
(from Eq.(6))
cos  i
2 
1.430 2 1/ 2
[sin 2 (85 )  (
) ]
1.450

 1.61447
cos  i
1
  tan 1 (1.61447)  58.226
2
  116.45
130
EXAMPLE 1.6.2 Reflection of light from
a less dense medium (internal reflection)
•
Solution
The phase change in E r,// is given by //
1  [sin 2  i  n 2 ]1/ 2 1
1
1
tan  //    

tan(
 )
2
2
2 
n cos  i
n
2
2
n1 2
1
1.450 2
 ( ) tan(  )  (
) (1.61447)  1.65995
n2
2
1.430
(from Eq. (7))
1
1
 //    tan 1 (1.65995)  58.934
2
2

 //  2  (58.934  )  62.13
2

131
EXAMPLE 1.6.2 Reflection of light from
a less dense medium (internal reflection)
•
Solution
For i = 90
2
2 1/ 2
 1  [sin  i  n ]
tan    
(from Eq.(6))
cos  i
2 
1  1
1
1
tan  //     2 tan(  ) (from Eq. (7))
2  n
2
2
1.43 2 1/ 2
[12  (
) ]
2
2 1/ 2
[sin


n
]
1.45
i
  2 tan 1[
]  2 tan 1[
]  180
cos  i
0
1
1

//  2[tan [ 2 tan(  )]  ]  0
n
2
2
1
132
EXAMPLE 1.6.2 Reflection of light from
a less dense medium (internal reflection)
• Solution
c. The evanescent wave in medium 2
Et , ( y, t , z )  t  Ei 0, exp(  2 y ) exp j (t  kiz z )
2n2
•
n1 2 2
 2
[( ) sin  i  1]1/ 2

n2
For i = 85
2 (1.430) 1.450 2 2 
1/ 2
6
-1
 2
[(
)
sin
(
85
)

1
]

1
.
28

10
m
1.0 10 6 m 1.430
1
1
7



7
.
8

10
m  0.78m
6
-1
 2 1.28 10 m
133
EXAMPLE 1.6.2 Reflection of light from
a less dense medium (internal reflection)
•
Solution
For i = 90
 2 1.5 10 m   
6
-1
1
2
 0.66m
  ( i 85 )   ( i 90 )
 The penetration is greater for small incidence
angles.
• This will be an important consideration later in
analyzing light propagation in optical fibers.
134
EXAMPLE 1.6.3 Reflection at normal
incidence. Internal and external reflection
• Normal incidence
a. If light : air (n = 1)  glass (n = 1.5)
Reflection coefficient r//, r ?
Reflectance R?
b. If light : glass (n = 1.5)  air (n = 1)
r//, r, R ?
c. What is the polarization angle in the external
reflection in a above ?
How would you make a polarized device that
polarizes light based on the polarization angle?
135
EXAMPLE 1.6.3 Reflection at normal
incidence. Internal and external reflection
Solution
a. Normal incidence i = 0
air (n1 = 1 )  glass (n2 = 1.5) (external reflection)
n1  n2 1  1.5
r   r // 

 0.2 ( 0, phase shift  180 )
n1  n2 1  1.5
Reflectanc e R  |r // |  0.04  4%
2
136
EXAMPLE 1.6.3 Reflection at normal
incidence. Internal and external reflection
Solution
b. Normal incidence i = 0
glass (n1 = 1.5 )  air (n2 = 1) (internal reflection)
n1  n2 1.5  1
r   r // 

 0.2( 0, no phase shift)
n1  n2 1.5  1
Reflectanc e R  |r // |  0.04  4%
2
137
EXAMPLE 1.6.3 Reflection at normal
incidence. Internal and external reflection
Solution
c. For i = p (polarization angle)
n2 1.5
r//  0, tan  p  
n1
1
  p  tan 1 (1.5)  56.3 Brewster' s angle
•
If i = p
 The reflected is polarized with E  plane of incidence.
 The transmitted light is partially polarized with the
field greater in the plane of incidence.
•
Pile-of-plates polarizer
 It can increase the polarization of the transmitted light
138
EXAMPLE 1.6.4 Antireflection coating on
solar cells
•
•
air: n1  1, Si: n2  3.5 @  = 700 ~ 800 nm
When light is incident on the surface of a
semiconductor
– Reflectance
2
 n1  n2   1  3.5 
  
R  
  0.309
 n1  n2   1  3.5 
2
 31% of light is reflected and is not available
for conversion to electric energy.
139
EXAMPLE 1.6.4 Antireflection coating on
solar cells
Q: How to reduce the reflected light intensity?
A: Coat the surface with a thin film layer of
dielectric material such as Si3N4 (silicon nitride),
that has an intermediate refractive index.
140
EXAMPLE 1.6.4 Antireflection coating on
solar cells
•
•
•
wave A  phase change = 180 (n2 > n1)
wave B  phase change = 180 (n3 > n2)
Phase difference between waves A and B
  2
2d
c
 k c ( 2d )
(kc  2 / c is the wave number in the coating )
c   / n2 ( c is the wavelengt h in the coating ))
2n2
   (
)( 2d )

141
EXAMPLE 1.6.4 Antireflection coating on
solar cells
•
To reduce the reflected light, waves A and B must
interfere destructively.
  (
2n2

 d  m(

)( 2d )  m

4n2
)m
c
4
(m  1,3,5,)
(m  1,3,5,)
The thickness of the coating
must be multiples of the
quarter wavelength in the
coating.
142
EXAMPLE 1.6.4 Antireflection coating on
solar cells
•
To obtain a good degree of destructive interference
between A and B, the two amplitude must be comparable.
It turns out we need
n2  n1n3
n1  n2 n1  n1n3 1  n3 / n1
r (air - coating ) 


n1  n2 n1  n1n3 1  n3 / n1
r (coating - semiconduc tor ) 
n n  n3
n / n  1 1  n3 / n1
n2  n3
 1 3
 1 3

n2  n3
n1n3  n3
n1 / n3  1 1  n3 / n1
 r (air - coating )  r (coating - semiconduc tor )

The reflection coefficient between air and coating is
equal to that between coating and semiconductor.
143
EXAMPLE 1.6.4 Antireflection coating on
solar cells
• n1 (air ) =1, n3(Si) = 3.5
n2  n1n3  1 3.5  1.87
• n (Si3N4)  1.9
 Si3N4 is a good choice as an antireflection
coating material on Si solar cells.
• Taking  = 700 nm, the thickness of coating

700 nm
d  m(
)  m(
)  m(92.1 nm), m  1,3,5,
4n2
4 1.9
144
EXAMPLE 1.6.5 Dielectric mirrors
• A dielectric mirror consists of a stack of dielectric layers
of alternating refractive indices.
Index : n1 < n2 , Thickness: d = layer/4
 Reflective waves from the interfaces interfere
constructively and give rise to a substantial reflected light.
 If there are sufficient number of layers, the reflectance
can approach unity at the wavelength 0.
n1 < n2
145
EXAMPLE 1.6.5 Dielectric mirrors
• Reflection coefficients
n1  n2
r12 
 0 (phase change  180 )
n1  n2
n2  n1
r21 
 r12  0 (phase change  0 )
n2  n1
• Wave B travels an additional distance
  2(2 / 4)  2 / 2
 Phase difference between waves A and B
      2
due to reflections at different boundaries
due to wave B travels an additional distance
146
EXAMPLE 1.6.5 Dielectric mirrors
• Waves A and B interfere constructively
• Waves B and C interfere constructively
• 
After several layers, R  1.
• Dielectric mirrors are widely used in modern
vertical cavity surface emitting semiconductor
lasers (VCSEL).
147
1.7
MULTIPLE INTERFERENCE
AND OPTICAL RESONATORS
148
Charles Fabry and Alferd Perot
• Charles Fabry (1867-1945), left, and Alfred Perot (18631925),right, were the first French physicists to construct an
optical cavity for interferometry.
149
Fabry-Perot optical cavity
• Two flat mirrors are perfectly aligned to be
parallel with free space between them.
• Waves reflected from M1 interfere with waves
reflected from M2.
A series of allowed stationary or standing EM
wave in the cavity.
150
Cavity modes
•
E = 0 at the mirrors (assume metal coated), we can only fit in a integer number of halfwavelength into cavity length:
L = m (/2), m = 1,2,3 …
•
Each particular allowed m for a given m defined a cavity mode:
m = 2L/m, m = 1,2,3 …
•
Resonant frequencies:
m = c/m = m(c/2L) = mf , f = c/2L : the lowest frequency (fundamental mode)
•
Frequency separation of two neighboring modes:
m = m+1  m =f  free spectral range.
151
Fabry-Perot optical resonator
• If the mirrors are perfectly reflecting, no losses
from the cavity.
The peaks at frequencies m would be sharp.
• If some radiation escapes from the cavity
 Peaks have a finite width.
• This cavity with its mirrors (etalon) store
radiation energy only at certain frequencies and
is called a Fabry-Perot optical resonator.
152
• Wave B has one round-trip phase
difference (k (2L)) and a magnitude r2 with
respect to A.
• When A and B interfere, the result is
A  B = A +Ar2exp(j2kL)
M1 and M2 are identical with a
reflection coefficient of magnitude r
153
Resultant field
• After infinite round-trip reflections
Ecavity  A  B  
 A  Ar 2 exp(  j 2kL)  Ar 4 exp(  j 4kL)
 Ar 6 exp(  j 6kL)  
• Sum of geometric series
a  ar  ar    ar
2
• The resultant field
s 1

    ar
k 1
s 1
a

1 r
A
Ecavity 
1  r 2 exp(  j 2kL)
154
Cavity Intensity
• Reflectance R = r2
•
A
A
| Ecavity | 

2
1  r exp(  j 2kL) 1  r 2 exp(  j 2kL)
2
A2

1  r 4  r 2 exp(  j 2kL)  r 2 exp(  j 2kL)
A2
A2


2
1  R  R[2 cos( 2kL)] 1  R 2  R[2  4 sin 2 (kL)]
A2

(1  R) 2  4 R sin 2 (kL)
cos2 = 12sin2
• Intensity Icavity  |Ecavity|2
I cavity 
I0
(1  R) 2  4 R sin 2 (kL)
155
Maximum cavity intensity
• I cavity 
I0
(1  R)  4 R sin (kL)
2
2
• If sin2(kL) = 0  Icavity = Imax
sin (kL)  0  kL  m
m
m

L

 m( ) m  1,2,3,
k
2 / 
2
I0
• I max 
;
k m L  m
2
2
(1  R)
 If R  then I max 
156
Cavity Intensity
• I cavity 

I0
(1  R) 2  4 R sin 2 (kL)
I0
2L
(1  R)  4 R sin (
)
c
2
2
Smaller R values result in broader mode
peaks and a smaller difference between
the minimum and maximum intensity.
• Spectral width m = the full width at half
maximum (FWHM)
157
Spectral width and finesse
•  m 
f
R
1/ 2
/(1  R)

f
F
f
R1/ 2
F

 m 1  R
• F : the finesse of the resonator
F  the ratio of mode separation (m= f) to
spectral width (m)
• R increases (losses decrease)
F increases  m decreases
sharper mode peaks
158
Interference filters
• If the incident beam has a wavelength
corresponding to one of the cavity modes,
it can sustain oscillations in the cavity and
hence lead to a transmitted beam.
159
Transmitted intensity
•
I transmitted  (1  R)I cavity  (1  R)
I0
(1  R) 2  4 R sin 2 (kL)
(1  R) I incident

2
2
(1  R)  4 R sin (kL)
2
I0 = (1R) Iincident
which is maximum just as for Icavity whenever kL = m
160
EXAMPLE 1.7.1 Resonator modes
and spectral width
• Fabry Perot optical cavity of air
L = 100m, R = 0.9
(a) The cavity mode nearest to 900 nm?
(b) The separation of the modes?
(c) The spectral width of each mode?
161
EXAMPLE 1.7.1 Resonator modes
and spectral width
• Solution
Cavity modes
m( / 2)  L,
m  1,2,3,
2(100 10 6 m)
m

 222.22
9

900 10 m
Take m  222
2L
6
2 L 2(100  10 m)
222 

 900.90 nm
m
222
c
3  108 m/s
  222 

 3.33 1014 Hz
222 900.90 nm
162
EXAMPLE 1.7.1 Resonator modes
and spectral width
• Solution
Separation of the modes
c
3 10 m/s
12
 m   f 

 1.5 10 Hz
-6
2 L 2 100 10 m
8
163
EXAMPLE 1.7.1 Resonator
modes and spectral width
• Solution
Separation of the modes
c
3 10 m/s
12
 m   f 

 1.5 10 Hz
-6
2 L 2 100 10 m
8
164
EXAMPLE 1.7.1 Resonator
modes and spectral width
• Solution
Finesse
F 
R1/ 2

 (0.9)1/ 2
1 R
Spec tral width
1  0.9
 29.8
f
1.5 1012 Hz
 m 

 5.03 1010 Hz
F
29.8
The co rrespondin g spectral wavelengt h width
3  108 m/s
10
m  |  ( ) | |  2 |  m 
5
.
03

10
Hz
14
2
m
m
(3.33 10 Hz)
c
c
 0.136 nm
165
1.8
GOOS-HÄNCHEN SHIFT AND
OPTICAL TUNNELING
166
GOOS-HÄNCHEN SHIFT
• IF n1 > n2 and i > c then TIR
• Simply ray trajectory analysis:
– the reflected ray emerges from the
point of contact of the incident ray
with the interface.
• Careful optical experiments:
– the reflected wave appears to be
laterally shifted from the point of
incidence at the interface
n1 > n2
167
GOOS-HÄNCHEN SHIFT
• The reflected beam appears to be reflected from a
virtual plane inside the optically less dense medium.
• Phase change ( 0 <  < 180) and penetration into the
second medium for TIR  Shifting of the reflected wave
along the direction of the evanescent wave.
i = r
168
GOOS-HÄNCHEN SHIFT
• z = 2 tani
• Ex.,  = 1 m, i = 85, glass-glass (n1 = 1.45, n2 =
1.43) interface

1
2

2n2
1
(n1 / n2 ) 2 sin 2  i  1
 0.78 μm

Δz  2 tan  i  2(0.78 μm)tan(85  )  18μm
169
Optical tunneling
• If B is sufficiently thin, an attenuated light emerges on
the other side of B in C.
 An incident wave is partially transmitted through a
medium where it is forbidden in terms of simple
geometrical optics.
• Optical tunneling is due to the fact that the field of the
evanescent wave penetrates into B and reaches the
interface BC.
170
Frustrated total internal
reflection (FTIR)
• The proximity of medium C frustrated TIR.
The transmitted beam in C carries
some of the light intensity
The intensity of the reflected
beam is reduced.
171
Beam splitters
• FTIR at the hypotenuse face of A leads to
a transmitted beam and hence to the
splitting of the incident beam into two
beams.
172
1.9
TEMPORAL AND SPATIAL
COHERENCE
173
Perfect coherence
• Pure sinusoidal wave
E x  E0 sin( 0t  k0 z )
(1)
angular frequency 0  20
wave number k0
• The wave extends infinitely over all space and
exists at all times.
• This sine wave is perfect coherent because we
can predict the phase of any portion of the wave
from any other portion of the wave.
174
Temporal coherence
• Temporal coherence measures the extent to which
two points, such as P and Q, separated in time at a
given location in space can be correlated.
• At a given location, for a pure sine wave, any two
points such as P and Q separated by any time
interval are always correlated because we can
predict the phase of Q from the phase of P for any
temporal separation.
175
• Any time dependent arbitrary function f(t) can be
presented by a sum of pure sinusoidal waves:

f (t )   f () sin( t )d
0
f (): the spectrum of f(t), presents the
amplitudes of various sinusoidal oscillation.
• We need only one sine wave at 0 = 0/2 to
make up Ex  E0 sin( 0t  k0 z )
176
Finite wave train
• In practice a wave can exits only over a finite time
duration t  wave train of length l = c t
• We can only correlate points in the wave train within the
duration t or over the spatial extent l = c t.
t
coherence time

coherence length l  ct
• It contains a number of different frequencies in its
spectrum  center frequency 0 = c t,
• spectral width
1
 
t
177
Coherence time and coherence length
• Sodium lamp
–  = 589 nm (orange), spectral width   5 1011 Hz
1
1

12
t 


2

10
s  2 ps
coherence time
11

 5 10 Hz
8
12
4

coherence
length
l

c

t

(
3

10
m/s)(
2

10
s)

6

10
m

• He-Ne laser (in multimode)
– spectral width  1.5109 Hz
1
1

10
t 


6
.
67

10
s
coherence time
9

 1.5 10 Hz
8
10

coherence
length
l

c

t

(
3

10
m/s)(6.67

10
s)  0.2m  200 mm

178
Continuous wave (CW) laser
• A continuous wave (CW) laser operating in a
single mode
very narrow linewidth 
long coherence length:
l = ct = c/ ~ several 102 m
• It can be widely used in wave-interference
studies and applications.
179
White light signal
• Given a point P on this waveform, we can not
predict the phase of the signal at any other point
Q.
• No coherence  White noise
• The spectrum contains a wide range of
frequencies.
• White light is an idealization because all
frequencies are present in the light beam.
180
Mutual temporal coherence
• The waves A and B
– same frequency 0
– interfere only over the time interval t
• They have mutual temporal coherence over t.
• Ex. Two identical wave-trains travel different
paths and then arrive at the destination. They
can interfere only over a space portion ct.
181
Spatial coherence
• Spatial coherence describes the extent of coherence
between waves radiated from different locations on a
light source.
• If the waves emitted from locations P and Q on the
source are in phase, then P and Q are spatially coherent.
• A spatially coherent source emits waves that are in
phase over its entire emission surface.
182
Incoherent beam
• A incoherent beam contains waves that have
very little correlation with each other.
• A incoherent beam contains waves whose
phase change randomly at random time.
• However, there may be a very short time interval
over which there is a little bit of temporal
coherence.
183
1.10
DIFFRACTION PRINCIPLES
184
1.10 DIFFRACTION PRINCIPLES
A. Fraunhofer Diffraction
185
Diffraction effects
• An important property of waves.
• It can be understood in terms of the
interference of multiple waves
emanating from the obstruction.
• e.g.,
– Sound waves are able to bend (deflect
around) corners.
– A light beam can similarly “bend” around an
obstruction.
186
Diffraction phenomena
• Fraunhofer diffraction
– the incident light beam is a plane wave (a
collimated light beam)
– the observation or detection of the light
intensity pattern is done far away from the
aperture so that the waves received also look
like plane waves.
• Fresnel diffraction
– the incident light beam and the received light
waves are not plane waves but have
significant curvatures.
187
Diffraction pattern
• A collimated light beam incident on a small circular
aperture becomes diffracted.
• Its light intensity pattern after passing through the
aperture is a diffraction pattern with circular bright rings
• Airy rings
• If the screen is far away from the aperture, this would be
a Fraunhofer diffraction pattern.
188
Huygens-Fresnel principle
• Every unobstructed point of a wavefront,
at a given instant in time, serves as a
source of spherical secondary waves (with
the same frequency as that of the primary
wave).
• The amplitude of the optical field at any
point beyond is the superposition of all
these wavelets (considering their
amplitudes and relative phases).
189
Huygens-Fresnel principle
• When the plane wave
reaches the aperture, points
in the aperture become
sources of coherent spherical
secondary waves,
• These spherical waves
interfere to constitute the new
wavefront.
• The new wavefront is the
envelope of the wavefronts of
the secondary waves.
190
Huygens-Fresnel principle
• These spherical waves can interfere
constructively not just in the forward
direction as in (a) but also in other
appropriate directions as in (b), giving
rise to the observed bright patterns on
the observation screen.
191
Single slit diffraction
• Aperture is divide into N point sources: y  a/N
• Amplitude of each point source  y
• Wave emitted from point source at y:
E  y exp(jkysin)
• All of those waves from
point sources from y =
a/2 to y = a/2 interfere at
the screen.
192
Resultant field at the screen
E ( )   Cy exp(  jky sin  )
y a / 2
C
 exp(  jkysin  )dy
y  a / 2
Let x   jky sin   uy  u   jk sin  , dy  dx / u
E ( )  C
au / 2

 au / 2
ex
1
C
2C
au
dx  [e au / 2  e  au / 2 ] 
sinh( )
u
u
u
2
2C
 jka sin 
sinh(
)
 jk sin 
2
2C
 ka sin 

j sin(
) ( sinh( jx)  j sin x)
 jk sin 
2
Ca sin[( ka / 2) sin  ]

(ka / 2) sin 

193
Single slit diffraction equation
• The light intensity I at the screen:
I ( )  |E ( ) |2
 C ' a sin[( ka / 2) sin  ] 

 I ( )  
(ka / 2) sin 


 sin (  ) 

 I (0)
  
 I (0)sinc 2 (  )
2
2
where C'  constant,
  (ka / 2) sin  ,
sinc (  )  sin (  ) / 
194
Diffraction pattern from a single slit
• Bright regions  constructive interference
• Dark regions  destructive interference
• Width of the center bright region c > slit width a
 the transmitted beam must be diverging.
• Zero intensity occurs when
(ka / 2) sin   m , m  1,2, 
 sin   2m / ka  m / a
 divergence 2  /a
• Ex.,  = 1300 nm, a = 100 m
 sin = /a = 0.013
  = sin-1(0.013) = 0.75
 divergence 2 = 1.5
 sin[( ka / 2) sin  ] 

I ( )  I (0)
 (ka / 2) sin  
2
195
Diffraction pattern from a circular aperture
• Intensity
 2 J1 (kr sin  ) 
I ( )  I (0)

 kr sin  
r : radius of the aperture
J1 : Bessel function
2
• Diffraction pattern Airy rings
• The central white spot is called Airy disk
196
Airy disk
• Intensity
 2 J (kr sin  ) 
I ( )  I (0) 1

kr
sin



I ( )  0  J1 (kr sin  )  0
2
 kr sin   3.83  the first root of J1
 sin  
3.83
3.83
3.83 


kr
(2 /  )r
 2r
 sin   1.22

D
r
D
b

R
aperture
 Angular radius of Airy disk
Airy disk
• Radius of Airy disk, b
b


 tan     sin   b  1.22 R  b 
R
D
D
197
Diffraction pattern from a rectangular aperture
• Intensity
2
 sin( kaZ / 2 R)   sin( kbY / 2 R) 
I (Y , Z )  I (0)
 

 kaZ / 2 R   kbY / 2 R 
R : distance of screen from aperture
2
• a<bA>B
Y
B
A
Z
198
EXAMPLE 1.10.1 Resolving power
of imaging system
• Two neighboring coherent sources are examined an
imaging system with an aperture of diameter D.
•  : angular separation
• As the points get closer,  becomes narrower and
the diffraction patterns overlap more.
199
EXAMPLE 1.10.1 Resolving power
of imaging system
• Rayleigh criterion:
– Two spots are just resolvable when the principal
maximum of one diffraction pattern coincides with the
minimum of the other.
• Angular limit of resolution
sin(  min )  1.22

D
S1
min
D
b
S2
aperture
200
EXAMPLE 1.10.1 Resolving power
of imaging system
•
•
•
•
•
Human eye
pupil diameter D  2 mm
Two objects are 30 cm from the eye.
Green light of 550 nm
Minimum angular separation of two points?
Minimum separation?
201
EXAMPLE 1.10.1 Resolving power
of imaging system

550 10 9 m
sin(  min )  1.22
 1.22
 2.522 10  4 Eye (n  1.33 (water))
3
nD
(1.33)( 2 10 m)
S1
  min  0.0145
S /2
L
 S  2 L tan(  min / 2)  2(300 mm)tan(0.0 145 / 2)
tan(  min / 2) 
 0.076 mm  76 μm  thickness of a human hair
min
S
D
b
pupil
Retina
S2
L = 30 cm
202
1.10 DIFFRACTION PRINCIPLES
B. Diffraction grating
203
Diffraction grating
• A diffraction grating is an optical device that has a
periodic series of slits in an opaque screen.
• An incident beam is diffracted in certain well-defined
directions that depend on  and the grating properties.
204
Diffraction grating
• There are “strong beams of diffracted light” along certain
directions () and these are labeled according to their
occurrence: zero-order (center), first order, either side of
the zero order, and so on.
205
Grating equation
•
•
•
•
Width of each slit : a
Separation of the slits: d
a << d
Optical path difference for
waves emanating from two
neighboring slits: d sin
• All such waves from pairs of
slits will interfere
constructively when
d sin = m; m = 0, ±1, ±2, 
– m = 0, zero-order
– m = 1, first-order
Grating equation
(Bragg diffraction condition)
– etc.
206
The diffraction light pattern
• The amplitude of the diffracted beam is
modulated by the diffraction amplitude of
a single slit since the latter is spread
substantially.
• d sin = m; m = 0, ±1, ±2, 
 for give d and m, sin  
 The diffraction grating provides a means
of deflecting an incoming light by an
amount that depends on its wavelength.
• The reason for their use in spectroscopy.
207
Diffraction gratings
• Transmission grating
– The incident and diffracted beams are on opposite sides of the grating.
– Ex., ruled periodic parallel thin grooves on a glass plate.
•
Reflection grating
– The incident and diffracted beams are on the same side of the grating.
The surface of the devices has a periodic reflecting structure.
– Ex., etching parallel grooves in a metal film.
208
Grating equation
• When the incident beam is not normal to the
diffraction grating,
d (sinm – sini) = m; m = 0, ±1, ±2, 
– i : the angle of incidence with respect to the grating
normal
– m : the diffraction angle for the m-th mode
209
Shift energy to a higher order
• The undiffracted light that corresponds to the
zero-order beam is not desirable because it
wastes a portion of the incoming light intensity.
• Is it possible to shift this energy to a higher order?
210
Blazed grating
• d (sinm – sini) = m applies with respect to the
normal to the grating plane.
• The first order reflection corresponds to
reflection from the flat surface, which is at an
angle .
• It is possible to “blaze” one of
the higher orders (usually m
= 1) by appropriate .
• Most modern diffraction
gratings are of this type.
211
a
212