OPTICAL INTERFEROMETERS AND OSCILLATORS
6.2 FABRY–PEROT RESONATORS
6.2.2 Fabry–Perot Equations
The equations for a Fabry–Perot resonator are derived by considering reflection and transmission at a single mirror. Partially reflecting mirrors allow input to and output from the cavity between them. Figure 6.10a and b shows for an incident field from the left the reflection and transmission at a mirror interface without attenuation and with power attenuationA, respectively. A negative power attenuation can be used to represent power gain. Note that the reflected power isRtimes the input power. To preserve power, the output power must then be 1−Rtimes the input power. Hence, from Figure 6.10b, for an incoming field Ei, √
R is reflected and√
1−A−R is transmitted by the mirror.
From Figure 6.10, we can hypothesize light bouncing back and forth be- tween the mirrors, as shown in Figure 6.11. We assume a phase change in cross- ing the etalon of βx, where β in radians per second is the propagation con- stant and x is the etalon width. Hence, on reaching the right-hand side mirror in Figure 6.11, the field is√
1−R−AEi exp{jβz}and on reflection from the mirror,
√R√
1−R−AEi exp{jβz}. With perfect mirror alignment, the light reflects back and forth along the same path. For clarity in Figure 6.11, we shifted the beams down on reflection.
Ei 1 −R−A Ei
Ei R
√
√ Ei 1 −REi
Ei R
√
√
(a) (b)
FIGURE 6.10 Reflection and transmission at a mirror: (a) without attenuation and (b) with attenuation.
FABRY–PEROT RESONATORS 111
Ei exp{jβ x}
exp{jβ x}
R exp{j2β x}
exp{j2β x}
R R exp{j3β x}
exp{jβ x}
Ei (1−R−A) Right mirror Left mirror
Input field
Output field
x
(1−R−A)
R Eiexp{j3 x}β 1−R−A Ei
√
1−R−A Ei
√
1−R−A Ei
√ √1−R−A Ei
1−R−A Ei
√
1−R−A Ei
√ √R√
FIGURE 6.11 Illustration for deriving Fabry–Perot transfer function equations.
From Figure 6.11, the output at the right of the etalon is the infinite sum of the outputs transmitted through the output mirror on each bounce of the light:
Eo=(1−A−R)Ei exp{jβx} ∞ m=0
R exp{j2βx}m
(6.19)
where for each cycle of the light, we add in the sumRfor two reflections and exp{j2βx} for two propagations across the cavity. Similarly, we could determine the reflection from the resonator by summing the infinite reflections to the left.
If incoming broadband light has an electric fieldEi only those wavelengths will resonate for which a whole number of half-wavelengths fit between the mirrors (Figure 6.12).
1/2
6/2 6 ×
5 ×
4/2 4 ×
3/2 3 ×
2/2 2 ×
Mirror 1 Mirror 2
Number of half- wavelengths that fit between mirrors provides mode number λ
5/2 λ
λ λ λ λ
FIGURE 6.12 Electric fields for modes in Fabry–Perot resonator.
112 OPTICAL INTERFEROMETERS AND OSCILLATORS
Wavelength, λ
n n+1
Δλ 1.0
0.5
0.0
0 5 10
λ λ
FIGURE 6.13 Longitudinal modes in wavelength spectrum out of Fabry–Perot resonator.
Consequently, if the bandwidth ofEiin wavelength isλ, as shown in Figure 6.13, there will be many resonance peaks corresponding to different numbers of half-cycles, λn,λn+1, . . ., as shown in Figure 6.13.
Conversion between frequencyf and wavelengthλand variations in frequency f and wavelengthλcan be made from the respective equations,
f = c
λ and by differentiation |f| = c
λ2λ (6.20)
where air (refractive indexn=1) is assumed in the cavity.
6.2.2.1 Fabry–Perot Transfer Function Propagation of a plane wave can be written in exponential form as expj(βx−ωτ). So, in propagating once across etalon, the phase angle changes by
βx= −ω
cx= −ω x
c = −ωτ= −2πfτ (6.21) wherexis the distance between mirrors across the air-filled etalon andτ is the time for a wave to traverse the etalon once. Equation (6.21) allows us to write the phase delay across the etalon, instead of in distancex, in terms of frequencyf, which is more convenient for studying frequency filters.
From electromagnetic theory, the electric field at the conducting mirror is zero, Emirror=0, so the field across the etalonE(x)∝sin(βx), which requiresβxto be an arbitrary integer multiple ofπ. Hence, forman arbitrary integer,
βx=mπ or x= mπ
β (6.22)
FABRY–PEROT RESONATORS 113
We conclude that resonance occurs only when the distance across the etalonx is a multiple number of half-wavelengthsλ/2 as shown in Figure 6.12 and by using β=2π/λin equation (6.22),
x=mλ/2 (6.23)
Transmission through the etalon is the output field, equation (6.19), divided by the input field,Ei, withβxfrom equation (6.21):
H(f)=Eo(f)
Ei(f) =(1−A−R)e−j2πfτ ∞ m=0
Re−j4πfτ m
(6.24)
From tables or by using Maple, an infinite summation such as this can be written in closed form:
∞ m=0
am= 1
1−a for a <1 (6.25) Using equation (6.25) with equation (6.24) to condense the infinite series gives
H(f)=(1−A−R)e−j2πfτ 1
1−Re−j4πfτ (6.26)
The transfer function for power transmission through the cavity can be written as T(τ)= |H(f)|2=H(f)H(f)∗
= (1−A−R)2 (1−Re−j4πfτ)(1−Rej4πfτ)
= (1−A−R)2
1+R2−2Rcos 4πfτ from ejθ+e−jθ/2=cosθ
= (1−A−R)2
1+R2−2R(1−2 sin2 2πfτ) from cos 2θ=1−2 sin2θ
= (1−A−R)2 (1−R)2+R(22 sin2 2πfτ)
= (1−A/(1−R))2 1+((2√
R/(1−R)) sin 2πfτ)2 dividing top and bottom by (1−R)2 (6.27)
114 OPTICAL INTERFEROMETERS AND OSCILLATORS
2 1
2 0 2
2 1
2 1
Frequency
τ τ τ τ
FIGURE 6.14 Longitudinal frequency modes out of Fabry–Perot resonator.
6.2.2.2 Free Spectral Range Equation (6.27) peaks for frequencies at which sin(2πfτ)=0, computed from
2πfτ=0, π,2π,3π, . . . or 2πfτ=mπ, f =0, 1 2τ, 2
2τ, 3
2τ, . . . (6.28) whereτ is one-way time to travel across etalon. Each peak corresponds to a laser longitudinal mode in Figure 6.14. Free spectral range (FSR) is the range of frequencies over which tuning is possible by moving one mirror away from the other until phase wrapping occurs to the next half-wavelength. Hence, from Figure 6.14,
FSR= 1 2τ =1
2 c
x, for n=1 (6.29)
Previously, we represented modes in terms of the number of half-cycles of wave- length that fit into the distance across the cavity width x (equation (6.23) and Figure 6.12). In equation (6.28) and Figure 6.14, we represent modes as resonant frequencies separated by half the reciprocal of the travel timeτacross the etalon.
Maximum. From equation (6.27), the maximum occurs at sin 2πfτ=0 and is T(f)=
1− A
1−R 2
(6.30) Laser power transfer functionT(f) is very large because gainAis high relative to 1−R, where 0< R <1.
Half-maximum power. Half-maximum power occurs at frequencyf1when denom- inator of equation (6.27) is two or long bracket is one.
2√ R
1−Rsin 2πf1τ=1 or sin 2πf1τ= 1−R 2√
R (6.31)
FABRY–PEROT RESONATORS 115
For a resonator, we generally have a small 2πf1τ, so sin(2πf1τ)→2πf1τ, or f1=
1 2τ
1−R π2√
R (6.32)
Half-power bandwidth. Half-power bandwidth (HPBW) is defined as HPBW=2f1=
1 2τ
1−R π√
R (6.33)
Finesse. Finesse is defined as F = FSR
HPBW = 1/2τ
((1/2τ)/(1−R)/(π√
R)) = π√ R
1−R (6.34)
Finesse indicates how many unique frequency channels may be placed inside the free spectral range before mode overlap arises. The number of available channels is one less thanF as deduced from examination of Figure 6.15, where although the finesse is three, the number of unique frequency channels is two because one of the three channels is split half at each end and therefore ambiguous in frequency. Hence, if we desire 100 channels, we need to design for 101. This is accomplished by increasing FSR by a factor 101/100. Furthermore, as the HPBW allows too much crosstalk between channels, it is normal to assume that we need 50–65% more channels and then use only the desired number of channels spaced out. Finesse also represents the average number of times a photon bounces back and forth in a cavity before the power in the cavity has fallen by 1/e.
HPBW FSR
Frequency f Power
transfer function T(f )
FIGURE 6.15 The number of unique channels from finesse.
116 OPTICAL INTERFEROMETERS AND OSCILLATORS
(a) (b)
Distance x
Left mirror
Right mirror
Distance x
Mirrors
FIGURE 6.16 Tuning a Fabry–Perot filter with piezoelectric material: (a) piezoelectric ma- terial inside etalon and (b) with mechanical gain.
Contrast. Contrast may be defined as max/min for the Fabry–Perot resonator out- put. From equation (6.27), the maximum and minimum values are
Max=
1− A
1−R 2
Min= (1−(A/(1−R)))2
1+(4R/(1−R)2), where sin(2πfτ)=1 Contrast = Max
Min =1+ 2√
R 1−R
2
Using equation (6.34), Contrast =1+
2F π
2
(6.35)