Download The Double Etalon filter in Vernier configuration

The Double Etalon filter in Vernier configuration
Outline



The ideal Etalon
The Double Etalon filter
The back mirror
1. The ideal Etalon
1.1 The pure refractive Etalon
Let us suppose to have a thin slice of homogeneous non absorbing material whose refractive index
is n. Let d be the thickness of the slice.
Let a plane optical wave of unit amplitude and frequency  impinge on the slice from left.
c

Its wavelength 0 in vacuum will be 0  , while inside the slice it will be   0 .

n
The corresponding wavenumbers will be k0 and k, respectively, with k=nk0.
Let the left surface of the slice be located in x=0, so that its right surface is in x=d.
Because of reflections, a reflected wave will appear on the left of the slice, while a pure transmitted
wave will propagate forwards on its right. Inside the slice, both a progressive and a regressive wave
will exist and interfere.
Dividing the space into three regions, as in fig.1, the waves will have the following form:
I
II
III
n>1
n0=1
0
n0=1
d
x
e ik 0 x  re  ik 0 x
 ikx
 ikx
ae  be
teik 0 x

in I
in II
in III
The coefficients r and t are related to the intensity reflection and transmission coefficients R and T
by
2
2
R r , T  t
The boundary condition of continuity of the solutions and their first derivatives holds on both
surfaces x=0 and x=d. This means:
1  r  a  b
1  r  a  b
1  r  k a  b 
k 1  r   k a  b 

k0
 0
that may be written as:  ikd
 ikd
ik0d
ikd
 be ikd  teik0d
ae
ae  be  te
k
 ae ikd  be ikd   teik0d
k aeikd  be ikd   k0teik0d
 k 0
Adding the first two equations and subtracting the second two ones leads to


k  
k 
2  a 1    b1  

 k0   k0 

aeikd 1  k   be ikd 1  k   0
 k 
 k 

0 
0 


which allows to find a and b :
2k0 k  k0 
 ikd
ikd
ae

e

k  k0 2  e 2ikd k  k0 2


2k0 k  k0 
be ikd  eikd

k  k0 2  e 2ikd k  k0 2

that, because of the third continuity condition, leads to the coefficient t:
4kk0
4kk0
teik0d  e ikd


2
2
2 ikd
2 ikd
k  k0   e k  k0  k  k0  e  eikd k  k0 2

2kk0
2kk0 coskd   i k 2  k 02 sin kd 
The intensity reflection coefficient is then:
2kk0 2
2kk0 2
2
Tt 

2kk0 2 cos2 kd   k 2  k02 2 sin 2 kd  2kk0 2  k 2  k02 2 sin 2 kd 

In conclusion:
T
1
k k
1  
 2kk0
2
2
0
2

 sin 2 kd 




2
2
 k 2  k02   n 2  1 
  
 can be put into a more useful form, when we consider the
The term F  
 2kk0   2n 
power reflection coefficient R0 of a single interface (that means to consider d infinitely large, so that
the regressive wave inside region II vanishes, leading b=0).
In that case,
1  r0  a
1  r0  a
k k

k
or also 
from which r0  0
and then

1

r

a
k0  k
0
k 0 1  r0   ka

k0
R0  r0
2
 k  k0 

 
 k  k0 
2
It is matter of direct substitution to verify that
F
4 R0
1  R0 2
so that
T
1
1  F sin 2 kd 
The obtained result refers to an ideal Etalon where reflection and transmission are completely
defined by the sole refractive index of the medium.
1.2 The coated Etalon
It is clear that the term
2
2
 k 2  k02   n 2  1 
  

F  
 2kk0   2n 
cannot exceed some moderate values. For instance, when n=3.5 (as for Silicon) one has F  2.6 .
On the other hand, F has an important role in defining the sharpness of the periodic resonances of T:
the higher F, the sharper T.
In order to achieve higher values of F the Etalon facets are usually coated with specific reflective
coatings, transforming the device into a multi-slice element.
The theoretical treatment would become cumbersome, although not impossible.
A shorter way is to consider the total power reflection coefficient R for a single interface, also when
properly coated, and put
4R
F
1  R 2
We can then definitively accept for the Etalon transmission coefficient the expression
T
4R
1
with F 
2
1  F sin kd 
1  R 2
provided we remember that T refers to the Etalon (two surfaces), while R refers to a single surface.
In general, when the two Etalon surfaces are coated with different coatings, of respective power
reflection coefficients R1 and R2, one has R  R1R2 .
The term F is commonly indicated as the finesse parameter.
It is evident that T is a periodic function of the product kd.
Let us recall that here d is the physical thickness of the Etalon slice, and k 
2
n
2
 nk0 is the

0
wavenumber as it appears inside the Etalon itself. This wavenumber is different than in vacuum,
and also very probably different from its value inside the gain medium that generates the wave.
On the contrary, the frequency  is the same everywhere, which strongly prompts to define the
Etalon transmission function T as a function of .
1
T
 2 
1  F sin 2 
nd 
 c

In this way, the performances of different Etalons can be compared.
Now, T is a periodic function of the frequency  depending on the two device-specific parameters n
and d.
It is clear that, despite of the finite (and often high) reflectivity of the two surfaces, perfect
transmission (T=1) occurs when kd  m , where m is an integer.
In terms of frequency , this maximum transmission corresponds to the equally spaced values for
2
nd  m , that is
which
c
c
m  m
2nd
c
, independent of the value of F.
2nd
The minimum of T is achieved when the sine function achieves the value  1 . In that case
1
Tmin 
1 F
This states that the separation of the maxima is  m 
Even more important is the linewidth for the transmission peaks. It may be measured for T=0.5,
(half maximum, hm) which obviously means
 2  m   m  
 2

F sin 2 
nd   F sin 2 
nd   1
c


 c

and then
c
 1 
 hm  
arcsin 

2nd
 F
The next figure plots T in the range 191-196 THz for a 160 m thick Si (n=3.5) Etalon with
reflection coefficient values of 0.5, 0.7 and 0.9.
1
1
T    0.5
T    0.7 0.5
T    0.9
0
0
191
191.5
192
192.5
193
193.5
194
194.5
195
195.5

191
196
196
2. The double Etalon filter
Let us suppose now to set up a cascade of two Etalons (Vernier configuration) differing for their
thickness d and their refractive index n.
Provided no reflections are allowed within the distance separating the two facing surfaces, the total
transmission function T is the product of the two single transmission functions T1, T2.
T  T1T2 
1
1
 2

 2

1  F sin 2 
n1d 1  1  F sin 2 
n2 d 2 
 c

 c

The mathematical analysis of the function is cumbersome, but a graphical insight can be
illuminating. In the following image two Etalons with identical finesse have been coupled allowing
some 5-6% difference between their respective values of the product nd.
The two separate transmission functions have slightly different periodicity, so that if two peaks
“accidentally” coincide, they progressively separate for the neighbouring modes. The product is
then maximum at the coincidence and then quickly decreases.
1
1
0.8
T  
0.6
T1  
T    T1  
0.4
0.2
5
2.32810
0
191
191.5
192
192.5
193
193.5
194
194.5
195

191
195.5
196
196
One of the critical points is to define how to select a specific peak. It is clear, indeed, that the most
probable result when casual values are allowed for nd, is no overlap in the region of interest.
1
1
0.8
T  
0.6
T1  
T    T1  
0.4
0.2
5
2.58210
0
191
191
191.5
192
192.5
193
193.5

194
194.5
195
195.5
196
196
When a specific frequency 0 has to be selected, the two following conditions must hold:
 0  m1
c
c
 m2
2n1d1
2n 2 d 2
where m1 and m2 are integers, not necessarily identical.
Moreover, also the extinction ratio of the side modes is important, and its maximization can address
the choice of the practical setup.
This ratio can be evaluated by considering first a perfectly selected frequency, as from the formula
above, and then looking for the height of the side modes of, say, the first Etalon.
In this case, the transmission function of the first Etalon is, by choice, unitary
1
1
1


2m1
2m1
2m1



2
2
2
1  F sin 
n1d1  1  F sin 
n2d 2  1  F sin 
n2d 2 
 c

 c

 c

1

 2 

c 
 m1
n2d 2 
1  F sin 2 
 c  2n1d1 

Tm1  T1T2 
Let m1peak be that value of m1 for which 0 is selected:
1
1
1
Tm peak 


1
1


 1  F sin 2 m2peak 
2  2
c 
2 2 
peak
 0 n2 d 2 
 m1
n2 d 2  1  F sin 
1  F sin 
c


c
2
n
d
1
1




when we move to a different maximum we add or subtract an integer j to or from m1peak :
1
Tm peak  j 

1




2

c
 m1peak  j 
n2 d 2 
1  F sin 2 
c
2n1d1 



1
1



2



2  2


2

c
1  F sin   0n2 d 2 
jn2 d 2  1  F sin 2  m2peak 
j
n2 d 2 

c
 c

c  2n1d1 



1
1

 nd 
 n d nd 
1  F sin 2  j 2 2  1  F sin 2  j 2 2 1 1 
n1d1
 n1d1 


The last step, that uses the periodicity of the square sine function (any addition or subtraction of
multiples of  leaves the result unchanged), brings the argument in the range centred at 0, that
clearly shows that two identical Etalons have a perfect transmission for all peaks.
Now, we are able to see that the j-th side peak is displaced, with respect to 0 to the frequency:
c
 0 j   0  j
2n1d1
and its height is
1
1
 n d nd 
1  F sin 2  j 2 2 1 1 
n1d1


We can recover an expression that deals with frequency, substituting for j from the equation of the
peaks:
1
1
Tm peak  j 

1
 2  0   0  j 
 2 0  j
n d n d 
n d n d 
1  F sin 2 
n1d1 2 2 1 1  1  F sin 2 
n1d1 2 2 1 1 
c
n1d1
n1d1


 c

This is the equation of a sampled function, whose envelope is
1
T envelope 
 2
n d nd 
1  F sin 2 
n1d1 2 2 1 1 
n1d1
 c

and is centred on the perfectly transmitted peak.
Tm p ea k  j 
The function is the same as for the single Etalon, but with a “scale factor”
n2 d 2  n1d1
.
n1d1
This means that the width at half maximum of the envelope is
hm
 envelope


c
n1d1
n1d1
 1 
   hm
.
arcsin 

2n1d1
n2d 2  n1d1
 F  n2d 2  n1d1 
The practical consequence is that for any device, for which side mode suppression is highly
recommended, the difference n2d2  n1d must be as large as possible.
When using identical Etalons, this difference cannot be practically achieved by simple changes in
the refractive index: if, for instance, temperature is used to modulate n, too large thermal control
should be required to achieve the proper conditions.
It follows the need to setup since the beginning some difference in the thickness d. This can be
obtained even from identical devices by simply rotating them with respect to perfect parallelism.
This would also have the highly beneficial result to destroy the possible parasitic resonating cavity
that would setup between the two devices.
In that case, a rotation by an angle  would introduce an effective optical pattern, parallel to the
d
optical axis, given by d eff 
.
cos 
The back mirror
The two Etalons in Vernier configuration provide a nice and effective solution for frequency
selection inside the external cavity laser.
Nevertheless, one more condition must be satisfied: the total length L of the cavity (internal and
external) must equate an integer number of half wavelengths.

Lm
2
For a tunable device, this means that, even if the cavity is carefully trimmed to, say, the central
value of the tuning range, when tuning is allowed, the selected frequency changes, and so does the
wavelength both in vacuum and inside matter.
In order to recover the cavity resonance, the back mirror must be able to adjust its position or by
mechanically moving, or introducing one more tunable refraction-controlled path.
It is important to estimate what is the amount of adjustment that is required.
Going back to the Etalon spectra, that have been calculated in a practical case, corresponding to
commercial devices, the frequency range spans the interval 190-196 THz.
This corresponds, in vacuum, to a wavelength range between 1.530 to 1.578 m, that is a range of
about 0.05 m.
Supposing the untuned cavity well trimmed to the central value of the range, this means a required
interval for mechanical shift of  0.025m .
This can be easily achieved in practical cases by means of simple control of thermal expansion of
some high reflecting element.
Referring once again to
practical solution that have
been found on real devices, let
us suppose to have a thick slice
of, say, Germanium. Its thermal
linear expansion coefficient is
5.9x10-6 per °C. Let H=500 m
be the thickness of the slice,
and also let its front surface be
coated with a highly reflecting
layer (this means that the
optical properties of Ge do not
enter the problem).
AR coating
Ge
When we consider that the transverse dimension changes, under thermal control, by
H
 5.9  106 per C , that is
H
we have that we must allow a total expansion range of H  0.05m to allow each lateral surface
to move by  0.025m .
With H=500 m, we have
H
0.05
T 

 17C
6
H  5.9  10
500  5.9  10 6
This is in the range achievable by a Peltier-effect based Thermo Electric Cooler (TEC).