Download 5. GAUSSIAN BEAMS 5.1. Solution to the wave equation in

5. GAUSSIAN BEAMS
5.1. Solution to the wave equation in Cartesian coordinates
• Recall the Helmholtz equation for a scalar field U in rectangular coordinates
∇ 2U ( r, ω ) + β 2 (r, ω )U ( r, ω ) =
0,
(5.1)
• β is the wavenumber, defined as
=
β 2 ( r, ω ) ω 2 µε ( r, ω ) − iωµσ ( r, ω )
= n ( r, ω )
2
ω2
c
2
(5.2)
− iωµσ ( r, ω )
• Assuming lossless medium (σ = 0 ) and decoupling the vacuum contribution
( n = 1) from β ( r, ω ) , we re-write Eq. 2 to explicitly show the driving term,
∇ 2U ( r, ω ) + k0 2U ( r, ω ) =
− k0 2  n 2 ( r, ω ) − 1 U ( r, ω ) , (5.3)
• k0 = ω .
c
1
• Eq. 3 preserves the generality of the Helmholtz equation. Green’s function, h,
(an impulse response) is obtained by setting the driving term to a delta function,
∇ 2 h ( r , ω ) + k0 2 h ( r , ω ) =
−δ (3) ( r )
= −δ ( x ) δ ( y ) δ ( z )
(5.4)
• To solve this equation, we take the Fourier transform with respect to r,
− k 2 h ( k , ω ) + k0 2 h ( k , ω ) =
−1,
(5.5)
• k 2= k ⋅ k .
• This equation breaks into three identical equations, for each spatial coordinate,
h ( k x , ω ) =
1
k x 2 − k0 2
(5.6)
1  1
1 
=
−


2 k0 x  k x − k0 k x + k0 
2
• To calculate the Fourier transform of Eq. 6, we invoke the shift theorem and the
Fourier transform of function 1 ,
k
f ( k − a ) → eiax ⋅ f ( x )
(5.7)
1
→ i sign ( x )
k
• Thus we obtain as the final solution
1
h ( x, ω ) =⋅ eik0 x , x > 0
ik0
(5.8)
• The procedure applies to all three dimensions, such that the 3D solution reads
h ( r, ω ) ∝ eik0 k ⋅r ,

(5.9)
• k is the unit vector, k = k / k .
3
• Equation 9 describes the well known plane wave solution, which is
characterized by the absence of amplitude modulation upon propagation. This is
an infinitely broad wavefront that propagates along direction k̂ (Figure 1).
y
r
k
x
Figure 1. Plane wave.
4
5.2. Solution of the wave equation in spherical coordinates
• For spherical symmetry, such as emission from a point source in free space, the
problem becomes one dimensional, with the radial coordinate as the only
variable,
h ( r, ω ) = U ( r , ω )
h ( k , ω ) = U ( k , ω )
1
(3)
δ (r ) =
2π r 2
(3)
δ (r ) → 1
(1)
(5.10)
δ (r )
• Recall that the Fourier transform pair is defined in this case as
∞
h ( r ) ∝ ∫ h ( k ) ⋅ sinc ( kr ) k 2 dk
0
(5.11)
∞
h ( k ) ∝ ∫ h ( r ) ⋅ sinc ( kr ) r 2 dr
0
5
• The Fourier properties of
(3)
δ ( r ) and ∇ 2 extend naturally to the spherically
symmetric case as
(3)
δ (r ) →1
∇ → −k
2
(5.12)
2
• Thus, by Fourier transforming Eq. 4, we obtain the frequency domain solution,
h ( k , ω ) =
1
k 2 − k0 2
(5.13)
• Not surprisingly, the frequency domain solutions for the Cartesian and spherical
coordinates (Eqs. 6 and 13, respectively) look quite similar, except that the
former depends on one component of the wave vector and the latter on the
modulus of the wave vector.
6
• The solution in the spatial domain becomes
sin ( kr ) 2
1
⋅
⋅ k dk
h ( r , ω )= ∫ 2
2
k − k0
kr
0
∞
∞
1  1
1  eikr − e − ikr
=
+
dk (5.14)


∫
2r 0  k − k0 k + k0 
2i
∞
eikr
1
=
dk .
∫
4ir 0 k − k0
• We recognize in Eq. 15 the Fourier transform of a shifted 1
k
function, which
we encountered earlier (Eqs. 7). Evaluating this Fourier transform, we finally
obtain Green’s function for propagation from a point source,
eikr
h ( r,ω ) ∝
, r >0
r
(5.15)
7
• This solution defines a (outgoing) spherical wave.
z
cos 2π r 10
r
;r = x 2 + y 2 + z 2
y
x
Figure 5-2. Spherical wave
8
5.3. Solution of the wave equation in cylindrical coordinates
• The most common situation from an experimental point of view is light
propagation in cylindrical coordinates with radial symmetry, the z coordinate
defines the optical axes and the radial coordinate, r, defines the transverse
coordinate.
• The Fourier pair in this case can be written as
=
h ( r, z )
∞ ∞
∫
 ( k , k ) J ( k ⋅ r ) ⋅ eik z ⋅z ⋅ k ⋅ dk dk
h
⊥
⊥
z
∫ ⊥ z 0 ⊥
−∞ 0
=
h ( k⊥ , k z )
(5.16)
∞ ∞
∫
− ik z ⋅ z
⋅
⋅
⋅ r ⋅ drdz
h
r
z
J
k
r
e
,
(
)
(
)
⊥
0
∫
−∞ 0
• Due to the angular symmetry in both the spherical and cylindrical coordinates,
the spatial frequency domain solution of the Helmholtz equation is identical in
form with Eq. 13 and depends only on the modulus of the wave vector
9
h ( k , ω ) =
1
,
2
2
k − k0
(5.17)
2
k=
k⊥ 2 + k z 2
• U ( k , ω ) was obtained from the procedure used in the spherically symmetric
case, replacing the source term with
(3)
δ ( r , z ) and using the property ∇ 2 → −k 2
that holds for cylindrical coordinates with circular symmetry.
• We now Fourier transform Eq. 13 back to the spatial domain,
=
h ( r, z )
∞ ∞
1
ik z ⋅ z
⋅
J
k
⋅
r
⋅
e
k⊥ dk⊥ dk z
(
)
⊥
0
∫−∞ ∫0 k 2 − ( k 2 − k 2 )
z
⊥
0
(5.18)
• Eq. 18 can be interpreted as the Hankel transform (over k⊥ ) of a 1D Fourier
transform (over k z ). In evaluating the integral over k z first, we recover the 1D
case discussed in Section 5.1, the solution in Eq. 8, where we replace k0 with
k0 2 − k⊥ 2 and x with z.
10
• The remaining integral over k⊥ becomes
∞
eiqz
h ( r , z=
) ∫ ⋅ J 0 ( k⊥ ⋅ r ) ⋅ k⊥ dk⊥ ,
q
0
=
q
(5.19)
k0 2 − k ⊥ 2
• Equation 19a can be regarded as the exact Green’s function in cylindrical
coordinates with circular symmetry. The integral can be evaluated analytically
in the limit of low transverse frequencies, i.e. k⊥  k0 .
• In this case we expand
eiqz e

q
k0 2 − k⊥ 2 in Taylor series to obtain
 k 2 
ik0 z 1− ⊥ 2 
 2 k0 
k0
(5.20)
.
• The amplitude term varies slowly with k⊥ , so the zeroth order term is sufficient,
but the phase term is more sensitive, thus the second order term is necessary.
11
• Following this expansion, the Hankel transform remains to be calculated for a
−
Gaussian phase function e
iz
⋅k ⊥ 2
2 k0
. Recall that Hankel transforms of Gaussian
functions can be easily calculated and follows a similar transformation rule with
−
that of 1D functions, e
ix 2
2 b2
ib 2 k 2
2
2
→b e
.
• By combining Eqs. 20 and 19a, we obtain
ik
eik0 z 2 z0 ⋅r 2
h ( r, z ) ∝
.
⋅e
z
(5.21)
• This solution is characterized by a slow attenuation with distance, ∝ z , a plane
wave-like phase delay vs. z, and a Gaussian transverse phase, of width that also
grows as
z.
12
5.4. Propagation of Gaussian beams
• Often, experiments involve light beams. A light beam can be defined as a
distribution of field that fulfills the approximation in Eq. 20, is characterized by
a dominant wave vector component, k z  k x , k y . A beam is the spatial
equivalent of quasi-monochromatic light, where the field is characterized by a
dominant (temporal) frequency component.
• A Gaussian beam, such as that delivered by a single mode laser, has a field
distribution described by a Gaussian function in the transverse coordinate.
Green’s function in Eq. 21 provides a way to propagate this type of field.
• Consider a Gaussian beam that at z=0 has the transverse distribution
U 0 ( r ,0 ) =
1
e
2
w0
−
r2
w0 2
(5.22)
,
• w0 defines the minimum beam waist.
13
• Upon propagating a distance z, the field becomes a convolution between U 0 and
h,
=
U
( r , z ) U 0 ( r ,0 ) ∗ h ( r , z )
(5.23)
• We take the Fourier transform of Eq. 23, make use of the convolution theorem
and use the approximation in Eq. 20 to obtain the frequency domain solution
ik0 z
e
⋅e
U ( k⊥ , z ) =
k0
−
izk⊥ 2
2 k0
⋅e
−
w0 2 k⊥ 2
4
(5.24)
• The spatial domain solution is obtained by taking the inverse Hankel transform
of U ,
U ( r, z )
=
∞
∫U ( k , z ) ⋅ J ( k r ) k
⊥
0
⊥
⊥
(5.25)
dk⊥ .
0
14
• After propagation over distance z, the field remains Gaussian in the transverse
direction and has the form
ik0 z
e
U ( r, z ) = 2
e
c (z)
c 2 (=
z ) w0 2 + i
−
r2
c2 ( z )
(5.26)
2z
k0
• The solution has a particularly simple form, thanks to the properties of Gaussian
functions and their transforms. All the information regarding the beam change
due to propagation is contained in function c(z). This function can be rearranged
z 
2
2
c=
z
w
+
i
1
( ) 0

z
0 

k0 w0 2 π w0 2
=
z0 =
2
λ
(5.27)
15
• We can write explicitly the absolute value and phase of c2(z), express it in polar
coordinates, as
2
c=
( z ) w2 ( z ) ⋅ e
iψ ( z )
(5.28)
,
• where
2


z
2
2
w=
( z ) w0 1 + 2 
 z0 
(5.29)
 z 

z
 0
ψ ( z ) = arg 
• The distance z0 is known as the Rayleigh range, over which the beam waist
increases by
2 . We can re-write U(r, z) explicitly,
1
U ( r, z ) =
⋅e
w2 ( z )


z r2
i  k0 z +
−ψ ( z ) 
2
z0 w ( z )


= A( r, z ) ⋅ e
⋅e
−
r2
w2 ( z )
(5.30)
iφ ( r , z )
16
• Equation 30 represents the fundamental solution of the Gaussian beam
propagation. It can be seen that both the broadening in r and propagation in z
requires lowering the amplitude, such that the energy is conserved,
−
=
A( r, z )
1


w0 2 1 + z 2 
z0 

⋅e
r2
2


w0 2 1+ z
2
z0 

(5.31)
.
2
2
z
• The new beam waist is w=
( z ) w0 1 +
z0 2
.
• The phase dependence reveals more features
 z 
k0 r 2
φ ( r, z ) =
k0 z +
− arg   .
2R ( z )
 z0 
(5.32)
17
• Besides the plane wave phase delay, k0 z , the wavefront radius of curvature R(z)
induces additional phase shift, with R(z) defined as
 z0 2 
R (=
z ) z 1 + 2 
z 

(5.33)
• The wavefront is centered a distance d ( z ) = 2 z0 / z behind the z=0 plane.
r
w (z )
w0 {
z
R (z )
d(z)
Figure 5-3. Propagation of Gaussian beams.
18
θ
z
• We can define a far field divergence angle,
θ = lim
w( z )
z →∞
=
=
z
w0
z0
(5.34)
λ
π w0
• In far field, R ( z )  z .
• Perhaps the most surprising result is the phase delay due to the last term in Eq.
33, ψ ( z ) = arg  z  . It can be seen that on propagating from z=0 to ∞ , the
 z0 
wavefront is advanced by π . Overall, from −∞ to ∞ , there is a total phase
2
shift of π . This peculiar phase shift is referred to as Gouy’s phase.
19
ψ (z )
π
π
−z0
2
4
z0
z
−π 4
−π 2
Figure 5-4. Gouy phase.
• This phase advancement contributes to the overall phase velocity v p along the z
axis, such that it exceeds the speed of light in vacuum,
vp =
ω
keff
(5.35)
,
20
• keff is the effective modulus of the wave vector, defined as
d
 k0 z −ψ ( z ) 
dz
1
z0
=
k0 −
< k0.
2
z
1+ 2
z0
keff
=
(5.36)
• Thus, the contribution of the Gouy phase is always to advance the phase and
bring the phase velocity to superluminal values, v p > c .
21
5.5 Matrix formalism for Gaussian beam propagation.
• We discussed that in the far zone, z  z0 , we can define a diffraction angle,
θ = w z . In this asymptotic regime, the propagation is well described by straight
lines, which recovers geometrical optics.
• We will develop a matrix formalism for retrieving the beam characteristics upon
propagation.
θ
y2
y1
z
Figure 5. Geometrical limit of Gaussian beam propagation.
22
• Find the transformation rules for the elevation y and angle θ associated with the
ray (Fig. 5). Investigate propagation in free space (translation), refraction, and
propagation through a thin lens.
23
5.5.1 Translation.
• In free space propagation, the angle stays constant and the elevation changes,
y=
y1 + dθ1
2
(5.37)
θ 2 =0 ⋅ y1 + θ1
• Equation 37 is valid for very small angles, tan θ  θ . This transformation can be
expressed in matrix form,
=
=
  M
, M d
d 
 y2 
 y1 
θ 
θ 
 2
 1


1 d 
0 1 


(5.38)
• Equation 38 establishes the characteristic matrix for translation.
24
5.5.2 Refraction at dielectric interface.
n1
n2
• At the interface between two dielectric media,
only the angle changes, according to Snell’s law,
y2 = y1
θ2 =
n1θ1
n2
(5.39)
θ2
θ1
• Thus, the refraction matrix has the form
MR = 

1 0 

n1 
1

n
2 

(5.40)
25
Figure 5-6. Refraction at a planar interface.
5.5.3 Propagation through a thin lens.
θ2
θ1
y

 f θ1

θ1
f
F
Figure 5-7. Propagation through a lens.
y2 = y1
f θ1 − y1
θ2 =
f
y
= θ1 − 1
f
(5.41)
26
• Thus the lens transfer matrix is
Mf =
 1
 1
−
 f

0

1

(5.42)
• These rules of transforming y and θ are capable of solving complicated
geometrical optics problems, combining multiple optical components reduces to
matrix multiplication.
27
• Consider the optical arrangement shown in Fig. 8.
L1
L2
θ1

y1 

}y2
θ2
n
d1
d2
d3
d4
d5
Figure 5-8. Cascading many optical elements.
• The ray starts in the configuration
( y1 ,θ1 ) ,
translates a distance d1 , passes
through lens L1 (focal distance f1 ), translates a distance
d 2 , refracts at the
interface with a piece of glass, translates a distance d3 , refracts back to air,
translates a distance d 4 , passes through lens L2 , and translates a distance L5 , to
end in configuration ( y2 ,θ 2 ) .
28
• The final configuration can be obtained via matrix multiplication in the proper
order,
 =M 
 y2 
 y1 
θ 
θ 
 2
 2
M = M d5 ⋅ M f2 ⋅ M d4 ⋅ M n1 ⋅ M d3 ⋅ M 1n ⋅ M d2 ⋅ M f1 ⋅ M d1
(5.43)
• The order of matrix multiplication is crucial. To remember the correct order,
think of it as “chronological,” the first matrix applied to ( y1 ,θ1 ) is the translation
matrix M d1 , followed by lens L1 , and so on.
• The transfer matrices associated with optical systems are sometimes referred to
as “ABCD matrices,” where the matrix elements are specifically labeled as

.
A B
C D


29
• Let us see how this matrix formalism applies to Gaussian beams.
Optical System
g2



W1 


g1

W1

R1
R2
Figure 5-9. Propagation of Gaussian beams through an optical system.
• The goal is to find the transformation rules for Gaussian beam parameters, such
as the radius of curvature R and waist W.
• Recall that the Gaussian beam is described as (Eq. 26)
U ( r, z ) ∝ e
−
r2
c2 ( z )
(5.44)
30
• We ignore the factor 1
z 
2
2 
,
with
=
+
c
z
W
1
i
( )
.
0
c2 ( z )
z0 

2
r
• The z dependence of the exponent −
g ( z )= z − iz0 .
c2 ( z )
is carried by the complex quantity,
(5.45)
• g(z) is the complex beam parameter and its transformation rules are of interest
here. g(z) can be expressed in terms of both R and W, as follows
1
1
λ
=
+i
g ( z ) R ( z ) πW 2 ( z )
(5.46)
• A Gaussian beam is fully characterized by the complex beam parameter g(z).
• For an optical system characterized by matrix M = 
 , such as shown in
A
B


C D


Fig. 9, g1 and g2 are related by
31
g2 =
Ag1 + B
.
Cg1 + D
(5.47)
• The transfer matrices can be efficiently used to propagate Gaussian beams
through complicated optical systems. The overall matrix that characterizes a
cascade of optical elements is obtained by simple matrix multiplication.
32
Example 1
L
W0
W0
2
d
Figure 10. Focusing down a Gaussian beam by a convergent lens.
• A Gaussian beam of θ1 = 1mrad divergence beam is focused down by a lens to
half its minimum waist at a distance d=1m. What is the focal distance of the
lens?
33
• First, we find that the overall matrix of the system is
M =

A
B


C D


=

1 − d1 d + d − d1d 2 
1
2

f
f 


d
1
 −

1− 2
 f

f


(5.48)
• The radius of curvature is infinite at both planes, such that using Eq. 46, we
obtain
π W0 2
=
g1 j= jz0
λ
π W0 2
=
g 2 j= jz0
4λ
1
(5.49)
2
34
• Using that g 2 =
( Ag1 + B ) / ( Cg1 + D ) , we obtain
z01 z02
f
= d1 + d 2 −
d1d 2
f
(5.50)
 d2 
 d1 
z02 1 −  = z01 1 − 
f 
f 


• Using the information that d1 + d 2 =
1m , after some manipulations, we obtain
f = 0.19m .
35