Transcript Information Rate

Chapter 3
Lightwave Fundamentals
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Contents
Electromagnetic waves
Dispersion, pulse distortion, information rate
Polarization
Resonant cavities
Reflection at a plane boundary
Critical-angle reflection
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3.1 Electromagnetic Waves
Wave Properties
•velocity
•power
•polarization
•interference
•refraction
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Electromagnetic Waves
Wave traveling in the z direction
On the figure, t1 < t2 < t3
Electric Field
t1
t2
t3
Position (z)
Electric field:
(3.1)
At t1 < t2 < t3 , peak amplitude E0 is fixed, Ф = wt - kz
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Electromagnetic Waves
This is a solution of the wave equation:
2 E  k 2 E  0
 E  
  E  0
2
z v

Propagation factor: k 
v
Frequency: f = v/l
2
2
Radian frequency:  = 2pf (rad/s)
Wave peak amplitude: E0
Wave phase: f = t-kz
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(3.2)
Electromagnetic Waves
Relationships for the Propagation Factor k
k = /v = /(c/n) = n/c
In free space, n = 1, so that k = ko = /c
In general, then k = kon

2p f
2p
k 

v
v
l
lo: wavelength in free space
l: wavelength in the medium
v c / n lo
l 

f
f
n
Then,
l  lo / n
(3.6)
c
lo 
f
f is fixed, f = f0
(3.7)
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Power
Power in a resistor: Pr = V2/R
Power is proportional to the voltage (V) squared.
In an optical beam, define Intensity: I = E2
Since P ∝ E2, P ∝ I.
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Power
Irradiance S = Power Density (watts/m2)
For a plane wave, the irradiance and intensity are given
by:
:材料的磁导率
: 材料的介电常数
Now
We conclude that the
intensity is proportional to
the irradiance I ∝ S.
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Electromagnetic Waves
Recall the plane wave given by
(3.1)
This expression represents a wave traveling with zero
loss.
If loss occurs, the field is represented by
E  Eo e z sin(t  kz )
(3.8)
 is the attenuation coefficient for E.
The frequency and phase do not vary with loss, only
the amplitude of the wave Eoe-z changes with loss.
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Electromagnetic Waves
P∝E2, for a path length L in a lossy medium, the power
diminishes by a factor of: e 2 L
2 is the attenuation coefficient for P
The corresponding P (or I) reduction in dB is:
dB  10lg e
2 L
This will be a negative number for propagation through a
lossy medium.
Define: g (dB/km) in terms of the attenuation coefficient .
g = -8.685  (proof)
If L is in unit of km, then  is in units of km-1.
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Electromagnetic Waves
Wave Traveling in a Lossy Medium
t2
Electric Field
t1
Distance (z)
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3.2 Dispersion, Pulse Distortion, Information Rate
When we write E = Eosin (t – kz), we imply a single
frequency source.
Frequency
Radio oscillators approximate single f pretty well.
Optical sources do not produce single f.
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Dispersion, Pulse Distortion, Information Rate
Example： Emission Spectrum of an Optical Source
1
Normalized
Power
0.5
f1
f
f
f2
Frequency
f = source bandwidth (range of frequencies emitted by
the source).
f is the central frequency.
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Dispersion, Pulse Distortion, Information Rate
Alternatively, we can plot the wavelength emission
spectrum as follows:
Normalized 1
Power
0.5
l1
l
l
l2
l = linewidth or spectral width
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Wavelength
Dispersion, Pulse Distortion, Information Rate
Example: If l = 0.82 m, l = 30 nm
so we have 3.7% bandwidth.
The conversion between wavelength and frequency is:
f l

f
l
(3.9)
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Dispersion, Pulse Distortion, Information Rate
Proof:
Define the mean wavelength as:
Then,
The mean frequency is: f = c/l
Now, we have
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l  l1l2
2
Dispersion, Pulse Distortion, Information Rate
Spectral Widths for Typical Light Sources (table 3.1)
Source
LED
Spectral Width l (nm)
20-100
Laser Diode
1-5
Nd:YAG-Laser (固态钇铝石榴石) 0.1
He-Ne Laser
0.002
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Dispersion, Pulse Distortion, Information Rate
If l = 0, (f = 0), the source is perfectly coherent.
It is monochromatic（单色）.
Laser diodes are more coherent than LEDs, but are
not perfectly coherent.
We will see how source bandwidth limits the
information capacity of fiber transmission lines.
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3.2.1 Material Dispersion and Pulse Distortion
Recall that v = c/n.
For glass, n varies with wavelength. Thus, waves
of different wavelengths (frequencies) travel at
different speeds.
Dispersion:
Wavelength dependent propagation velocity.
Material Dispersion: Dispersion caused by the material.
Waveguide Dispersion: Dispersion caused by the
structure of the waveguide.
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Material Dispersion and Pulse Distortion
Consider a pulse of light emitted by a source which
contains a range of wavelengths (say l1, l2, l3).
Input Power
T
Fastest
wavelength
Output Power
t
l1
l2
t
t
l3
t
Slowest wavelength
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T + 
t
Arrives first
l1
t
l2
l3
Arrives last
t
t
Because of dispersion, the components of the input
pulse at l1, l2, and l3 travel at different speeds and
thus arrive at the receiver at different times. The
previous slide displayed how this phenomenon spreads
pulses as they travel along a dispersive medium. The
output is widened by an amount we label as .
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Dispersion also distorts an analog signal waveform.
Input Power
Pac,in
Output Power
Pac,out
l1
l1
l2
l2
t
Slower wavelength t
Pac,out < Pac,in
Information is contained in the amplitude variation.
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DISPERSION
Refractive Index Variation for SiO2
n
Inflection Point
1.45
0
lo
Second Derivative
n’’
dn2/dl2
First Derivative
0
lo
dn/dl
0
n’
lo
Inflection point for SiO2 glass occurs near wavelength:
l  1300 nm
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Find the amount of pulse spread due to material dispersion.
Let  = time of travel of a pulse over path length L.
/L
(/L)2
/L
(/L)1
l
With No Dispersion
 
 
L
l
l
l1
l2
With Dispersion Present
The source linewidth is taken to be (with l2 > l1):
l = l2 - l1
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The pulse spread per unit length is then:
(3.10)
where l1 is the fastest and l2 is the slowest wavelength.
(/L)/l = d(/L)/dl （slope of the curve）
Pulse spread per unit length: (/L) = [d(/L)/dl] l
Actual spread would be:
(3.12)
/L
(/L)2
 
 
L
(/L)1
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l
l1
l2
l
(/L) = [d(/L)/dl] l  (/L)’ l
(3.12)
Two distinct terms determine the pulse spread
1. the slope of the /L curve
2. the linewidth of the source.
The linewidth will be available from manufacturer's
data or must be measured.
Further analysis shows that:
(3.13)
The prime and double prime denote first and second
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derivatives.
Proof:
Pulses travel at a speed called the group velocity u.
The group velocity is given by:
The pulse travel time is thus:
This is the pulse travel time per unit of path length.
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(l is the free space value)
If n  (l), then (/L)’ = 0 and there is no dispersion and no pulse
spread.
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Define material dispersion M :
Combining (3.12) and (3.13):
(3.14)
M (ps/nm/km) is in picoseconds of pulse spread per nanometer of
source spectral width and per kilometer of fiber length.
SiO2
110
M
(ps/(nm.km))
-20
1.3
1.55
l(m)
0.82
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1. For M > 0 (wavelengths < 1.3 m)
Wavelength l2 arrives before wavelength l1.
Energy at l2 travels faster than energy at l1. (l2 > l1)
2. For M < 0 (wavelengths > 1.3 m)
So that l 1 travels faster than wavelength l2.
3. At l 1.3 m, M = 0 , and there is no material
dispersive pulse spreading.
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Example: Consider an LED at l = 0.82 m, L = 10 km,
and l = 20 nm. Find (/L).
From the graph, at 0.82 m, M =110 ps/(nm·km).
Change the wavelength to l = 1.5 m, l = 50 nm.
At 1.5 m, M = -15 ps/nm·km. Then
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Example: l = 0.82 m, l = 1 nm. M = 110 ps/(nm·km)
ps
 


  L   10km 110
1 nm   1100 ps  1.1 ns
nm  km
L


  1.1 ns
Example: l = 1.5 m, l = 1 nm. M = -15 ps/nm·km
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Between 1200 nm and 1600 nm(near the inflection
point), M is given by
Mo = -0.095 ps/(nm2•km) and lo is the zero dispersion
wavelength ( 1300 nm).
Conclusion:
•The longer the path the greater the pulse spread.
•The greater the source spectral width, the greater
the pulse spread.
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3.2.2 Solitons
A soliton is a pulse that travel without spreading.
The refractive index of glass depends upon the pulse
intensity.
This fiber nonlinearity is used to counter the effects
of dispersion. The leading edge of the pulse can be
slowed down, and the trailing edge speeded up to
reduce spreading. Thus, the pulse must be properly
shaped.
The nonlinearity is such that solitons are only
produced at wavelengths longer than the zerodispersion wavelength in glass fibers. Compensation
to overcome pulse broadening is only possible in the
longer wavelength region range 1300 to 1600 nm.
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x 10
-3
5
4
3
2
1
0
800
-10
600
0
400
200
10
0
单脉冲传输800km时孤子脉冲波形的演变
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基态和高阶孤子沿光纤传输时的变化特点
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Solitons overcome the bandwidth limitations of the
fiber, but not the attenuation. Optical amplifiers are
needed along the transmission path to maintain the
pulse energy above the minimum required for
soliton production.
Amplifier
Fiber
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3.2.3 Information Rate
Consider sinusoidal modulation of the light source with
modulation frequency f. Modulation period T = 1/f.
Sinusiodal Modulation Of The Light Source
4
3.5
PT
Optic Power Trasmitted
Power
3
T
2.5
Pavg
2
1.5
1
0.5
0
0
1
2
3
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4
5
6
7
Information Rate
Information Rate
Maximum Allowable Pulse Spead At The Receiver
2
1.8
Blue: l1
PR
1.6
Optic Power Receiver
1.4
c
o
n
s
d
e
r
T
1.2
T/2
1
0.8
0.6
0.4
0.2
0
0
1
T/2
2

Red: l2
3
4
5
6
7
time
Time
This spread reduces the total power variation to zero.
Modulation is canceled.
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Information Rate
The limit on the allowable pulse spread will be
taken to be:
T
 
2
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(2)
Information Rate
From (2) we have the requirement that
1/T < 1/(2)
(3)
so that the modulation frequency has the limits:
1
f 
2
(4)
The maximum modulation frequency is then:
f max
1

2
This modulation frequency turns out to be the 3-dB
bandwidth. The signal is actually reduced by half (3-dB)
at this modulation frequency.
3-dB optical bandwidth:
f
上海师范大学电气信息系 3  dB
1

2 
(5)
Information Rate
The total signal loss has two parts and can be expressed
by the equation:
(6)
La = Loss due to absorption and scattering (fixed loss).
Lf = Modulation (message) frequency dependent loss.
The modulation frequency dependent loss is given by:
(7)
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Information Rate
Example: Suppose f = f3-dB. Compute the loss.
Example: Suppose f << f3-dB. Compute the loss.
The equation predicts no
modulation frequency
loss for modulation
frequencies well below
the 3-dB frequency.
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Information Rate
Example: Suppose f = 0.1 f3-dB. Compute the loss.
Maximum frequency length product is calculated
from (5) as follows:
1
0.5
f3 dB 

(5)
2 
L
f3 dB  L 
(8)
2
f3 dB  L 
1
 
2  
 L
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(3.16)
Information Rate
Find the frequency at which Lf = 1.5 dB.
Use (7) for Lf
2
L f  10lg e
 f 

 ln 2
 f3 dB 
(7)
2
1.5  10lg e
 f 

 ln 2
f
 3 dB 
Solving for the frequency at which the loss is 1.5
dB, we obtain
f1.5dB  0.71 f3 dB
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Information Rate
Now consider the photodetection circuit:
Photodetector
Optical
Power
P
i
RL
P = incident optic power
i = P detector output current
 = detector responsivity (A/w)
The electrical power in the load resistor RL is:
Pe  RL i 2
Pe  RL (  P )2  RL  2 P 2
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Information Rate
Consider two optical power levels P1 and P2 and their
corresponding electrical power levels Pe1 and Pe2.
dBelectrical  2dBoptical
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Information Rate
Examples:
•A loss of 3 dB in optical power yields a loss of 6
dB in the corresponding electrical power.
•A loss of 1.5 dB in optical power yields a loss of 3
dB in the corresponding electrical power.
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Information Rate
We found that the modulation frequency at which
the optical loss is 1.5 dB was:
(3.18)
Electrical 3-dB bandwidth length product is:
0.35
f 3 dB ( electrical )  L 
 
 
 L
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(3.19)
Information Rate
Consider a Return-to-Zero (RZ) digital signal.
Power
1
1
1
0
1
tp
t
0
T
2T
3T
4T
Each bit is allotted a time T.
tp = T/2
pulse width
R = 1/T
data rate, b/s
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5T
6T
7T
Information Rate
Spectrum of the RZ Signal
Power
Spectral
Density
(Watts/Hz)
0
1
T
2 1

T tp
Frequency
Most of the signal power lies below 1/T Hz, so the
required transmission bandwidth by a system is:
BRZ
1
 R
T
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Information Rate
If the system passes this band of frequencies the
pulses will be recognizable. To be conservative,
use the 3-dB electrical bandwidth.
The RZ rate length product is then:
0.35
RRZ  L 
 
 
 L
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(3.20)
Information Rate
We obtain the same result by allowing a pulse
spread of 70% of the initial pulse duration.
As on the preceding slide.
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Information Rate
Consider the Non-Return-to-Zero (NRZ) digital signal.
1
1
1
0
1
tp
t
0
T
2T
Power
Spectral
Density
(Watts/Hz)
0
3T
4T
5T
6T
7T
Spectrum of the (NRZ) Signal
1
2T
1 1

T tp
Frequency
Required transmission bandwidth: BNRZ
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1 R


2T 2
Information Rate
The allowed data rate is:
Use the electrical 3-dB bandwidth:
NRZ rate length product is:
RNRZ  L 
0.7
 
 
 L
(3.21)
Comparing the results for the RZ and NRZ data rates:
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Information Rate
BANDWIDTH DATA RATE SUMMARY
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3.3 Polarization
Linearly polarized:
An electric field points in just one direction, it always
points along a single line.
a. Linearly polarized in x direction and traveling in the
z direction.
b. linearly polarized in y direction and traveling in the
z direction.
E
v
x
x
E
v
z
z
y
y
(a)
(b)
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Polarization
1. The two orthogonal linear polarizations are the
plane wave modes of an unbounded media.
2. They can exist simultaneously.
3. The actual polarization is determined by the
polarization of the light source and by other
polarization sensitive components in the optical
system.
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Polarization
If the direction of electric field E varies randomly (as
shown) the wave is unpolarized.
y
E
x
Most fibers depolarize the input light. Only special
fibers maintain the light polarization.
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3.4 Resonant Cavities
镜
子
镜
子
1、让任意一个光波从左边的镜子传向右边的
镜子，如A图所示。绿波在右边的镜子处发生
发射，因此这个波经历了一次180度相移。从
A图我们可以看出，这个波在其相位上发生了
中断，在这里应该是不可能的，也就是说，这
个谐振器不支持这个波。
L
A
镜
子
镜
子
2、在图B中，在右边的镜子处，这个波也发生
了一个180度相移，然后继续传播，在左边的
镜子处，同样经历了一个180度相移，然后继
续传播。因此，图B所示的波有着一个稳定的
模式，我们称之为驻波
L
B
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Resonant Cavities
2p
l
L  mp
ml
L
2
(3.22)
L
Standing-wave pattern
in a cavity (m = 4)
Design:
The cavity must be an integral number of half
wavelengths long to support a wave.
The wavelength is that in the medium filling the cavity.
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Resonant Cavities
The resonant wavelengths are:
2L
l 
m
(3.23)
The corresponding resonant frequencies are:
f 
v
l
c
f 
nl
c  m  cm
f   
n  2L  n2L
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Resonant Cavities
Cavity Resonant frequencies
c
2nL
fc
fm1
fm
fm1
fm2
Frequency
This picture shows the longitudinal modes of the cavity.
The resonant frequency spacing is:
f c  f m1  f m
(m  1)c mc
c
f c 


2 Ln
2 Ln 2 Ln
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(3.25)
Resonant Cavities
The free space wavelength spacing corresponding
to fc is lc calculated from:
l c
lo
f c

f
l o f c l o 2 f c
l c 

c
c
lo
(3.26)
This equation refers to the free space wavelengths.
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Resonant Cavities
Example: Consider an AlGaAs laser cavity.
L = 0.3 mm = 300 m; n = 3.6; lo = 0.82 m.
Find the cavity resonant wavelength spacing lc.
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Resonant Cavities
Example: Suppose the AlGaAs, LD has a spectral width of 2
nm. Determine the number of longitudinal modes in the
output.
Gain (AlGaAs)
0.82 m
l
2 nm
Cavity Resonances
lc
0.82 m
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l
Resonant Cavities
Laser Output
2 nm
lc
0.82 m
l
The laser emits 6 longitudinal modes.
A laser emitting only one longitudinal mode is a
single-mode laser.
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3.5 Reflection at a Plane Boundary
Consider normal incidence of light at a boundary.
This is referred to as Fresnel Reflection.
n1
n2
Incident Wave
Transmitted Wave
Reflected Wave
Boundary
Reflection Coefficient:
reflected

incident
electric
electric
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field
field
Reflection at a Plane Boundary
Reflection Coefficient:
n1  n2

n1  n2
Reflection at a Plane Boundary
Define Reflectance R (反射比) as:
power reflected
R
power incident
R 
2
 n1  n2 
R

 n1  n2 
2
This result is valid for normal incidence.
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Reflection at a Plane Boundary
For air-to-glass, compute the transmitted power.
4% power reflected. 96% power transmitted.
In dB, the transmitted power is:
10 lg (0.96) = -0.177 dB
Typically we round this off to 0.2 dB (omitting the
minus sign). This is called the Fresnel loss.
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Reflection at a Plane Boundary
Consider arbitrary incidence:
Er
n1

 Et
n2
t
r
i
Ei 
Perpendicular Polarization (s)
垂直偏振
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Reflection at a Plane Boundary
Consider arbitrary incidence:
n1
Er
n2
Et
t
r
i
Ei
Parallel Polarization (p)
平行偏振
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Reflection at a Plane Boundary
Plane of Incidence(入射平面)
Defined by the normal to the boundary and the ray
direction of the incident beam.
x
z
Incident
Boundary
The xz plane is the plane of incidence in this example.
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Reflection at a Plane Boundary
Fresnel’s Law of Reflection
For parallel polarization, the reflection coefficient:
p 
n2 2 cos i  n1 n2 2  n12 sin 2 i
n2 cos i  n1 n2  n sin i
2
2
2
1
2
(3.29)
For perpendicular polarization, the reflection coefficient:
s 
n1 cos i  n2 2  n12 sin 2 i
n1 cos i  n2 2  n12 sin 2 i
Note that  may be complex.
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(3.30)
Reflection at a Plane Boundary
Plots of p and s for n1 = 1 (air), n2 = 1.48 (glass)
p
Perpendicular (s)
s
Parallel (p)
Angle of incidence (i)
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Reflection at a Plane Boundary
From equation (3.29) for parallel polarization, we can
get total transmission (no reflection) if
n22 cos i  n1 n22  n12 sin 2 i
The angle satisfying this equation is the Brewster
angle B. The solution is:
n2
tan  B 
n1
Compute B for air-to-glass and glass-to-air:
For n1 = 1, n2 = 1.5
For n1 = 1.5, n2 = 1
For perpendicular polarization there is no Brewster
angle. No  i s.t. Equ.3.30 = 0.
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Reflection at a Plane Boundary
Antireflection Coatings
We have just seen that we can transmit a beam from one material to
another without reflection under Brewster angle conditions. We can also
transmit with no (or very little) reflection by placing a coating between the
two materials.
n1
n2
n3
l/4
The thickness of the coating is a quarter wavelength. The
reflectance R for this configuration is:
2 2
2
2 2
 n1n3  n 
R
 n1n3  n2 
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Reflection at a Plane Boundary
Clearly, the reflectance becomes zero if:
n2  n1n3
A coating that reduces the reflectance is called an
antireflection (AR) coating .消反射涂覆
Example: Compute the reflectance when a quarter
wavelength of magnesium fluoride (氟化镁n = 1.38) is coated
onto a piece of glass (n = 1.5).
Solution:
The reflectance is:
4% → 1.4%
1.5  1.38 )
(
R
(1.5  1.38 )
2 2
2 2
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 0.014
3.6 Critical Angle Reflection
Fresnel’s Law of Reflection
For parallel polarization:
p 
n2 2 cos i  n1 n2 2  n12 sin 2 i
n2 cos i  n1 n2  n sin i
2
2
2
1
2
(3.29)
For perpendicular polarization:
s 
n1 cos i  n2 2  n12 sin 2 i
n1 cos i  n2 2  n12 sin 2 i
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(3.30)
Critical Angle Reflection
From equations (3.29) and (3.30), we find that
The incident angle satisfying this equation is the angle
whose sine is given by:
n2
(3.34)
sin c 
n1
Call the solution c, the critical angle.
c exists only if n1 > n2. That is, travel from a high index
to a low index material. This result is valid for both
polarizations.
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Critical Angle Reflection
If
then
is purely imaginary.
Under this condition, equations (3.29) and (3.30) can
be written in the form:
where A, B, C, and D are real and j is the imaginary
term
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Critical Angle Reflection
Then:
We conclude that there is complete reflection (called
critical angle reflection) for all rays which satisfy the
condition:
i   c
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Critical Angle Reflection
Consider waves undergoing critical angle reflections:
n1
n2
i  c
i
In region n1 we have a standing wave due to the
interference of the incident and reflected waves.
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Critical Angle Reflection
In region n2 the electric field is not zero. The
boundary conditions require the electric field to be
continuous at the boundary. The field in n2 termed as
evanescent is a decaying exponentially carrying no
power. E  e  z
where the attenuation coefficient is given by
  k0 n12 sin 2   n2 2
k0 
2p
l0
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Critical Angle Reflection
Consider a wave where
Envelope
n1
i  c
n2
e-z
E
Standing Wave
Evanescent Wave
这里衰减因子和前面提到的衰减系数不同，
z
衰减系数是指功率的实际损耗，
这里衰减因子并不具有这样的含义，仅仅指
电磁波回到入射区之前，场在第二种介质中
要传播多远。
The decaying wave carries no power in the z-direction.
At the critical angle,
In this case, there is no decay. The wave penetrates
deeply into the second medium.
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Critical Angle Reflection
As i increases,  increases and the decay becomes
greater.
|E|
 = 0,  = c
e-z
i >  c
0
z
As i increases from c towards 90o,  increases
and the evanescent field penetrates less and less
into the second medium.
i
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LIGHTWAVE FUNDAMENTALS
•Pulse spread
•3-dB bandwidths
•Rate-length products
•Reflectance
•Critical angle reflections
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