Download Chapter 3 Passive resonators

Chapter 3
Passive resonators
3.1
Introduction
The first role of the laser is to provide a positive feedback in order to enhance the gain1 .
Obviously, spectral properties of both the gain medium and the cavity will affect the final
properties of the laser beam. In this chapter, we will look at the spectral properties of
the cavity, but also see how the cavity can shape the spatial structure of the laser beam.
The cavity will affect
1. Direction of emission. The photons created by spontaneous emission will nearly all
be eliminated by the cavity expect the one needed for the laser itself. Frequency
response of the cavity itself will influence the operating wavelength of the laser.
2. Directivity. The phenomenon of diffraction in combination with the different elements, which constitute the laser cavity will lead to the formation of modes. These
modes have a spatial structure, with a distribution of the amplitude and the phase.
3. Frequency response of the cavity itself will influence the operating wavelength of
the laser.
4. Polarization. The presence of a Brewster-cut element will strongly enhance the loss
of one of the polarization state of the generated beam.
In general, the more severe the selections, the higher the coherence properties of the
beam, in term of directivity (spatial modes), degree of polarization and purity of the
emitted frequency. By contrast with a real cavity, we will limit our study to empty
resonators. Real cavity includes an active medium and its complete study is usually
rather difficult. However experiments show that the general behavior obtained with empty
resonator can be applied to real laser cavity. Note that nonlinear processes can happen
due to the presence of the active medium such as competition between modes, frequency
detuning, or thermal lens.
3.2
Fabry-Pérot cavity
Let’s consider a Fabry-Pérot cavity. It is made of two parallel interfaces. The coefficients
for the reflected and transmitted field at the interfaces are respectively ri and ti . These
coefficients are given for the amplitudes. We assume that the incident wave is plane, with
a wavevector ~k.
1
Note that the cavity is particularly necessary in the case of low-gain. However there exists laser for
which the gain is so high that a single pass can be enough. We then talk about superradiance. We can
cite copper-vapor laser, N2 laser working in the ultraviolet, and free-electron lasers.
2
Passive resonators
Figure 3.1: Schematic of a Fabry-Pérot cavity.
If the wave enters at normal incidence, then
φ = 2ℓk =
4πLν
2πν
=
c
∆νF
(3.1)
where ∆νF is the free spectral range for the cavity. The output field is the superposition
h
i
iφ n
1
−
r
r
e
2
1
2
Et = E1 t1 t2 1 + r1 r2 eiφ + r1 r2 eiφ + . . . = E1 t1 t2
1 − r1 r2 eiφ
(3.2)
We can then write the power transmission of the Fabry-Pérot interferometer as TFP =
|Et |2 /|E0 |2 :
TFP =
t21 t22
t21 t22
=
(1 − r1 r2 eiφ ) (1 − r1 r2 e−iφ )
1 − 2r1 r2 cos φ + (r1 r2 )2
(3.3)
Using the trigonometric identity cos 2θ = 1 − 2 sin2 θ, we can then write this equation as
TFP =
t21 t22
(1 − r1 r2 )2 + 4r1 r2 sin2
φ
2
(3.4)
Since reflection and transmission coefficients are Ri = ri2 = 1 − t2i = 1 − Ti , the eq. (3.4)
becomes:
(1 − R1 ) (1 − R2 )
TFP =
(3.5)
2
√
√
2 φ
1 − R1 R2 + 4 R1 R2 sin
2
In the case of identical mirrors eq. (3.5) simply becomes
TFP =
1
1
=
4R
4R
πν
2 φ
2
sin
1+
1+
sin
(1 − R)2
2
(1 − R)2
∆νF
(3.6)
Fig. 3.2 shows the transmission of a Fabry-Pérot interferometer for two different values
of the coefficient of reflection (R). As the figure shows, the larger R, the sharper the
resonance. This also corresponds to a cavity, with a much longer lifetime. There are
two ways to characterize the cavity whether we only consider one resonance (fig. 3.2b) or
the overall response (fig. 3.2a). In the first case, we are talking about quality factor by
analogy with resonance phenomena that exist in other domain of physics. In the second
case, we use the finesse.
3
3.2 Fabry-Pérot cavity
(a)
(b)
∆νF
TFP
R=0.5
∆νC
∆νC
R=0.8
νn-1
νn
νn+1
ν
νn+2
ν0
Figure 3.2: (a) Transmission of a Fabry-Pérot interferometer for R = 0.8 and R = 0.5
and (b) one isolated peak of the overall response function.
3.2.1
Quality factor and photon lifetime
definition
If we only consider one resonance peak of the frequency comb that describe the transmission of the Fabry-Pérot interferometer (fig. 3.2b), we can simply define the quality factor
as
ν0
Energy stored
Q=
= 2π
(3.7)
∆νc
Energy lost per cycle
where ν0 is the central frequency of the resonance. As we will see the bandwidth of the
resonance ∆νc is directly linked to the lifetime of the photons in the cavity τc and therefore
Q can also be written as Q = 2πν0 τc . For the present case, the energy is proportional to
the number of photons and the energy lost to the change of photon number. Therefore2
Q = 2π
2πνI
hνI
= − 1
dI
dI
×
hν −
dt
ν
dt
This can be rewritten as
dI
ω
=− I
dt
Q
(3.8)
(3.9)
which is readily solved as
t
ω
− 
− t
I(t) = I0 e Q = I0 e τc


(3.10)
This equation clearly shows the link between the lifetime of the photon in the cavity,
and the quality factor. We remind that we defined previously the lifetime of the photon
inside the cavity as k = 1/τc .
case of the Fabry-Pérot cavity
In the case of a Fabry-Pérot cavity with R1 and R2 the reflection coefficient of the mirror
(Fig. 3.3). The length of the cavity is ℓ. The loss per unit of length inside the cavity is α.
Suppose that we have an initial intensity I0 inside the cavity. After one round trip
the intensity is then
I = I0 R1 R2 (1 − αℓ)2
(3.11)
2
Note that the cycle is 1/ν , where ν is the frequency of the light. This is actually the only thing
oscillating here (electric field) since there is only one peak, and so the cavity is totally absent in this
description.
4
Passive resonators
Figure 3.3: Fabry-Pérot resonator
After p round-trip, this intensity is now
t
2 p
I = I0 R1 R2 (1 − αℓ) = I0 exp −
τc
(3.12)
where p can be expressed by
p=
(c/n) t
2ℓ
(3.13)
From eq. (3.12) and (3.13), we get
c ω
1
ln R1 R2 (1 − αℓ)2 =
=−
τc
2nℓ
Q
and therefore
Q=
order of magnitude
1
ν0
−2nℓω
= 2πντc
2 =
c ln R1 R2 (1 − αℓ)
∆νc
(3.14)
(3.15)
Let a 90 cm-long He-Ne laser (λ = 633 nm) composed of 2 mirrors with equal coefficient
of reflection (R = 0.98) and no additional loss (α = 0) and n = 1. Let’s calculate the
cavity lifetime τc as well as the quality factor.
From eq. (3.14), we have
1
2 ln R
= − 2ℓ ≈ 1/150 ns−1
(3.16)
τc
c
The optical frequency of the laser is ν0 = c/λ = 5 × 1014 Hz and therefore the quality
factor Q = 2πν0 τc = 2π × 5 × 1014 × 150 ns = 4.7 × 108 .
3.2.2
Finesse
definition
For optical resonator, the Q-factor is usually large. It is then more convenient to define
another factor: the finesse. It takes into account both the resonance bandwidth and the
spacing between peaks and is define as
F =
∆νF
∆ν
(3.17)
case of Fabry-Pérot
From eq. (3.5), we can calculate the maximum and minimum of the transmission function
(fig. 3.4).
5
3.3 Stability of a resonator
∆νF
∆νC
Tmax
Tmin
νn
νn+1
ν
Figure 3.4: Transmission of a Fabry-Pérot interferometer.
These correspond respectively to sin2 φ = 0 or 1, and are readily given by
(1 − R1 )(1 − R2 )
Tmin = 2
√
1 + R1 R2
(1 − R1 )(1 − R2 )
Tmax = 2
√
1 − R1 R2
(3.18a)
(3.18b)
To calculate the width of the transmission peak, we can see that the transmission is
reduced by half when
i2
h
p
p
1 − R1 R2 = 4 R1 R2 sin2 φ
On the other hand, the transmission is minimal when φ = nπ. So considering that the
width of the peak ∆φ ≪ π we can approximate sin ∆φ ≈ ∆φ. This leads to the FWHM
of the peak:
√
√
1 − R1 R2
c 1 − R1 R2
√
∆φc = √
⇒ ∆νc =
4
2ℓ 4 R1 R2
R1 R2
Since the finesse is defined as F = ∆νF/∆νc we deduce:
√
√
π 4 R1 R2
π R
√
F =
=
1−R
1 − R1 R2
(3.19)
when R1 = R2 = R.
3.3
3.3.1
Stability of a resonator
General case
Consider the spherical-spherical cavity (fig. 3.5). where R1 and R2 are the radii of curvature of the mirrors. At this stage, the main question is to know whether such a cavity if
optically stable or not. The cavity can be describes (one return) by the following ABCD
matrix:
A B
M=
C D
And as a consequence, wondering whether the cavity is stable is equivalent to questioning
if an optical beam will remain inside the cavity or not. From the ABCD matrix of the
cavity, we can evaluate the characteristics of the beam (ray optics) after one round (r1 , θ1 )
from an initial beam (r0 , θ0 ), and repeat this for N return inside the cavity. Note that
such cavity is equivalent to the one presented on fig.
6
Passive resonators
Figure 3.5: Spherical-Spherical resonator
Figure 3.6: Expanded view of a spherical-spherical cavity
r0
θ0
→
r1
θ1
=M
r0
θ0
→
r2
θ2
=M
r1
θ1
After N round trip, this beam is then:
r0
rN
N
=M
θ0
θN
=M ×M ×
r0
θ0
(3.20)
The cavity is stable if and only if
rN
lim
remains finite
θN
N →∞
In the present case, one way to figure out the stability of the problem is to look at its
eigenvalue. Indeed, when λ1 , λ2 the eigenvalues of M, then
N
λ1 0
λ1
0
N
M=
=⇒ M =
0 λ2
0 λN
2
From the eigenvalues, we can say that the system is stable if and only if both |λ1 | and
|λ2 are both less that 1. To find the eigenvalue we need to solve det (M − λI) = 0. Note
that det(M) = 1.
A−λ B = (A − λ)(D − λ) − CB = 0
C
Dλ λ2 − (A + D)λ + 1 = 0
(3.21)
7
3.3 Stability of a resonator
From this equation, we can express the product and the sum of the solution (λ1 , λ2 ):
λ1 + λ2 = A + D
λ1 λ2 = 1
(3.22a)
(3.22b)
and (λ1 , λ2 ) are:
λ1,2 =
A+D
2
±
s
A+D
2
2
−1
(3.23)
• if (A + D)2 > 4 then (λ1 , λ2 ) ∈ R2 . However, because of eq. (3.22b), then one of
the eigenvalue is larger than 1: the cavity is not stable.
• if (A + D)2 < 4 then (λ1 , λ2 ) ∈ C2 . The eigenvalues are then
s
2
A+D
A+D
λ1,2 =
±i
−1
2
2
and Re(λ) < 1: the cavity is stable
conclusion: the cavity is stable if and only if
A + D 2 ≤1
3.3.2
(3.24)
Spherical-spherical cavity
Let calculate the matrix for the return in the spherical-spherical cavity (Fig. 3.7)
Figure 3.7: spherical-spherical cavity

M =
 
1
1
L
1
L   1
1−
−
−
f2
f2
f1

L
−
1−
L

f1
L =

L
L2  (3.25)
2L
1−
−
+
−
1−
f1
f2
f1 f1 f2


Then we can write the stability condition:
1
L
L2
L
−
+
=
tr(M) = 1 −
2
f2 f1 2f1 f2
L
L
1−
1−
2f1
2f2
(3.26)
The condition of stability is:
L
L
0≤ 1−
1−
≤1
R1
R2
(3.27)
8
Passive resonators
Figure 3.8: Graphical representation of the stability of a cavity. It is stable for g1 g2 ≤ 1.
Figure 3.9: Semi-spherical cavity
3.3.3
Semi-spherical cavity
Such cavity is composed by a flat mirror and a spherical one, with R its radius of curvature.
(R > 0 for concave mirror). Distance between both mirrors is L. On the flat mirror we
have:
"
2 #
λL
w 2 (L) = w02 1 +
(3.28a)
πw02
"
2 2 #
πw0
R(L) = L 1 +
(3.28b)
λL
Dividing eq. (3.28a) by eq. (3.28b) we obtain:
πw 2
λL
=
λR
πw02
(3.29)
which leads to
2
λ
L(R − L)
=
π
2 2
R L
λ
4
w0 =
π
R−L
w04
(3.30a)
(3.30b)
9
3.3 Stability of a resonator
3.3.4
General of Fabry-Pérot cavity
We can consider any Fabry-Pérot cavity as two Semi-spherical cavity with z1 and z2
their respective length. Since the waist is the same for both Semi-spherical cavity, from
Figure 3.10: spherical-spherical resonator
eq. (3.30a), we have:
z2 (R2 − z2 ) = z1 (R1 − z1 )
(3.31)
and since z1 + z2 = L we can express z1 and z2 with the parameter of the cavity L, R1
and R2 :
z1 =
L(R2 − L)
R1 + R2 − 2L
(3.32a)
z2 =
L(R1 − L)
R1 + R2 − 2L
(3.32b)
Since the radius of curvature is determined by both mirror, from eq. (3.30b, 3.32a, 3.32b)
we have:
w14
w24
=
=
λR1
π
λR2
π
2
2
R2 − L
L
R1 − L R1 + R2 − L
R1 − L
L
R1 − L R1 + R2 − L
(3.33a)
(3.33b)
When cavity is stable, we can express the waist w0 by:
w04
3.3.5
2
L(R1 − L)(R2 − L)(R1 + R2 − L)
λ
=
π
(R1 + R2 − 2L)2
(3.34)
confocal resonator
In the particular case of a confocal resonator, we have R1 = R2 = L and then eq. (3.33a)
becomes:
r
λL
(3.35)
w=
π
10
3.4
Passive resonators
ABCD law for a cavity
By definition of the q-parameter and after a return-trip inside the cavity we can write:
q(z + 2L) =
A q(z) + B
= q(z)
C q(z) + D
(3.36)
that we can rewrite as:
2
A−D 1
1
+2
−C =0
B
q
2
q
(3.37)
for which roots are given by
1
A−D
=−
±
q
2B
s
A−D
2
2
+ BD
(3.38)
Moreover, since AD − BC = 1 and the cavity is stable (then the discriminant of eq. (3.37)
is negative) we have:
s
2
D−A
i
iλ
1
1
A+D
=
−
(3.39)
= −
1−
q
2B
B
2
R πw 2
As a result we have:
R(z) =
w 2 (z) =
2B
D−A
λB
s
π
(3.40a)
1
1−
A+D
2
2
We can also express the Gouy phase over one round trip:
A+D
−1
Φ = cos
2
(3.40b)
(3.41)