2.4 wave equation in quadratic index media

1
2.4 wave equation in quadratic index media

• The most widely encountered beam in quantum
  electronics is one where the intensity
  distribution at planes normal to the propagation
  direction is Gaussian. To derive its
  characteristics we start with the Maxwells
  equations in an isotropic charge-free medium.

distribution ???? isotropic ???? charge
??, ??, ?? curl ????
Taking the curl of the second of (2.4-1) and
substituting the first results in
2
where
quantity ?, ??

• 3This neglect is justified if the fractional
  change of ein one optical wavelength is ltlt 1.

3
thus allowing for a possible dependence of e on
position r. We have also taken k as a complex
number to allow for the possibility of losses
(sgt0) or gain (slt0) in the medium

• 4If K is complex (for example, KriKi), then a
  traveling electromagnetic plane wave has the form
  of

4

• We limit our derivation to the case in which
  k2(r) is given by
• where, according to (2.4-4)
• so that k2 is some constant characteristic of
  the medium. Furthermore, we assume a solution
  whose transverse dependence is on r
  only, so that in (2.4-3) we replace by

2.4-5
2.4-6
transverse ???, ???
5

• The kind of propagation we are considering is
  that of a nearly plane wave in which the flow of
  energy is predominantly along a single( for
  example, z) direction so that we may limit our
  derivation to a single transverse field component
  E. Taking E as
• We obtain from (2.4-3) and (2.4-5) in a few
  simple steps,
• where and where we assume that the
  variation is slow enough that

2.4-7
2.4-8
Where and where we assume
that the variation is slow enough that
predominantly ???, ???, ????
transverse ???
6
coefficient ???
7
Mental deficiency????

• "Would you mind telling me, Doctor," Bob asked
  ...
• "how you detect a mental deficiency in somebody
  who appears completely normal?"
• "Nothing is easier," he replied.
• "You ask him a simple question which everyone
  should answer with no trouble.
• If he hesitates, that puts you on the track.
• " Well, What sort of question?"
• "Well, you might ask him, 'Captain Cook made
  three trips around the world and died during one
  of them. Which one?'
• Bob thought for a moment, and then said with a
  nervous???laugh, "You wouldn't happen to have
  another example would you? I must confess(??)I
  don't know much about history."

8

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• ?????????,???????????,?????????,????????????
• ???????????
• ?,??????,???????????,??????????????,??????
• ???????,??????,?????????????????,????????????

9
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10
4????? ????????????????????????,?????????,????
?????????????????,?????? ???TEM00??????,????
11
??E0???,????????
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????????Z???????????????
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12
????????? (1)????????????????????????????
???(?????)???????,??1-6????????????1/e???????,????
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13
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14
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15
??,????????Z?????????,?
??1-7???
?Z0?,?(z)?0?????,???????
16
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(2)????????????
????????????????? ??,kz????????????arctan(z/f
)??????????????z?,??????????????kr2/(2R(z))??????
??r?????,??????????????R(z)???????
18
R(z)?Z?????
?? a)?Z0?,R(z)?8,???????????????? b)
?Z?8?,R(z)z?8??????????????????,??????????
c)?zf?,R(z)2f,????? ?
19
d) ?0ltzltf?,R(z)gt2f,????????????(-8,-f)???
? e)?zgtf?,zlt R(z)ltzf,????????????(-f,0)??
??
20
(3)???????????,???????,??????????????
??????1/e2???z?8?,??????1/e2????????,?
??????????????0???
21
????,??????????????,??????????????,???????????
??????,?????????????????
22
2.5 Gaussian beams in a homogeneous??medium
From 2.4, the kind of propagation is a nearly
plane wave in which the flow of energy is
predominantly along a single direction, so
2.5-1
In a homogeneous medium the quadratic
coefficient(??) k2 of the equation above is
zero, so that
2.5-2 How to solve it? Have a try! please
23
Where q0 is an arbitrary integration constant.
From (2.4-11) and (2.5-4) we have
24
then
2.5-6
25
2.5-8
in which
vacuum (??)wave length
Let us consider, one a time, the two factors
in 2.5-7, the first one becomes
2.5-9
26
imaginary ??? confine ??, ??
substitution ??, ??? exponent ?? ,???, ??
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Define the important following parameters
29

• Combine equations above in 2.5-4 , and obtain

2.5-14
the distance r at which the field
amplitude is down by a factor 1/e compare to its
valve on the axis the minimum spot
size, it is the spot size at the plane z0 R
the radius of curvature of the very nearly
spherical wavefronts at z. and we identify R as
the radius of curvature of the Gaussian beam.
30

• That is the basic result. we refer to
  it as the fundamental Gaussian-beam solution,
  since we have excluded the more complicated
  solutions by limiting ourselves to transverse
  dependence involving r(x2y2) only. The
  higher-order modes will be discussed separately.
• The form of the fundamental Gaussian beam is
  uniquely determined once its minimum spot size
  w0 and its location ---that is, the plane z0---
  are specified, the spot size w0 and radius of
  curvature R at any plane z are then found .

31

• From (2.5-14) the parameter w(z) evolves
  according to (2.5-11), is the distance r at which
  the field amplitude is down by a factor 1/e
  compared to its value on the axis. We will
  consequently refer to it as the beam spot size.
  The parameter w0 is the minimum spot size. It is
  the beam spot size at the plane z0. The
  parameter R IN (2.5-11) is the radius of
  curvature of the very nearly spherical wavefronts
  at z. We can verify this statement by deriving
  the radius of curvature of the constant phase
  surfaces (wavefronts) or , more simply, by
  considering the form of a spherical wave emitted
  by a point radiator placed at z0. It is given by

evolve (?)??, (?)??
32

• Since z is equal to R, the radius of curvature
  of the spherical wave. Comparing (2.5-15) with
  (2.5-14), we identify R as the radius of
  curvature of the Gaussian beam. The convention
  regarding the sign of R(z) is that it is negative
  if the center of curvature occurs at zgtz and
  vice versa

convention ??, ??, ??
33
The form of the fundamental Gaussian beam is,
according to (2.5-14),uniquely determined once
its minimum spot size w0 and its location that
is , the plane z0 are specified. The spot size
w and radius of curvature R at any plane z are
then found from (2.5-11) and (2.5-12). Some of
these characteristics are displayed in Figure
2-5. The hyperbolas shown in this figure
correspond to the ray direction and are
intersections of planes that include the z axis
and the hyperboloids.
2.5-16
uniquely ???,??? hyperbola ??? hyperboloid
????
34

• These hyperbolas correspond to the local
  direction of energy propagation. The spherical
  surfaces shown have radii of curvature given by
  (2.5-12). For large z the hyperbolas x2y2?2 are
  asymptotic to the cone

2.5-17
hyperboloid ???? asymptotic ????,
??? cone ???, ???
35
Some of these characteristics are displayed as
following figure
36
1,for K20
1,for K20
0,for K20
1,for K20
37

• Whose half-apex angle, which we take as a measure
  of the angular beam spread, is
• This last result is a rigorous manifestation of
  wave diffraction according to which a wave that
  is confined in the transverse direction to an
  aperture of radius w0 will spread (diffract) in
  the far field (zgtgtp w0 2n/?) according to (2.5-18)

for
2.5-18
apex ?? ?? rigorous ???, ??? hyperboloid
????
38
Happy Birthday!!!
39

• 2.6 Fundamental Gaussian beam in a lenslike
  medium the ABCD Law
• We now return to the general case of a
  lenslike medium so that k2 ?0. The P and q
  functions of (2.4-9) obey, according to (2.4-11)

2.6-1
40

• If we introduce the functions defined by
• we obtain from (2.6-1)

2.6-2
turn to P46 2.3-4
41
2.3-4

• If at the input plane z0 the ray has a radius r0
  and slope r0, we can write the solution of
  (2.3-4) directly as

2.3-5
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• Using (2.6-3) in (2.6-4) and expressing the
  result in terms of an input value q0 gives the
  following result for the complex beam radius q(z)
  
• The physical significance of q(z) in this case
  can be extracted from (2.4-9). We expand the part
  ?(r,z) that involves r. The result is

2.6-4
Significance ??, ??? Extract ??, ??, ??, ??,
??, ??, ?? Expand ???, ??, ??
44

• If we express the real and imaginary parts of
  q(z) by means of
• We obtain
• so that w(z) is the beam spot size and R its
  radius of curvature, as in the case of a
  homogeneous medium, which is described by
  (2.5-14). For the special case of a homogeneous
  medium (k20), (2.6-4) reduces (2.5-4)

2.6-5
45
1,for K20
1,for K20
0,for K20
1,for K20
46

• Transformation of the Gaussian Beamthe ABCD Law
• We have derived above the transformation law of a
  Gaussian beam (2.6-4) propagation through a
  lenslike medium that is characterized by k2, we
  not first by comparing (2.6-4) to Table 2-1(6)
  and to (2.3-5) that the transformation can be
  described by
• where A,B,C,D are the elements of the ray matrix
  that relates the ray (r,r) at a plane 2 to the
  ray at plane 1.It follows immediately that the
  propagation through

2.6-6
47
Ray matrix a medium with a quadratic index
profile
48
2.6-6
where A,B,C,D are the elements of the ray matrix
that relates the ray (r,r) at a plane 2 to the
ray at plane 1.It follows immediately that the
propagation through.
49
plane 2 to the ray at plane 1. It follows
immediately that the propagation through, or
reflection from, any of the elements shown in
Table 2-1 also obeys (2.6-6), since these
elements can all be viewed as special cases of a
lenslike medium. For future reference we not that
by applying (2.6-6) to a thin lens of focal
length f we obtain from (2.6-6) and Tbale 2-1(2)
2.6-7
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Consider next the propagation of a Gaussian beam
through two lenslike media that are adjacent to
each other. The ray matrix describing the first
one is (AT,BT,CT,DT) while that of the second one
is (AT,BT,CT,DT) . Taking the input beam
parameter as q1 and the output beam parameter as
q3, we have from (2.6-6)
Adjacent ???, ???
52

• for the beam parameter at the output of medium 1
  and
• and after combining that last two equations,
• where (AT,BT,CT,DT) are the elements of the ray
  matrix relating the output plane (3) to the input
  plane (1), that is ,

53

• It follows by induction that (2.6-9) applies to
  the propagation of a Gaussian beam through any
  arbitrary number of lenslike media and elements.
  The matrix (AT,BT,CT,DT) is the ordered product
  of the matrices characterizing the individual
  members of the chain

2.6-10
induction ??, ????, ??
54

• The great power of the ABCD law is that it
  enables us to trace the Gaussian beam parameter
  q(z) through a complicated sequence of lenslike
  elements. The beam radius R(z) and spot size
• ?(z)at any plane z can be recovered through the
  use of (2.6-5). The application of this method
  will be made clear by the following example.

55
Most wanted autograph??

• Our university newspaper runs a weekly question
  feature. Recently, the question was "Whose
  autograph would you most want to have, and why?"
  As expected, most responses mentioned music or
  sports stars, or politicians. The best response
  came from a freshman, who said, "The person who
  signs my diploma."
• ???????????????????????????????????????????????
  ,???????????????????????,???????????????,???????
  ????????

56
Example Gaussian beam focusing
As an illustration of the application of the
ABCD law, we consider the case of a Gaussian beam
that is incident at its waist on a thin lens of
focal length f , as shown
57
2.6-1
58
The transformation can be discribed by
2.6-2
2.6-3
59
At plane 3
2.6-5
from 2.6-5,we have
2.6-6
60
2.6-7
as the location of the new waist, and to
2.6-8
for the output beam waist.
61
The confocal (????)beam parameter
is, according to (2.5-11), the distance from the
waist in which the input beam spot size increases
by and is a convenient(???,
???)measure of the convergence (??)of the input
beam. The smaller Z01, the stronger the
convergence.
62
2.7 A Gaussian beam in lens waveguide

• As another example of the application of the ABCD
  law, we consider the propagation of a Gaussian
  beam through a sequence of thin lenses, as shown
  in Figure 2-2. The matrix, relating a ray in
  plane s1 to the plane s1 is
• where (A,B,C,D) is the matrix for propagation
  through a single two-lens, unit cell (?s1) and
  is given by (2.1-6). We can use a well-known
  formula for the sth

P42
2.7-1
63

• we can use a well-known
  formula for the sth power of a matrix with a
  unity determinant (unimodular) to obtain

unity ??, ??, ??, ?? determinant ??? ????
Unimodular ??????
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• The condition for the confinement of the Gaussian
  beam by the lens sequence is, from (2.7-4), that
  ?be real otherwise, the sine functions will
  yield growing exponentials. From (2.7-3), this
  condition becomes cos?1,or
• That is, the same as condition (2.1-16) for
  stable-ray propagation.

2.7-5
confinement (?)??, (?)?? determinant ??? ????
Unimodular ??????
66
Schoolworks
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Chapter 3 Propagation of optical beams in fibers
Introduction Since the realization of
low-loss fibers in 1970s, fibers become the most
important medium for optics communication.
The technology of optical communication in fibers
can be stated as that of feeding(??, ?, ??)
optical pulses at a maximal rate into one end of
a fiber and retrieving(??) them at the other end.
The main goal of a communication system is to
receive the pulses at the output end with minimal
loss of energy, minimal spread, and minimal
contamination(??, ??) by noise. For example, the
silica(???)glass fiber ,which has the character
of low-loss propagation of confined optical
modes, has become the most important transmission
medium for long distance.
69

• In this chapter, we will study
• the subject of optical guided modes in fibers
• The problem of pulse spreading due to group
  velocity dispersion and various strategies of
  combatting it

70
3.1 wave equation in cylindrical coordinates(???)

• In Chaoter 2 we have shown that optical
  waveguides with a quadratic index profile (See
  Equation 2.9-1a) can support guided
  nondiffracting modes. The effect of diffraction
  spreading is counterbalanced by the lensing
  effort of the index profile of the
  guide.Commercial(???) silica-based optical fiber
  use a step index profile with a high index core
  and a low index cladding(??). These fibers form
  the backbone(??, ??)of most modern communication
  systems, and the study of their modes of
  propagation is the subject at hand.

71
Figure 3-1 Structure and index profile of a
step-index circular waveguide
72

• Since the unit vector ar and are not
  constant vectors, the wave equation involving the
  transverse(???) components are very complicated.
  The wave equation for the z component of the
  field vectors, however, remains simple

73
Since we are concerned with the propagation along
the waveguide, we assume
3.1-2
74
Maxwells curl equations are now written in
terms of the cylindrical components and are given
by
3.1-3
3.1-4
75
From above equation, we get
3.1-5
3.1-6
From the results, we know that these relations
show it is sufficient to determine Ez and Hz
in order to specify uniquely the wave solution.
The remaining components can be calculated from
(3.1-5) and (3.1-6)
76

• With the assumed z-dependence of (3.1-2), the
  wave equation (3.1-1) becomes
• This equation is separable, and the solution
  takes the form
• where l0,1,2,3.., so that E and H, are
  single-valued functions of . Then

3.1-7
3.1-8
77

• (3.1-7) becomes
• where
• Equation (3.1-9) is the Bessel differential
  equation, and the solutions are called Bessel
  functions of order l. If , the
  general solution of
• (3.1-9) is

3.1-9
3.1-10
78

• where , and are
  constants, and , are Bessel functions
  of the first and second kind, respectively, of
  order l. If , the general solution
  of (3.1-9) is
• where , and are
  constants, and are the modified Bessel functions
  of the first and second kind ,respectively ,of
  order l.

3.1-11
79

• To proceed with our solution, we need the
  asymptotic forms of these functions for small and
  large arguments. Only leading terms will be given
  for simplicity.
• For xltlt1

asymptotic ?????, ???
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Motivation ??

• MY ENGLISH PROFESSOR once launched into a lecture
  on "motivation." "What pushes you ahead?" he
  asked. "What is it that makes you go to school
  each day? What driving force makes you strive to
  accomplish?" Turning suddenly to one young woman,
  he demanded "What makes you get out of bed in
  the morning?" The student replied "My mother."
• ?????????????????????????????????????,???????
  ???????????????????????,??????????????????????
  ????

83
3.2 the step-index circular waveguide

• The geometry of the step-index circular waveguide
  is shown in Figure 3-1. It consists of a core of
  refractive index and radius a, and a
  cladding of refractive index and radius b. The
  radius b of the cladding is usually chosen to be
  large enough so that the field of confined modes
  is virtually zero at rb. In the calculation
  below we will put b this is a legitimate
  assumption in most waveguides, as far as confined
  modes are concerned.

legitimate???, ???
84
Figure 3-1 Structure and index profile of a
step-index circular waveguide
85

• The radial dependence of the fields Ez. and Hz.
  ,is given by(3.1-10) or (3.1-11) depending on the
  sign of .For confined propagation ,
  must be larger than (i.e.
  ). This ensures that the
  wave is evanescent in the cladding region, rgta.
  The solution is thus given by(3.1-11) with c10
  .This is evident from the asymptotic behavior for
  large r given by (3.1-13). The evanescent decay
  of the field also ensures that the power flow is
  along the direction of the z axis, i.e., no
  radial power flow exists. Thus the fields of a
  confined mode in the cladding (rgta) are given by

evanescent????, ????
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• For the fields in the core , rlta, we must
  consider the behavior of the fields as
  ,According to (3.1-12), Yl and Kl are
  divergent as .Since the fields must
  remain finite at r0, the proper choice for the
  fields in the core (rlta) is (3.1-10) with c20.
  This becomes evident only when matching, at the
  interface ra, the tangential components of the
  field vectors E and H in the core with the
  cladding field components derived from (3.2-1)
  we are unable to accomplish this if the radial
  dependence of the core fields is given by Il.
  Thus the propagation constant must be less
  than , and the core fields are given by

divergent??? tangential???
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• In the field expressions (3.2-1) and (3.2-3), we
  have taken a sign in front of in the
  exponents. A negative sign would yield a set of
  in dependent solutions, but with the same radial
  dependence. Physically, l plays a role similar to
  the quantum number describing the z component of
  the orbital angular momentum of an electron in a
  cylindrically symmetric potential field. Thus, if
  the positive sign in front of
• corresponds to a clockwise circulation of
  photons about the z axis, the negative

orbital ???, potential???, ???, ??, ??
90

• sign would corresponds to a clockwise
  circulation of photons around the axis. Since
  the fiber itself does not possess any preferred
  sense of rotation, these two states are
  degenerate.
• Equations (3.2-1) and (3.2-3) together
  require that and , which
  translates to
• which can be regarded as a necessary condition
  for confined modes to exist. This is

3.2-5
preferred???
91

• which can be regarded as a necessary condition
  for confined modes to exist. This is identical to
  the condition discussed in Section 13.1 for the
  slab dielectric waveguide and can be expected on
  intuitive grounds from our discussions of total
  internal reflection at a dielectric interface.
• Using (3.2-) and (3.2-3) in conjunction
  with (3.1-5) and (3.1-6), we can calculate all
  the field components in both the cladding and the
  core regions. The result

slab???, ??
92
The result is
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Nontrivial ???? transcendental ???
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New business was opening ????

• A new business was opening ... and one of the
  owner's friends wanted to send him flowers for
  the occasion. They arrived at the new business
  site and the owner read the card,.... "Rest in
  Peace." The owner was angry and called the
  florist to complain. After he had told the
  florist of the obvious mistake and how angry he
  was, the florist replied, "Sir, I'm really sorry
  for the mistake, but rather than getting angry,
  you should imagine this somewhere, there is a
  funeral taking place today, and they have flowers
  with a note saying, ... 'Congratulations on your
  new location!'"
• ??????,?????,??????????????????????????????????
  ???,?????????????????????????,???????????????????,
  ?????????????????,??????????????,?????????,????
  ??????,???????????????!

100
3.2 the step-index circular waveguide
Figure 3-1 Structure and index profile of a
step-index circular waveguide
101

• n1 core of refractive index
• n2 cladding of refractive index
• a, b radius. the radius b of the cladding is
  usually chosen to be large enough so that the
  field of confined modes is virtually zero at rb.
  in the calculation below we will put b8 this
  is a legitimate assumption in most waveguides, as
  far as confined modes are concerned.

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• Schoolwork
• Derive the equation 3.1-3 3.1-6
• Translate section 3.2 into Chinese

? ? ??
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106
Chapter 4 Optical Resonators
Introduction Optical resonators, like their
low-frequency, radio-frequency?????, and
microwave counterparts,??, ???????, ??? are used
primarily in order to build up large intensities
with moderate ???, ??? power inputs. They consist
in most cases of two, or more, curved mirrors
that serve to trap, by repeated reflecticons
and refocusing, an optical beam that thus becomes
the mode of the resonator.
107

• A universal (???, ???) measure of optical
  resonators property is the quality factor Q of
  the resonator, Q is defined by the relation
• dissipated ??????, ???

108
Consider the case of a simple resonator
formed by bouncing (?)??, ?? a plane TEM wave
between two perfectly conducting ?? planes of
separation l so that the field inside is
the average electric energy stored in the
resonator is
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Thus, recognizing (??, ??)that in steady state
the input power is equal to the dissipated power,
and designating the power input to the resonator
by P, we obtain from 4.0-1
and the peak field is given by
4.0-6
111
Mode density of optical resonators
The main challenge in the optical frequency
regime is to build resonators that possess a very
small number, ideally only one, high Q modes in a
given spectral(???) region. The reason is that
for a resonator to fulfill this condition, its
dimensions need to be of the order of the
wavelength.
112
Example One-Dimensional Resonator
113
Mode control in the optical regime would thus
seem to require that we construct resonators with
volume(??)--
. This is not easily
achievable. An alternative is to build large
resonators but to use a geometry
that endows only a small fraction of these modes
with low losses (a high Q). In our two-mirror
example any mode that does not travel normally to
the mirror will walk off after a few bounce and
thus will possess a low Q factor.
114
We will show later that when the resonator
contains an amplifying (inverted(???)population)
medium, oscillation(??) will occur preferentially
(??)at high Q modes, so that the strategy of
modal discrimination(??)by controlling Q is
sensible(???), we shall also find that further
model discrimination is due to the fact that the
atomic(???)medium is capable of amplifying
radiation only within a limited frequency region
so that modes outside this region, even if
possessing high Q, do not oscillate. One
question asked often is the following given a
large optical resonator, how many of
its modes will have their resonant frequencies in
a given frequency interval ,say,
115
To answer this problem, consider a large,
perfectly reflecting(???) box resonator with
sides,a , b, c along the x, y, z directions.
Without going into modal details, it is
sufficient for our purpose to take the amplitude
field solution in the form For the field
to vanish at the boundaries, we thus need to
satisfy In the equation above, the triplet
(????)r, s, t is any integers ,and they define a
mode.
(4.0-7)
(4.0-8)
116
We will restrict, without loss of generality,
r,s,t to positive integers. It is convenient to
describe the modal distribution in K space. Since
each (positive) triplet r, s, t generate an
independent mode, we can associate with each mode
an elemental volume in K space. V is the
physical volume of the resonator. We recall that
the length of the vector K satisfies equation
4.0-8, rewrite here as
(4.0-9)
(4.0-10)
117
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118

• to find the total number of modes with K
  values between 0 and k, we divide the
  corresponding volume in K space by the volume per
  mode
• We next use 4.0-10 to obtain the number of
  modes with
• resonant frequencies between 0 and v .
• The mode density, that is, the number of modes
  per unit
• near v in a resonator with volume V, is thus

(4.0-11)
119
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120
This objection (??) is overcome to a large
extent by use of open resonators, which consist
essentially(???)of a pair of opposing (???)flat
or curved reflectors. In such resonators the
energy of the vast majority of the modes does not
travel at right angles (??) to the mirrors and
will thus be lost in essentially a single
traversal. (??)???? These modes will consequently
possess a very low Q. if the mirrors are curved,
the few surviving (??????) modes will have their
energy localized (??)near the axis thus the
diffraction (??)loss caused by the open sides can
be made small compared with other loss mechanisms
( ??, ??)such as mirror transmission. ??, ???????
121
Problem we have a resonator which volume equal
and
(in atmosphere), please calculate
the number of modes that produce within the
interval centered on .
122
The Phone Call Goodbye ????

• When I was small, my Great-Aunt Nony was still
  alive. Her life had been a terrible ordeal for
  her as she had been plagued with lots of
  different types of cancer. One day, my parents
  took me to see her in the nursing home in which
  she lived. It was quite scary as the cancer had
  started to show and you could see tumors on her
  face and arms. She seemed to just be wasting
  away, but my parents wanted me to see her before
  she died.
• A few days after this, my parents were talking in
  the kitchen when I received a seemingly real
  phone call on my plastic toy phone. It was Nony.

123

• "Hello. It's Nony, Carrie. Don't worry.
  Everything will be alright. Tell your parents.
  Don't worry, everything's going to be fine."
• So I hung up and told my parents what Nony had
  said, but they didn't believe me. They just
  thought that she was playing on my mind when I
  was playing on my pretend telephone.
• One hour later, we received a phone call from my
  grandmother. She told us that Nony had died about
  an hour ago.

124

• ?????????????????,???????????,??????????????????,?
  ???????,????????????,?????????????,????????????,??
  ??????????????????????,?????????????????
• ????????,????????,????????????????????,??????????
• ??,??,????????,????????????????,????????
• ?????,?????????????????,?????????????????????
• ?????,??????????,??????????????????

125
(No Transcript)
126
2.10 propagation in media with a quadratic gain
profile

• In many laser media the gain is a strong
  fuction of position. This variation can be due to
  a variety of causes, among them (1) the radial
  distribution of energetic electrons in the plasma
  region of gas lasers 19, (2) the variation of
  pumping intensity in solid state lasers, and (3)
  the dependence of the degree of gain saturation
  on the radial position in the beam.

function ??, ??, ??, ??, ??, ??? among them
?????? radial ???, ????, ????, ??? energetic
electron ???????????? dependence ??, ??,??
127

• We can account for an optical medium with
  quadratic gain (or loss) vatiation by taking the
  complex propagation constant k(r) in (2.4-5) as
• where the plus (minus) sign applies to the case
  of gain (loss). Assuming k2r2ltlt k in (2.4-5), we
  have k2la2. Using this value in (2.4-11) to
  obtain the steady-state((1/q)0) solution of the
  complex beam radius yields6

2.10-1
2.10-2
account for ????,???? the complex propagation
constant ?????? gain(loss) ??(??) the
steady-state solution ???
128

• The steady-state beam radius and spot size are
  obtained from (2.6-5) and (2.10-2)
• We thus find that the steady-state solution
  corresponds to a beam with a constant spot size
  but with a finite radius of curvature.

2.10-3
129
The general (non-steady-state) behavior of
the Gaussian beam in a quadratic gain medium is
described by (2.6-4), where k2la2.
Experimental data showing a decrease of the beam
spot size with increasing gain parameter a2 in
agreement with (2.10-3) are shown in Figure 2-9.
130
Thank you

• Thank you for your answers are almost identical.
• Thus, you have saved me much time in review your
  schoolworks. Thanks again!
• Im deeply moved by your kindness. So If you have
  any trouble in your study, please dont hesitate
  to let me know, I will do my best to help you.
• For example, If you, unfortunately, failed in the
  final test, I will be glad to prepare the second
  time test for you.
• Even if more unfortunaly, you failed again in the
  second time test, I still would be glad to
  prepare the third time test for you, and so on.
• But, I do be more happy to test just one time, Do
  you think so?

131
4.1 Fabry-Perot etalon (???)

• The Fabry-Perot etalon or interferometer,
  named after its inventors (Fabry (1867.6-
  1945.12) ??????? ), can be considered as the
  archetype (??) of the optical resonator. It
  consists of a plane-parallel plate of thickness l
  and index n that is immersed(???)in a medium of
  index n'.

132
the internal???angle of incidence
the vacuum wavelength of the incident wave
133
(No Transcript)
134
D
F
E
135

• If the complex amplitude of the incidence
  wave is taken as Ai , then the partial
  reflections, B1 ,B2 , and so forth(??), are
  given by as shown by figure 4-2.
• r the reflection coefficient (????) (radio of
  reflected to incident amplitude)
• t is the transmission coefficient (????)
  for waves incident from n toward n, and r and
  t are the corresponding quantities for waves
  traveling from n toward n.

136

• The complex amplitude of the (total) reflected
  wave is given by
• for the transmitted wave
• For the complex amplitude of the total
  transmitted wave. We notice that the terms within
  the parentheses(???) in 4.1-2 and 4.1-3 form an
  infinite geometric progression (????), adding
  them, we get

4.1-2
4.1-3
4.1-4
4.1-5
137
where we used the fact that r-r, the
conservation-of-energy (????) relation that
applies to lossless mirrors
at the same time, we define
R the fraction of the intensity reflected T
the transmitted at each interface and will be
referred to in the following discussion as the
mirrors reflectance and transmittance.
138
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139
Consider the transmission characteristics of a
Fabry-Perot etalon, according to 4.1-7, the
transmission is unity whenever
4.1-8
For maximum transmission can be written as
4.1-9
140

• These are separated by the so-called free
  spectral range

• Theoretical transmission plots of a Fabry-Perot
  etalon are shown in Figure 4-3. The maximum
  transmission is unity, as stated previously. The
  minimum transmission, on the other hand,
  approaches zero as R approaches unity.

141
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142
If we allow for the existence of losses in the
elaton medium, the peak transmission is less than
unity. Taking the fractional intensity loss per
pass as (1-A), the maximum transmission drops
from unity to
4.1-11
143
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144
Are you a normal person????????

• During a visit to the mental asylum, a visitor
  asked the director ..., "What is the criterion
  that defines a patient to be institutionalized?"
  "Well..." said the director, "we fill up a
  bathtub, and we offer a teaspoon, a teacup, and a
  bucket to the patient and ask him to empty the
  bathtub." "Oh, I understand," said the visitor.
  "A normal person would choose the bucket as it is
  larger than the spoon or the teacup."
  "Noooooooo!" answered the director. "A normal
  person would pull the plug."
• ???????????????????,???????????????????????????
  ? ???,???,???????????,????????????,??????????
  ?????????? ?,????, ???????????????,
  ???????,??????? ??,????????????????

145
4.2 Fabry-Perot etalons as optical spectrum
analyzers
According to 4.1-8, the maximum transmission
of a Fabry-Perot etalon occurs when
?v the intermode frequency separation
146

• According to 4.2-2, we can tune the peak
  transmission frequency of the etalon by ?v by
  changing its length by half a wavelength. This
  property is utilized (??) in operating the etalon
  as a scanning interferometer. The optical signal
  to be analyzed passes through the etalon as its
  length is being swept (??).

147
If the width of the transmission peaks is
small compared to that of the spectral detail in
the incident optical beam signal, the output of
the etalon will constitute a replica (???) of the
spectral profile of the signal. In this
application it is important that the spectral
width of the signal beam be smaller than the
intermode spacing of the etalon so that the
ambiguity (??) due to simultaneous (????)
transmission through more than one transmission
peak is avoid. For the same reason the total
length scan is limited to
148
the operation of a scanning Fabry-Perot etalon
149
Intensity versus (???) frequency data
obtained by analyzing the output of a multimode
(???,????) He-Ne laser oscillating near 632.8nm
The peaks shown correspond to longitudinal (???)
laser modes, which will be discussed in section
4-5.
150

• It is clear from the foregoing (???) that
  when operating as a spectrum analyzer the etalon
  resolution---- that is, its ability to
  distinguish details in the spectrum---is limited
  by the finite width of its transmission peaks. If
  we take, somewhat arbitrarily, the limiting
  resolution of the etalon as the separation
  between the two frequencies at which the
  transmission is down to half of its peak value,
  from 4.1-7 we obtain

151
Assume
then
4.2-4
Defining the etalon finesse (???) as
Then
4.2-5
Fthe radio of the separation between peaks to
the width of a transmission bandpass (??). This
ratio can be read directly from the transmission
characteristics such as those of figure 4-4, for
which we obtain F26.
152

• Shcoolwork
• Study the part of 4.3 in page 132 by yourself.

quiet
153
Numerical Example Design of a Fabry-Perot Etalon
Consider the problem of designing a scanning
Fabry-Perot etalon to be used in studying the
mode structure of a He-Ne laser with the
following characteristics llaser100cm and the
region of oscillation??gain1.5109 Hz. The
free spectral range of the etalon ( that is, its
intermode spacing ) must exceed the spectral
region of interest, so from (4.1-10) we obtain
or
4.2-7
154
The separation between longitudinal modes of the
laser oscillation is c/2nllaser 1.5108 Hz
(here we assume n1). We choose the resolution of
the etalon to be a tenth of this value, so
spectral details as narrow as 1.5107Hz can be
resolved. According to (4.2-6), this resolution
can be achieved if ??1/2C/2nletalF 1.5107 Hz
or 2nletalF2103cm
155
To satisfy condition (4.2-7), we choose
2nletal20cm thus (4.2-8) is satisfied when
F100
(4.2-9) A finesse of 100 requires, according
to (4.2-5), a mirror reflectivity of
approximately 97 percent
As a practical note we may add that the finess,
as defined by the first equality in (4.2-6),
depends not only on R but also on the mirror
flatness and the beam angular spread. These
points are taken up in Problems 4-3 and 4-4.
156
Another important mode of optical spectrum
analysis performed with Fabry-Perot etalons
involves the fact that a noncollimated
monochromatic beam incident on the etalon will
emerge simultaneously, according to (4.1-8),
along many directions ?3, which corresponds to
the various order m. If the output is then
focused by a lens, each such direction? will give
rise to a circle in the focal plane on the lens,
and, therefore, each frequency component present
in the beam leads to a family of circles. This
mode of spectrum analysis is especially useful
under transient conditions where scanning etalons
cannot be employed. Further discussion of this
topic is include in Problem 4-6.
collimate ????????? Monochromatic ????,
??? transient ???,???
157
4.3 optical resonators with spherical (???)
mirrors
In this section we study the properties of
optical resonators formed by two opposing
spherical mirrors. We will show that the field
solutions inside the resonators are those of the
propagating Gaussian beams, which were considered
in chapter 2. It is, consequently, useful to
start by reviewing the properties of the beams.
158

• The field distribution corresponding to
  the (l,m) transverse mode (??) is given by
• The sign of R(z) is take as positive
  when the center of curvature is to the left of
  the wavefront, and vice versa (????).

4.3-1
4.3-2
159
The loci (??,locus?????) of the points at
which the beam intensity (watts per square meter)
is a given fraction of its intensity on the axis
are the hyperboloids (???).
The hyperbolas generated by the intersection of
these surfaces with planes that include the z
axis are shown in Figure 4-7. These hyperbolas
are normal to the phase fronts and thus
correspond to the local direction of energy flow.
The hyperboloid x2 y2 ?2 (z) is , according to
(4.3-1), the locus of the points where the
exponential factor in the field amplitude is down
to 1/e from its value on the axis. The quantity
?(z) is thus defined as the mode spot size at the
plane z.
160
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161
Given a beam of the type describe by 4.3-1, we
can form an optical resonator merely by inserting
at points z1 and z2 two reflectors with radii
of curvature that match those of the propagating
beam spherical phase fronts at these points.
Since the surface are normal to the direction of
energy propagation
162
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163
No rush ???

• AS STUDENTS in the college of veterinary medicine
  at Texas A M University, we frequently treated
  the farm animals at the state prison. While
  awkwardly performing a medical procedure on an
  unruly horse, a classmate said to the prisoner
  who was holding the animal, "Sorry I'm taking so
  long." "No problem," the prisoner replied, "I'm
  doing seven years.
• ??????AM????????????,????????????????????????????
  ?????????,?????????????????????,????????????,
  ????,????????????

164
Obituary ????

• The phone rang in the obituary department of the
  local newspaper. "How much does it cost to have
  an obituary printed"? asked the woman. "It's five
  dollars a word, ma'am," the clerk replied
  politely. "Fine," said the woman after a moment.
  "Got a pencil?" "Yes ma'am." "Got some paper?"
• "Yes ma'am." "Okay, write this down 'Cohen
  dead'." "That's all?" asked the clerk
  disbelievingly. "That's it." "I'm sorry ma'am, I
  should have told you - there's a five word
  minimum." "Yes, you should've," snapped the
  woman. Now let me think a minute... okay, got a
  pencil?" "Yes ma'am."
• "Got some paper?" "Yes, ma'am." "Okay, here goes
  'Cohen dead. Cadillac for Sale.'"

165

• ?????????????????????????????????????????,????
  ??????????,?????????,????????,????????,??
  ???,???????????????????????,???????,??,
  ????????,?????????????????,???????,???????,???
  ?????,?????????,????????,?????,????????
  ,???????????

166
1. Optical resonator algebra (???)
As mentioned in the preceding paragraphs, we
can form an optical resonator by using two
reflectors, one at z1, and the other at z2 ,
chosen so that their radii of curvature are the
same as those of the beam wave-fronts at the two
locations. The propagating beam mode (4.3-1) is
then reflected back and forth between the
reflectors without a change in the transverse
profile. The requisite radii of curvature are
167
from above, we can get
168
2. The symmetrical mirror resonator
The special case of a resonator with
symmetrically (about z0) placed mirrors merits a
few comments. The planar phase front at which the
minimum spot size occurs is, by symmetry, at z0.
Putting R2-R1R in (4.3-7) gives
169
Which yields the following expression for the
spot size at the mirrors
170
The value of R (for a given l ) for which
the mirror spot size is a minimum, is readily
found from 4.3-10 to be Rl. When this condition
is fulfilled we have what is called a symmetrical
confocal (??) resonator, since the two foci
(??),occurring at a distance of R/2 from the
mirrors, coincide. From 4.3-9, we obtain
4.3-11
whereas from 4.3-10, we get
4.3-12
171
3. Design of a Symmetrical Resonator
the spot size at the mirrors would have the value
172
where R400l800m
173
thus, to increase the mirror spot size from its
minimum (confocal) value of 0.0798cm to 0.3cm, we
must use exceedingly plane mirrors (R800m). This
also shows that even small mirror curvature (that
is, large R) give rise to narrow beams.
174
Results

175

• Shcoolwork
• Study section 4.4 by yourself.

quiet
176
4.4 Mode stability criteria(??????)
The ability of an optical resonator to
support low (diffraction) loss modes depends on
the mirrors separation l and their radii of
curvature R1 and R2. To illustrate (????) this
point, consider first the symmetric resonator
with R1R2R .
The ratio of the mirror spot size at a give
l/R to its minimum confocal (l/R1)
4.4-1
177
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178

• Shcoolwork
• Study section 4.5 by yourself.

quiet
179
4.6 Resonance Frequencies (????) of
Optical Resonator
180
If we consider a spherical mirror resonator
with mirrors at z1 and z2 , the resonance
condition for the l,m mode can be written as
4.6-1
in which dz2-z1 is the resonator length. It
follows that
4.6-2
4.6-3
for the intermode frequency spacing.
181

• Shcoolwork
• In the case of confocal resonator (Rd),
  please derive the ?v.

quiet
182
4.7 Losses in optical resonators
An understanding of the mechanisms by which
electro-magnetic energy is dissipated in optical
resonators and the ability to control them are of
major importance in understanding and operating a
variety of optical devices. For historical
reasons as well as for reasons of convenience,
these losses are often characterized by a number
of different parameters.
183
The decay lifetime (photon lifetime) tc of a
cavity mode is defined by means of the equation
4.7-1
184
So
4.7-2
So that, we have
4.7-3
185
Where the approximate equality applies when
R1R21 . The quality factor of the resonator is
defined universally as
4.7-4
4.7-5
186

• Shcoolwork
• please summarize the most common loss
  mechanisms in optical resnoator and write them in
  both English and Chinese.

quiet
187
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188
(No Transcript)
189
Laser Diode

• Chapter 7 Some Specific Laser Systems

Introduction Laser , that is light
amplification by stimulated emission of radiation
, have much predominance(??) compared with common
light , because laser come into being(??, ??) in
condition of stimulated by external light or
electricity , the process is that the stimulation
cause the electron in low energy jump to high
energy level .and when the electron in high
energy level much than in the low level , that
can be said reversion of electron .

190
????
191
???? continued
192
????
193
the characteristics of laser are that (1)?high
luminance(??) lasers luminance is
100000 much than sun lights . (2)?simplicity
of light (??) The laser is very pure , but
sun light or white light can be
decomposed and got seven pure light , there are
red , orange , yellow , green , blue ,
indigo(?, ??), purple .
194

• (3) ?high correlation (???)
• Lasers frequency , phase , direction of
  librations (??) are all coherent(???),. so its
  correlation is very high.
• (4)?strong direction
• Common light is emitted towards far and
  near (????) , but the laser beam transmit
  directed(???).and laser radiations(??) angle is
  very small , it close to parallel light .
  because laser radiations angle is very small ,
  it can transmit far from lamp-house (??) .

195

• Owing to the advantages of laser , on the
  high road of information in future , it can be
  applied far and wide (???)
• for example , fiber is primary channel
  to translate data in the high road of information
  . at the same time , laser beam can be made much
  thin than fiber , so the fiber can use laser beam
  translate information in it .

196
All kinds of lasers
197
7.1 the theory of excitated radiation
Laser , that is light amplification by
stimulated emission of radiation. So first, we
will learn the theory of excitated radiation.
2
E2
st
sp
st
1
E1
??????
198
Lasing Principle
199
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200
On the condition of balance, the distribution of
particle numbers in E2 and E1 has
7.1-1
From the viewpoint (??) of energy, we know that
when a particle transits to low energy from high
energy, it will release a photon
7.1-2
It is the same process form low energy to high
energy.
201

• The characteristics about spontaneous
  (???)radiation
• un-ruled , spontaneous, stochastic (???) ,
  in a word it is an un-correlative (???)
  radiation and a noise.
• The characteristics about excitated(??)radiation
• there are four uniform aspects, the same
  polarized direction (????), the same frequency,
  the same phase and the same radiated direction.

202
Off from work ???

• My personnel-management class consisted mainly of
  adult, working students. One night while
  discussing job enrichment, the teacher asked if
  any of us would be happy doing what we did that
  day for the rest of our lives. A student in the
  back raised his hand. Surprised, the teacher
  asked him, "What did you do today?" Smiling, he
  said, "I took off from work."
• ???????????????,??????????????????????????????????
  ??????????????????????????????????????????????????
  ???????????,???????,?????????,??????,???????

203
7.2 the technology of pumping
There are two common pumping technologies
optical pumping(be fit for solid and liquid
laser) and electric pumping.
204
7.3 pumping and laser efficiency
Absorption band or groups of levels
3
Laser transition
Pump transition??
hv21
hv30
Fast decay
Figure 7-1 pumping-oscillation cycle of a typical
laser
205

• It is evident from this figure that the
  minimum energy input per output photon is hv30 ,
  so the power efficiency of the laser cannot
  exceed
• to which quantity we will refer as the atomic
  quantum efficiency.

7.3-1
206

• The overall laser efficiency depends on
• (1)? the fraction of the total pump power
  that is effective in transferring atoms into
  level 3
• (2)the pumping quantum efficiency defined as the
  fraction of the atoms that once in 3, make a
  transition to 2.
• The product of the last two factors, which
  constitutes an upper limit on the efficiency of
  optically pumped lasers, ranges from about 1
  percent to about 30 percent.

207
7.4 laser systems

• Ruby (???)
• Nd3 YAG ?-?????
• Nd3glass (???)
• He-Ne
• CO2
• Ar
• Excimer???
• Organic-dye ????

208
1. Ruby laser(???)

• Ruby was the first material in which
  laser action was demonstrated and is still one
  of the most useful materials

materialAl2O3 crystal
Active particles Cr3 ? (chromium) , it
concentrations 0.05 by weight
Characters
(2) The pumping of ruby is usually performed by
subjecting it to the light of intense flashlamps
(???)(quite similar to the types used in flash
photography) (?????). And A portion of this light
that corresponds in frequency to the two
absorption bands 4F2 and 4F1 is absorbed,
thereby causing Cr3 ions to be transferred
into these levels. (3) three-level laser. The
lifetime of atoms in the upper laser level
E is t210-3 s . Each decay results in the
(spontaneous) emission of a photon, so t2tspont..
209
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210
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211
The pertinent(???)energy level diagram
212
An absorption spectrum of a typical ruby with two
orientations of optical field relative to the c
axis is shown as
1?The 300K data were derived from transmittance
measurements on pink ruby with an average Cr ion
concentration of 1.881019 cm-3 2? The two main
peaks corresponds to absorption into the useful
4F1 and 4F2 bands, which are responsible
for the characteristic (ruby) color.
213
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214
The absorption coefficient is
7.4-1

• ?????????????????(??)??????

215
Figure 1 The detailed plot of the absorption near
the laser emission wavelength is
Sample was a pink ruby laser rod having a 900
c-axis orientation with respect to the rod axis
and a Cr concentration of 1.581019cm-3
216
Figure 2 The width v of the laser transition as
a function of temperature
At room temperature ?V11cm-1
217
Figure 3 Typical setup of a pulsed ruby laser
using flashlamp puming and external mirrors.
The helical (????)flashlamp surrounds the ruby
rod (?). The flash excitation is provided by the
discharge (??) of the charge (??) stored in a
capacitor bank (????) across the lamp.
218
7.4-2
absorption coefficient of the crystal
219
7.4-3
If the lifetime of atoms in level 2 is
considerably longer than the flash duration( for
example t2310-3 s , tflash510-4 s), the
spontaneous decay out of level 2 during the time
of the flash pulse can be neglected, and then N2
represents the population of level 2 after the
flash.
220
2. NdYAG(Y3Al5O12) laser
materialYAG
Active particles Nd3
Characters
2. Four-level laser
3. Spontaneous lifetime tspont 5.510-4s

• 5. The optical gain constant is approximately 75
  times that of ruby, so the oscillation threshold
  is very low ,and that can explains the easy
  continuous operation of this laser compare to
  ruby.
• 6. The absorption responsible for populating the
  upper laser level takes place in a number of
  bands between 13000 and 25000 cm-1 .

221
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222
(No Transcript)
223

• ? 2 ??
• ???YAG???????????????

224

• ? 3 ???
• ?????????????,??????????????,??????????????,?
  ?????????????????????????,????????????????,???????
  ???????????,???????????????,?????????????????,????
  ?????????

225

• ? 4 ????????
• ??????????????E1??????E3,E3???????????????????E2,?
  E2??????????,???????E1??????,?????E2?E1??????????
  ????????,??????????????????????????????????????
• ?????????? Y3Al5O12 ??????????,????
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226

• ? 5 ??????????1

227

• ? 6 ??????????2

228

• ? 7 ?Q?????
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