Download Lecture Notes 9 - Northeastern University

Optics for Engineers
Chapter 9
Charles A. DiMarzio
Northeastern University
Mar. 2014
Gaussian Beams
• Applications
– Many Laser Beams
– Minimum–Uncertainty
– Simple Equations
– Good Approximation
– Extensible (e.g. Hermite–
Gaussian)
• Equations
– Solution of Helmholz
Equation
– Solution to Laser Cavity
– Kogelnik and Li, 1966
– Spherical Gaussian
Waves
– “Gaussians Are Forever”
Mar. 2014
• Imaginary Part of Field
– Gaussian Profile
– Spherical Wavefront
• Focusing and Propagation
– Simple Equations
– Relation to ABCD Matrix
c C. DiMarzio (Based on Optics for Engineers, CRC Press)
slides9r1–1
Derivation (1)
• Spherical Wave and Paraxial Approximation
√
√
√
2 +y 2 +z 2
x
jk
Psphere e
Psphere jkz jk x2+y2
2z
√
Esphere =
≈
e e
4π
4πz
x2 + y 2 + z 2
• Radius of Curvature = z: Substitute Complex q
√
√
√
2 +y 2 +q 2
jk
x
Psphere e
Psphere 1 jkq jk x2+y2
2q
√
e e
E=
≈
4π
4π
q
x2 + y 2 + q 2
• Separate Real and Imaginary Parts of q in Exponents
√
E≈
Mar. 2014
Psphere 1 jkq jk(x2+y2)< 1 −k(x2+y 2)= 1
2q e
2q
e e
4π
q
c C. DiMarzio (Based on Optics for Engineers, CRC Press)
slides9r1–2
Derivation (2)
• Curvature and Gaussian Profile
jk
e
(
x2+y 2
)
1
< 2q
=
x2 +y 2
jk 2ρ
e
−k
e
(
x2+y 2
)
1
= 2q
=e
2
2
−k x +y
0
2b
• Definitions
q = z + jb
z 2 + b2
ρ=
z
Mar. 2014
1
1
j
= − 0
q
ρ b
z 2 + b2
0
b =
b
c C. DiMarzio (Based on Optics for Engineers, CRC Press)
slides9r1–3
Gaussian Beam Parameters
• Gouy Phase
• Beam Radius, w
w2 =
2b0
b0 =
k
(
e
−k x2+y
)
2
1
= 2q
=
πw2
λ
=
πd2
4λ
2
2
− x +y
e w2
• Curvature (Prev. Pg.)
(
e
jk x2 +y
)
2
1
< 2q
=
2 +y 2
jk x 2ρ
e
z
1
1
1
−j
arctan
b
=
=√
e
q
z + jb
z 2 + b2
z
λ 1
−
arctan
b = e−jψ
e
= √
b πw2
• Linear Phase
ejkq = ejk(z+jb) = ejkz e−kb
• Normalized Complete Equation
√
E≈
Mar. 2014
2 +y 2
x2 +y 2
−
2ρ e
w2
2P − arctan z jkz jk x
be
e
e
2
πw
c C. DiMarzio (Based on Optics for Engineers, CRC Press)
slides9r1–4
Physical Meaning of Parameters
√
E≈
2P jkz jk x2 +y2 − x2 +y2 2 −jψ
e e 2ρ e w e
2
πw
• Radius of Curvature, ρ
Dashed Black Line
1
1
=<
ρ
q
• Distance from Waist, z
z = <q
• Rayleigh Range, b → d0
b = =q
Mar. 2014
b=
πw02
λ
=
πd20
4λ
• Beam Diameter, d
Black Diamonds
1
1
=
−=
b0
q
πw2
πd2
b =
=
λ
4λ
0
c C. DiMarzio (Based on Optics for Engineers, CRC Press)
slides9r1–5
Gouy Phase
• Phase Term
z
ψ = arctan
b
• See White Circle
• Plot is =E
Mar. 2014
c C. DiMarzio (Based on Optics for Engineers, CRC Press)
slides9r1–6
The Really Useful Equations
• Beam Diameter, d
√
z2
1+ 2
b
b2
ρ=z+
z
5
5
4.5
4
4
3
ρ/b, Scaled Radius of Curvature
d/d0, Scaled Beam Diameter
d = d0
• Radius of Curvature, ρ
3.5
3
2.5
2
1.5
1
0.5
0
−5
1
0
−1
−2
−3
−4
0
z/b, Scaled Axial Distance
• Near Field dg ≈ d0
• Far Field dd ≈ π4 dλ z
0
Mar. 2014
2
5
−5
−5
0
z/b, Scaled Axial Distance
• Near Field ρ ≈ b2/z → ∞
• Far Field ρ ≈ z → ∞
c C. DiMarzio (Based on Optics for Engineers, CRC Press)
slides9r1–7
5
Six Questions
• Four Parameters for q
– z, Distance from Waist
– b or d0, Waist Diameter
– ρ, Radius of Curvature
– b0 or d Local Diameter
(at z)
• Two Independent Numbers
for Complex q
• Pick Any Pair
• Solve for the Rest
• Practical Problems for
Each Pair
Mar. 2014
z
ρ
z
ρ
z
b
and
and
and
and
and
and
b known
b0
b0
b
ρ
b0
c C. DiMarzio (Based on Optics for Engineers, CRC Press)
slides9r1–8
Question 1: Known z and b
• Sample Application: Starting with a collimated (plane–wave)
Gaussian, what are d and ρ at distance, z?
• Example: Green Laser Pointer: λ = 532nm.
• Solution
πd2
0 = 5.91m
b=
4λ
ρ=z+
b2
z
√
z2
d = d0 1 + 2
b
(
z = 3m:
z = 30m:
z = 376, 000km:
Mar. 2014
ρ = 14.6m
ρ = 31.2m
ρ = 376, 000km
d = 2.2mm
d = 10.4mm
d = 127km
4λz
π d0
)
(1mm)
(10.2mm)
(127km)
c C. DiMarzio (Based on Optics for Engineers, CRC Press)
slides9r1–9
Question 2: Known ρ and b0
• Sample Application: What is the waist location for a focused
beam?
• Example: Laser Radar, λ = 10.59µm, d = 30cm, b0 = 6.6km
• Solution
ρ = −f
1
1
j
= − 0
q
ρ b
z = <q = −
Mar. 2014
f
(
1
j
1
= − 0
q
ρ b
b0 = πd2/ (4λ)
)
4λf 2
1 + πd2
1
ρ
j
b0
q= 1
+ 1
1
1
+
+
2
02
2
ρ
b
ρ
b02
b = =q =
b0
b0 + 1
f2
d2
0=
c C. DiMarzio (Based on Optics for Engineers, CRC Press)
d2
(
πd2
)2
1 + 4λf
slides9r1–10
Question 2 Results
4
30
3.5
d0, Waist Diameter, cm
−z, Distance to Waist, km
35
3
2.5
2
1.5
25
20
15
10
1
5
0.5
0
0
5
10
15
20
25
30
35
0
0
40
f, Focal Length, km
A. Waist Location
5
10
15
20
f, Focal Length, km
25
30
35
B. Waist Diameter
50
45
4
10
35
30
I0/P, m−2
d, Beam Diameter, cm
40
25
20
15
2
10
0
10
10
5
−2
0
0
0.5
1
1.5
z, Distance, km
2
2.5
C. Beam Diameter (f = 1km)
Mar. 2014
10 −2
10
−1
10
10
0
10
1
z, Distance, km
Axial Irradiance
c C. DiMarzio (Based on Optics for Engineers, CRC Press)
slides9r1–11
Question 2 Summary
• Maximum Distance to Waist: (−z)max = b/2 at f = b
• Two Ways to Make a Waist at any (−z) < b/2
• Near Field of d0; Strong Focusing:
– (−z) ≈ −ρ = f
– d0 = π4 λd
• Depth of Focus: Quadratic in f
8f 2λ
1
zc = 2b1 =
(
)2
2
4λf
πd
+1
πd2
• Collimated Beam as f → ∞
Mar. 2014
c C. DiMarzio (Based on Optics for Engineers, CRC Press)
slides9r1–12
Geometric and Diffraction
Diameters
• Geometric Optics
• Diameter at Any Distance
2
+
d
d2 = d2
g
d
• Simple Calculation
• Intuitive
• Valid for Gaussians
Mar. 2014
|z − f |
dg = d
f
• Diffraction Limit
4λ
dd =
z
πd
• Fraunhofer Zone
4λ
d0 ≈ dd =
z
πd
c C. DiMarzio (Based on Optics for Engineers, CRC Press)
slides9r1–13
Question 3: Known z and b0
• Sample Application: What lens to make a waist at −z for
Negative z? How big is the waist?
• Example: Endoscope: λ = 1.06µm, waist at −z = 1mm with
d never > 100µm
• Solution: b0 = πd2/(4λ)
b2 + z 2
0
b =
Solve for b only if
b
b=
b0 ±
√
b02 − 4z 2
2
(z < b0/2)
b2
ρ=z+
z
• Typically 2 Solutions
Mar. 2014
c C. DiMarzio (Based on Optics for Engineers, CRC Press)
slides9r1–14
Endoscope Focusing
• Fixed Diameter at Source d = 100µm, z = 1mm
πd2
0
b =
= 7.4mm
4λ
b=
b0 ±
√
b02 − 4z 2
2
z = −1mm
b = 3.7047mm ± 3.5672mm
• Options
b = 137µm
d0 = 13.6µm
or
b = 7.27mm
d0 = 99µm
• Probably Choose the First
– High Irradiance
– Small Depth of Field
Mar. 2014
c C. DiMarzio (Based on Optics for Engineers, CRC Press)
slides9r1–15
Question 4: Known ρ and b
• Sample Application: How long must the laser be to get a
given d0 with a mirror of curvature ρ?
• Example: Later
• Solution: Solve for z if d0 Small Enough
z 2 + b2
ρ=
z
z=
ρ±
√
(2b < ρ)
ρ2 − 4b2
2
√
z2
d = d0 1 + 2
b
Mar. 2014
c C. DiMarzio (Based on Optics for Engineers, CRC Press)
slides9r1–16
Question 5: Known z and ρ
• Sample Application: How large is the beam from a laser
cavity?
• Example: Later
• Solution: Solve for b
z 2 + b2
ρ=
z
√
√
b=
ρz − z 2
b=z
√
d0 =
Mar. 2014
(ρ > z)
ρ
−1
z
or
q = z + jb
√
4λb
π
z2
d = d0 1 + 2
b
c C. DiMarzio (Based on Optics for Engineers, CRC Press)
slides9r1–17
Question 6: Known b and b0
• Sample Application: How do we launch a beam of diameter,
d, into a fiber with diameter, d0?
• Example: Beam diameter, d = 15µm and desired diameter,
d0 = 5µm
• Solution: Solve for z
z 2 + b2
0
b =
(b0 > b of course: d > d0)
b
√
z=±
Mar. 2014
bb0 − b2
b2
ρ=z+
z
c C. DiMarzio (Based on Optics for Engineers, CRC Press)
slides9r1–18
Fiber Launch
• Equations
√
z=±
bb0 − b2
b2
ρ=z+
z
• Inputs for d = 15µm and d0 = 5µm
πd2
0 = 13µm
b=
4λ
πd2
0
b =
= 118µm
4λ
• Results
√
z=−
bb0 − b2 = −37µm
b2
= −41.7µm
ρ=z+
z
• Lens: f = −ρ = 41.7µm, with Dlens > 15µm, (<f/3)
• Distance: −z = 37µm (A Little Shorter than f )
Mar. 2014
c C. DiMarzio (Based on Optics for Engineers, CRC Press)
slides9r1–19
Gaussian Beam Characteristics
• The complex radius of curvature can be manipulated to solve
problems of Gaussian beam propagation.
• The complex radius of curvature and its inverse contain four
terms, which are related in such a way that only two of them
are independent.
• Given two parameters it is possible to characterize the Gaussian beam completely. Six examples have been discussed.
• Solutions are at worst quadratic, producing zero, one, or two
solutions.
• Other, more complicated problems can also be posed and
solved using this formulation.
Mar. 2014
c C. DiMarzio (Based on Optics for Engineers, CRC Press)
slides9r1–20
Gaussian Beam Propagation
• Propagation through Free Space (q = z + jb)
q (z 2 ) = q (z 1 ) + z 2 − z 1
• Propagation through a Lens ( 1q = 1ρ − bj0 )
1
1
1
+ 0=
s
s
f
1
1 1
−
=
0
q
q
f
Mar. 2014
1
1
1
+
=
0
ρ
−ρ
f
or
1
1 1
= −
0
ρ
ρ f
1
1
−P
=
0
q
q
c C. DiMarzio (Based on Optics for Engineers, CRC Press)
slides9r1–21
Matrix Optics Again
• General
qout =
Aqin + B
Cqin + D
• Translation
q + z12
q2 = 1
= q1 + z12
0+1
• Lens
q0 =
q+0
− nP0 q + nn0
1
n
P
=
− 0
0
0
q
qn
n
– In Air
1
1 1
1
−
P
=
−
=
0
q
q
q
f
Mar. 2014
c C. DiMarzio (Based on Optics for Engineers, CRC Press)
slides9r1–22
Refraction at a Surface
• General
q0 =
q+0
n−n0 q + n
n0 R
n0
n − n0
n
1
=
+
q0
n0R
qn0
• Planar Surface
n0
0
q =q
n
• Rayleigh Range in a Medium
πd2
0
b=
4λ/n
Mar. 2014
c C. DiMarzio (Based on Optics for Engineers, CRC Press)
slides9r1–23
Propagation Example with a
Lens: F to F 0
• Waist at Front Focal Plane, F
q0 = jb0
πd2
0
b0 =
4λ
• Three Steps
q 1 = q 0 + f1
1
1
1
=
−
q10
q1 f 1
q2 = q10 + f1
• Result: <z2 = 0 (Waist)
q1 = jb0 + f1
4λ
d2 =
f1
π d0
Mar. 2014
f12
0
q1 = −f1 + j
b0
jf12
q2 =
b0
(Fourier Transform)
c C. DiMarzio (Based on Optics for Engineers, CRC Press)
slides9r1–24
Propagation Example
Continued: A Second Lens (f3)
• Relay Telescope
d4 =
4λ
f
f3 = 3 d0
π d2
f1
• Matrix Optics: First Lens
(
M02 = T12L1T01 =
1
0
f1
1
)(
1
− f1
1
0
1
)(
1
0
0+f
f2
Aq0 + B
= 1
q2 =
=−
Cq0 + q0
q0
− f q0 + 0
f1
1
)
(
=
0
− f1
1
4λ
d2 =
f1
π d0
• Both Lenses (f3 = f1): Inverting 1:1 Telecentric System
(
M04 = M24M02 =
0
− f1
3
f3
0
)(
0
− f1
1
f1
0
)
f1
0
)
(
=
−1
0
0
−1
)
−q0 + 0
Aq0 + B
=
= q0
q4 =
Cq0 + D
0−1
Mar. 2014
c C. DiMarzio (Based on Optics for Engineers, CRC Press)
slides9r1–25
Two 1:1 Relays
• Telecentric (Previous Page)
M04 = M24M02 =
(
0
− f1
3
f3
0
(
)(
0
− f1
−1
0
1
0
−1
f1
0
)
• Afocal
q4 = q0 = jb0
– Plane Wave
– No Field Curvature
– ρ→∞
Mar. 2014
• Single Lens
(4–f Configuration)
)
Mac = TbcLbTab =
=
(
1
0
2f
1
)(
(
1
−1
f
−1
−1
f
qc =
0
1
0
−1
)(
1
0
2f
1
)
Aqa + B
=
Cqa + D
−1qa + 0
1
=
1+ 1
−1
q
−
1
a
f
f
qa
– ρ=f
c C. DiMarzio (Based on Optics for Engineers, CRC Press)
slides9r1–26
)
=
Summary of Gaussian Beam
Propagation
• Once two parameters of a Gaussian beam are known, then
it’s propagation can be analyzed using simple equations for
translation, refraction, and focusing.
• Translation changes the beam size and curvature. It is represented by adding the distance to q.
• Refraction through a lens changes the curvature, but keeps
the beam diameter constant. It is represented by subtracting
the optical power of the lens from the inverse of q.
• Refraction through a dielectric interface changes the curvature and scale factor of q.
• These equations can be generalized using the ABCD matrices
developed for geometric optics.
Mar. 2014
c C. DiMarzio (Based on Optics for Engineers, CRC Press)
slides9r1–27
The Collins Chart
Confocal Parameter
Lines of Constant z (vertical) and Constant b (horizontal)
Curves of Constant b0 (circles) and Constant ρ (semicircles)
Now Mostly a Visualization Aid
2
1.5
1
0.5
0
−2
−0.5
0
0.5
1
1.5
2
z, Axial Distance
• Focusing (Along Circles)
• Translation (Horizontal)
• ρ Changes, d & b0 Don’t
• z Changes, b Doesn’t
Mar. 2014
−1.5
−1
c C. DiMarzio (Based on Optics for Engineers, CRC Press)
slides9r1–28
5
5
4
4
Confocal Parameter
Confocal Parameter
Two 1:1 Relays on Collins Chart
3
2
1
0
−3
3
2
1
−2
−1
0
1
z, Axial Distance
A. Afocal Telescope
Mar. 2014
2
3
0
−3
−2
−1
0
1
z, Axial Distance
2
3
B. Single Lens
c C. DiMarzio (Based on Optics for Engineers, CRC Press)
slides9r1–29
Minimum Beam Diameter for
Distance z12
Confocal Parameter
2
1.5
1
0.5
0
−1.5
Mar. 2014
−1
−0.5
0
0.5
z, Axial Distance
1
c C. DiMarzio (Based on Optics for Engineers, CRC Press)
1.5
slides9r1–30
Waist at −z for diameter, d
1
0.5
0
−1.5
−1
−0.5
0
0.5
1
2
2
1.5
1.5
Confocal Parameter
1.5
Confocal Parameter
b, Confocal Parameter
2
1
0.5
0
−1.5
1.5
z, Axial Distance
B. No Solution
|z| > b0/2
−1
−0.5
0
0.5
z, Axial Distance
1
C. One Solution
|z| = b0/2
1.5
1
0.5
0
−1.5
−1
−0.5
0
0.5
z, Axial Distance
1
D. Two Solutions
|z| < b0/2
• Question 3
b2 + z 2
0
b =
Solve for b only if
b
b=
Mar. 2014
b0 ±
√
b02 − 4z 2
2
(z < b0/2)
b2
ρ=z+
z
c C. DiMarzio (Based on Optics for Engineers, CRC Press)
slides9r1–31
1.5
Stable Laser Cavity Design
A. Flat Output Coupler
B. Flat Rear Mirror
C. Confocal Resonator
Mar. 2014
c C. DiMarzio (Based on Optics for Engineers, CRC Press)
slides9r1–32
Steady State in Laser Cavity
• The amplitude after a round trip is unchanged. This means
that any loss (including power released as output) must be
offset by corresponding gain. (Gain Saturation)
• The phase after a round trip must be unchanged. We discussed, in our study of interference, how this requirement on
the axial phase change, ejkz , affects the laser frequency.
• The beam shape must be unchanged, so that the phase and
amplitude is the same for all x and y. This is the subject to
be considered in this section.
Mar. 2014
c C. DiMarzio (Based on Optics for Engineers, CRC Press)
slides9r1–33
Design Problem
• Carbon Dioxide Laser: P(20), λ = 10.59µm
• Beam Output: Collimated, 5mm Diameter
• Cavity Length: 1m (Probably because of Gain)
• Solution
– Collimated Output: Flat Output Coupler
– Rear Mirror to Match Curvature at z = −1m
b = 1.85m
ρ = −4.44m
d = 5.7mm
– Rear Mirror Concave, ρ = −4.44m
– Diameter Larger than d = 5.7mm (Typically 1.5X)
Mar. 2014
c C. DiMarzio (Based on Optics for Engineers, CRC Press)
slides9r1–34
Stable Cavity Examples
Output toward Bottom
Mar. 2014
A. Flat Output Coupler
B. Flat Rear Mirror
C. Confocal Resonator
D. Focusing Cavity
c C. DiMarzio (Based on Optics for Engineers, CRC Press)
slides9r1–35
Laser Cavity on the Collins
Chart
Given ρ1 ρ2 and z12
b, Confocal Parameter
3
2.5
2
1.5
1
0.5
0
−2
−1.5
−1
−0.5
0
0.5
1
1.5
2
z, Axial Distance
Mar. 2014
c C. DiMarzio (Based on Optics for Engineers, CRC Press)
slides9r1–36
More Complicated Cavities
• Ring Laser with Focusing Lens
• Intracavity Lens (Potential Loss)
• Many Other Configurations
Mar. 2014
c C. DiMarzio (Based on Optics for Engineers, CRC Press)
slides9r1–37
Matrix Optics for Stable Cavities
• Round–Trip Equation
qout = qin
• Need to Solve This:
Aq + B
q=
Cq + D
Cq 2 + (D − A)q − B = 0
• Result (Need Imaginary Solutions)
q=
A−D±
√
(A − D)2 + 4CB
2C
(A − D)2 + 4CB < 0
Mar. 2014
c C. DiMarzio (Based on Optics for Engineers, CRC Press)
slides9r1–38
Stability Condition
• Argument of Square Root Must be Negative
(A − D)2 + 4CB < 0
• Determinant Condition (n = n0 on Round Trip)
AD − CB = 1
• Result
(D − A)2 + 4DA < 4
Mar. 2014
(D + A)2 < 4
c C. DiMarzio (Based on Optics for Engineers, CRC Press)
slides9r1–39
Summary of Stable Cavity
Design
• Stable cavities or stable optical resonators have mirrors and
possibly other optical elements that cause a Gaussian beam
to replicate itself upon one round trip.
• Resonator design can be accomplished using Gaussian beam
propagation equations.
• Most cavities will support only one Gaussian beam, and the
beam parameters are determined by the curvature of the
mirrors and their spacing along with the behavior of any intervening optics.
• Matrix optics can be used to determine the round–trip matrix,
which can be used to find the Gaussian beam parameters.
• The Collins chart can help to evaluate the stability of a laser
cavity.
Mar. 2014
c C. DiMarzio (Based on Optics for Engineers, CRC Press)
slides9r1–40
Hermite–Gaussian Modes
• Mode Definitions (Solutions of Helmholz Equation)
hmn (x, y, z ) = hm (x, z ) hn (y, z )
• One–Dimensional Functions with Hermite Polynomials, H
hm (x, z ) =
( )1/4 √
2
π
• Hermite Polynomials
(
)
2
x2 jkx
1
x
w2 e 2ρ ejψm
H
e
m
2mm!w
w
H1 (x) = 1
H2 (x) = 2x
Hm+1 (x) = 2xHm (x) − 2 (m − 1) Hm−1 (x)
• Gouy Phase for Hermite–Gaussian Modes
(
)
1
z
ψm =
+ m arctan
2
b
Mar. 2014
c C. DiMarzio (Based on Optics for Engineers, CRC Press)
slides9r1–41
Some Hermite–Gaussian Modes
0:0
(0:1) + i (1:0) = Donut
0:1
1:0
Mar. 2014
1:1
1:3
2:0
2:1
2:3
5:0
5:1
5:3
c C. DiMarzio (Based on Optics for Engineers, CRC Press)
slides9r1–42
Expansion in Hermite–Gaussian
Modes
• Expand Field in Hermite–Gaussian Modes
• Ortho–Normal Basis Set
E (x, y, z ) =
∞
∑
∞
∑
Cmnhmn (x, y, z )
m=0 n=0
• May Take a Lot of Terms (but choose w wisely)
• Power in Each Mode (and Phase)
∗
Pmn = CmnCmn
Mar. 2014
φ = 6 Cmn
c C. DiMarzio (Based on Optics for Engineers, CRC Press)
slides9r1–43
Finding the Coefficients, C
for Given E(x, y, z)
∫
∫
∞
−∞
∫
−∞
∫
∞
∞
E (x, y, z) h∗m0 n0 (x, y, z) dx dy =
∞
∗
E (x, y, z) h (x, y, z) dx dy =
−∞
−∞
∫
∞
−∞
∫
∫
∞
∫
−∞
∞ ∑
∞
∑
m=0 n=0
∞
−∞
∞ ∑
∞
∑
−∞ m=0 n=0
∫
Cmn
∞
−∞
hmn (x, y, z) h∗m0 n0 (x, y, z) dx dy = δm,m0 δn,n0
∫
Cm0n0 =
Mar. 2014
∞
∫
−∞ ∞
∫
Cmn hmn (x, y, z) h∗m0 n0 (x, y, z) dx dy
∞
−∞
hmn (x, y, z) h∗m0 n0 (x, y, z) dx dy
Ortho–Normality
E (x, y, z ) h∗m0n0 (x, y, z ) dx dy.
c C. DiMarzio (Based on Optics for Engineers, CRC Press)
slides9r1–44
Picking the Best Waist Size
• H–G Expansion Works for Any w
• But it Might Not Converge Nicely
• Pick the Best w to Maximize Power in T EM00 Term
C00 = min [C00 (w)]
∫
C00 (w) =
Mar. 2014
∫
−∞ ∞
E (x, y, z ) h∗00 (x, y, z ) dx dy
c C. DiMarzio (Based on Optics for Engineers, CRC Press)
slides9r1–45
−3
x 10
6
−5
4
0
2
5
−5
0
0
5
C00, Coefficient
Expanding Uniform Circular
Wave in H–G Modes
0.26
0
0.25
−10
10
−20
0.24
−30
15
0.23
1
−40
1.5
2
2.5
w, Gaussian radius
B.
A.
5
20
5
10
15
20
−50
C.
7
6
5
0
4
3
2
1
5
−5
0
D.
Mar. 2014
5
0
−20
−40
−60
−5
0
5
x, Radial Distance
E.
Irradiance, dB
x 10
8
−5
Irradiance, dB
−3
0
−20
−40
−60
−10
0
10
x, Radial Distance
F.
c C. DiMarzio (Based on Optics for Engineers, CRC Press)
slides9r1–46
Coupling Equations and Mode
Losses for Finite Apertures
• Integration over Finite Aperture
∫ ∫
Kmnm0n0 =
hmn (x, y, z ) h∗m0n0 (x, y, z ) dx dy = δm,m0 δn,n0
aperture



0
C00
K0000
 0 

 C01 


 =  K0100
 0 

 C10 
 K1000
...
Mar. 2014
...
K0001
K0101
K1001
...
K0010
K0110
K1010
...


...
C00


 C01 
... 



. . .   C10 

...
...
c C. DiMarzio (Based on Optics for Engineers, CRC Press)
slides9r1–47
Transverse Mode Beating:
Mode Frequencies
• Round Trip (Additional Phase Shift beyond Chapter 7)
(
z1
z2
− arctan
2k` + 2 (1 + m + n) arctan
b
b
)
= 2πN
(
f (m, n, N )
1
z
z
2
` + (1 + m + n) arctan 2 − arctan 1
c
π
b
b
)
=N
arctan zb2 − arctan zb1
c
c
−
f (m, n, N ) = N
(1 + m + n)
2` 2`
π
Mar. 2014
c C. DiMarzio (Based on Optics for Engineers, CRC Press)
slides9r1–48
Transverse Mode Beating:
Mode Control
• Transverse Mode Beat Frequency
z
z
c arctan b2 − arctan b1
f (1, n, N ) − f (0, n, N ) =
2`
π
• Carbon Dioxide Laser b = 1.85m, z = 1m
(
1m − arctan 0m
arctan 1.85m
1.85m
c
2m
π
• Solution: Include Appropriate Aperture
– Suppress Higher Order Transverse Modes
– K0000gρ1ρ2 > 1
– Kmnmngρ1ρ2 < 1 for m 6= 0, n 6= 0
f (1, n, N )−f (0, n, N ) =
Mar. 2014
c C. DiMarzio (Based on Optics for Engineers, CRC Press)
)
= 24MHz
slides9r1–49
Summary of H–G Modes
• Hermite–Gaussian modes (Equation ) form an orthogonal basis set of
functions which can be used to describe arbitrary beams.
• The beam can be expanded in a series.
• Coefficients of the expansion can be calculated.
• Each mode maintains its own profile with a size w and radius of curvature,
ρ given by the usual Gaussian–beam equations.
• The choice of w in Equation is arbitrary, but a value chosen badly can
lead to poor convergence of the series expansion.
• Higher order modes have increased Gouy phase shifts as shown in Equation .
• Beam propagation problems can be solved by manipulating the coefficients with appropriate Gouy phase shifts.
• Apertures can be treated using coupling coefficients which can be computed.
• Higher–order transverse modes in a laser can mix with each other to
produce unwanted mixing, or beat, signals.
• The higher–order modes can be eliminated by placing an appropriate
aperture inside the laser cavity.
Mar. 2014
c C. DiMarzio (Based on Optics for Engineers, CRC Press)
slides9r1–50