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Discussion to the equivalent point realized by the two polarized beams in AOTF system

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Abstract

By an accurate theoretical analysis, an equivalent point is found where the filtered optic wave lengths of the two acoustic-optic reactions of “e in o out” and “o in e out” are the same with the same acoustic frequency. Two cases of δ=0 and δ≠0 are discussed and compared. The merits of the equivalent point is discussed in different points of view. The discussions conclude that the parameters set around the equivalent point leading to the optimum designing.

©1999 Optical Society of America

1. Overview of existing AOTF system

The Acoustic-Optic Tunable Filter (AOTF) is a new conception of mono-wave-length generator differs from traditional prism and grating. The wavelength selection is made by means of optic wave and acoustic wave reaction inside a birefringence crystal under a phase matching condition. Changing in acoustic frequency (in the range of several ten MHz to several hundreds MHz) can change the selected optic wavelength. There are no moving parts involved in the system so a quick scanning of spectrum can be realized electronically by computers. The compact size, no moving parts and the quick scanning ability are the main advantages that make AOTF a welcomed new member of the spectrum instrument family. The disadvantage of AOTF is the spectrum resolution depends on the size of the expensive crystal and which is normally lower than gratings.

The Acousto-Optic Tunable Filter (AOTF) was first made practice in 1967 in the form of collinear optic arrangement [1]. But the most usable and useful AOTF device was realized after the TeO2 was found to be the nearly ideal crystal in making AOTF cell and the noncollinear optic arrangement was put forward by I.C.Chang [2]. In the past 20 years, there were hundreds of patents and papers on the topic of developing and designing of new AOTF devices. Almost all the R&D work in the period of time was based on the early theoretical contributions of I.C.Chang [2] and T.Yano and A. Watanabe [3] in which the momentum matching and phase match conditions are commonly accepted. In the early theoretical work the physical model is perfect but the mathematics descriptions were always in approximate ways. In 1985, Mo Fuqin put forward the first accurate mathematics description to the parallel-tangent condition [4]. In 1987, Epikhin gave a more general equation set to describe the accurate relations between the acoustic and optic parameters which is one of the most valuable contributions to the AOTF designing [5]. In 1991, Gass selected an acoustic vector angle of -80.23° (with no reason) and calculated the so called optimum parameters for the system [6]. In 1992, Ren Quan etc. copied almost entire analysis of Gass’s paper and selected another acoustic vector angle of 105° (with no reason) and calculated a set of parameters for this particular angle [7]. Up to now, nobody has ever given an overall investigation to the entire parallel-tangent-phase-matching curve and then made the choice that which acoustic vector angle would be the best one. We are going to do this and select the optimum parameters with our reasons.

2. Equivalent Points in AOTF

Fig. 1 describes the basic relations of the vectors and angles. The sections of the wave vector surfaces of the uniaxis crystal for ordinary optic beam and extraordinary beam are plotted in a nearly round circle and an ellipse respectively. The Z axis denotes the crystal [001] axis. δ is a physical constant. Vector OC and OB denote the incident optic wave vectors of o-ray and e-ray with the same incident angle of θi respectively. By the interaction of acoustic wave of Kaeo at the direction of θaeo and the optic incident wave of Kie, the diffracted optic wave vector Kdo is generated at the direction of θdo. Similarly, the acoustic vector Kaoe at the direction of θaoe and the optic incident vector Kio generates the diffracted optic vector of Kde at the direction of θde. Vector triangles of OAB and OCD meet the parallel-tangent momentum matching condition. From Fig.1, the vector equation and the partial equations can be written as:

Kaeo=KieKdoKi=2πniλ
Kaeosinθaeo=KiesinθiKdosinθdoandKd=2πndλ
Kaeocosθaeo=KiecosθiKdocosθdoKa=2πfaVa

Where ni and nd are refractive indices respectively, fa is the acoustic frequency, Va is the acoustic velocity at the vector Va direction and λ is the vacuum optic wave length.

 figure: Fig. 1.

Fig. 1. The vectors and phasemat chcondition

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The physical model used in [4] and [5] is not so accurate because experiments and detail physical analyzing shows that δ≠0 and [8]:

δ(λ)=nie(0,λ)ndo(0,λ)2no(λ)4.55×104

The following equations are developed based on δ≠0 and the parallel-tangent phase matching condition. The refractive index surface for the extraordinary beam is described as:

nie(θi,λ)=(cos2θi(1+δ)2no2(λ)+sin2θine2(λ))12

where the subscript of “ie” means extraordinary beam incident.

The refractive index surface for the ordinary beam is no longer a circle. It becomes:

ndo(θi,λ)=(cos2[θde(θi,λ)](1+δ)2no2(λ)+sin2[θde(θi,λ)]no2(λ))12

where the subscript of “do” means the ordinary beam diffracting. The slope angle of the tangent of the ellipse of the index surface for the extraordinary beam at the locus decided by θi, and the slope angle of the tangent of the ellipse of the index surface for the ordinary beam at the locus decided by θdo are described as:

ke=(1+δ)2no2ne2tanθiko=(1δ)2no2no2tanθdo

Using the parallel-tangent phase matching condition of ke =ko , we have the diffracting angle for the arrangement of “e in o out”:

tan[θdo(θi,λ)]=[(1+δ)no(λ)(1δ)ne(λ)]2tanθi

Then we obtain the acoustic vector angle θaeo and the frequency faeo for “e in o out”:

tan[θaeo(θi,λ)]=nie(θi,λ)sinθindo(θi,λ)sin[θdo(θi,λ)]nie(θi,λ)cosθindo(θi,λ)cos[θdo(θi,λ)]
faeo(θi,λ)=Vaλ0[nie2(θi,λ)+ndo2(θi,λ)2nie(θi,λ)ndo(θi,λ)cos(θdo(θi,λ)θi)]12

Equations (1–13) to (1–17) are for the arrangement of “o in e out”:

nio(θi,λ)=(cos2θi(1δ)2no2(λ)+sin2θino2(λ))12
nde(θi,λ)=(cos2[θde(θi,λ)](1+δ)2no2(λ)+sin2[θde(θi,λ)]ne2(λ))12
tan[θde(θi,λ)]=[(1δ)ne(λ)(1+δ)no(λ)]2tanθi
tan[θaoe(θi,λ)]=nio(θi,λ)sinθinde(θi,λ)sin[θde(θi,λ)]nio(θi,λ)cosθinde(θi,λ)cos[θde(θi,λ)]
faoe(θi,λ)=Vaλ0[nio2(θi,λ)+nde2(θi,λ)2nio(θi,λ)nde(θi,λ)cos(θde(θi,λ)θi)]12

Where the subscripts of “aoe” means the acoustic vector angle or frequency for “o in e out”, “io” means ordinary beam incident and “de” means extraordinary beam diffracting.

Using equation (7) and (12), calculate the acoustic vector angle by scanning the optic incident angle from 0 to 90°. Fig. 2 gives the results in which both the arrangements of “o in e out” and “e in o out” with δ=0 and δ≠0 are plotted. Using equations (8) and (13), scanning the incident angle and changing the wave length we have Fig. 3 (δ≠0).

From Fig. 2 we see the parallel-tangent phase matching condition does not hold for both the ordinary incident and the extraordinary incident beams with a given incident angle except for the equivalent incident angle at about 56°.

In Fig. 3, again we find an equivalent incident angle at about 56° where the acoustic frequencies for a given filtered optic wave length is the same for both the ordinary and extraordinary incident beams. We also noticed that the position of the equivalent point is almost fixed with the variation of the filtered optic wave length or it is nearly independent to the filtered optic wave length.

 figure: Fig. 2.

Fig. 2. The acoustic wave vector via optic incident angle

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 figure: Fig.3.

Fig.3. The acoustic frequency via optic incident angle

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By making Equation (7) = Equation (12), Fig. 4 shows the behavior of the acousto-angle differences via the incident angle with conditions of δ=0 and δ≠0. It can be seen that the equivalent point is almost fixed with the changes in the optic wave length. We have found a resolution express for tan θaeo = tan θaoe:

(a2+1)2x64(a6+a4+a2)x24(a6+a4)=0

We will not discuss the mathematics here though we have found a real root for Equation (14). By making Equation (8) = Equation (13), Fig. 5 shows the behavior of the acoustic frequency difference via the incident angle with δ≠0. It can also be seen that the equivalent point is nearly independent to the filtered optic wave length. It is interesting when put the two curve sets together, as showed in Fig.6, that the two equivalent points discussed above are well coincided.

With numerical evaluation, we have the computing results listed in TABLE 1. It is seen that the wave length dependence of the equivalent incident angle is within 0.01° in the near infrared range and it is about 0.05° in the visible range. As far as the divergence of the incident optic beam and the acoustic wave coupled from the transducer, the influence of the variations can be neglected.

According to the parallel-tangent phase matching condition or equations (8) and (13), for θi=55.982°, θa=108.244°. An AOTF device is made with the parameters. The experiment result given in Fig. 7. proves the discussion presented in this paper. The data are collected by spectrometer of FT-IR System 2000 from PERKIN ELMER.

3. Conclusion

(a) The value of the equivalent point is evidence that if the incident signal is an un-polarized beam, the same wave length portion of both the horizontal polarizing part and the vertical polarizing part can be filtered by a single acoustic frequency from an AOTF cell at which the optic incident angle is in the angle of 55.9°. This arrangement is welcomed in the application that a weak spectrum incident signal is involved in the spectrum analyzing. Usually in spectrum analyzing the polarizing states are not so important as the signal intensity and the wave length resolution. Sum the diffracted output signals generated from e-ray and o-ray incident beams respectively (with the same wave length but different polarizing states), the output signal can be doubled.

(b) From Fig.2 we notice that around the equivalent incident angle of 56° the changes in the acoustic vector angle via the optic incident angle become mild which means the perfect phase matching condition is not quite critical to the optic incident angle and the acoustic wave vector angle. This feature gives the AOTF system a few advantages: less accuracy in orientating the angle to affix the acoustic transducer; large acceptable angular aperture for the acoustic wave; large acceptable angular aperture for the optic incident beam.

(c) From Fig. 3 we notice that around the incident angle of 56°, the curve of the parallel-tangent phase matching for the acoustic frequency via the incident angle becomes flat which means the parallel-tangent phase matching condition at this range is not quite critical so a large acceptable optic incident angular aperture can be obtained.

(d) From Fig. 3 we also notice that the vertical intervals between the curves with a given wave length (0.6μm, 0.7μm, 0.8μm, 2.4μm etc.) reach their maximum around the equivalent incident angle of 56° which means for a given optic wave length to be filtered, a higher acoustic frequency is needed while a higher spectrum resolution is obtained for a fixed band width of the electric driving signal. To obtain a higher driving frequency for the transducer is no longer a problem but for a given spectrum resolution the requirement for the purity of the electric signal is lowered or for a given purity of the electric signal, the spectrum resolution is increased.

(e) Every point of θi and θa on the curves deduced from the parallel-tangent phase matching condition such as that in Fig. 2 may be taken as the designing basis. But none of them can meet the perfect momentum matching condition for both the “e in o out” and “o in e out” simultaneously except the equivalent point or the cross point of the curve for “e in o out” and the curve for “o in e out”.

 figure: Fig. 4.

Fig. 4. The equivalent point of the acoustic vector angle

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 figure: Fig. 5.

Fig. 5. The equivalent point of the acoustic frequency

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 figure: Fig. 6.

Fig. 6. The two equivalent points are put together

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 figure: Fig. 7.

Fig. 7. Experiment result for comparing two AOTF cells : One is 23° cutt in gand the other is 56° cutting. Both of incident angle are perpendicular. Data collected by FT-IR System 2000 from PERKINELMER

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Tables Icon

TABLE:. The numeric computing results of the equivalent point

References:

1. R. W. Dixon, IEEE. J. Quantum Electron. QE-3, 85 (1967). [CrossRef]  

2. I. C. Chang, Appl. Phys. Lett. 25, 370 (1974). [CrossRef]  

3. T. Yano and A. Watanabe, Appl. Opt. 15, 2250 (1976). [CrossRef]   [PubMed]  

4. Mo Fuqin, Acta Optica Sinica, 6, 446 (1986).

5. V. M. Epikhin, F. L. Vizen, and L. L. Pal’tsev, Sov. Phys. Tech. Phys. 32, 1149 (1987).

6. P. A. Gass and J. R. Sambles, Opt. Lett. 16, 429 (1991). [CrossRef]   [PubMed]  

7. Ren Quan etc., Acta Optica Sinica, 13, 568 (1993).

8. A. W. Warner, D. L. White, and W. A. Bonner, J. Appl. Phys. 43,4489 (1972) [CrossRef]  

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Figures (7)

Fig. 1.
Fig. 1. The vectors and phasemat chcondition
Fig. 2.
Fig. 2. The acoustic wave vector via optic incident angle
Fig.3.
Fig.3. The acoustic frequency via optic incident angle
Fig. 4.
Fig. 4. The equivalent point of the acoustic vector angle
Fig. 5.
Fig. 5. The equivalent point of the acoustic frequency
Fig. 6.
Fig. 6. The two equivalent points are put together
Fig. 7.
Fig. 7. Experiment result for comparing two AOTF cells : One is 23° cutt in gand the other is 56° cutting. Both of incident angle are perpendicular. Data collected by FT-IR System 2000 from PERKINELMER

Tables (1)

Tables Icon

TABLE: The numeric computing results of the equivalent point

Equations (16)

Equations on this page are rendered with MathJax. Learn more.

K aeo = K ie K do K i = 2 πni λ
K aeo sin θ aeo = K ie sin θ i K do sin θ do and K d = 2 πnd λ
K aeo cos θ aeo = K ie cos θ i K do cos θ do K a = 2 π f a V a
δ ( λ ) = n ie ( 0 , λ ) n do ( 0 , λ ) 2 n o ( λ ) 4.55 × 10 4
n ie ( θ i , λ ) = ( cos 2 θ i ( 1 + δ ) 2 n o 2 ( λ ) + sin 2 θ i n e 2 ( λ ) ) 1 2
n do ( θ i , λ ) = ( cos 2 [ θ de ( θ i , λ ) ] ( 1 + δ ) 2 n o 2 ( λ ) + sin 2 [ θ de ( θ i , λ ) ] n o 2 ( λ ) ) 1 2
k e = ( 1 + δ ) 2 n o 2 n e 2 tan θ i k o = ( 1 δ ) 2 n o 2 n o 2 tan θ do
tan [ θ do ( θ i , λ ) ] = [ ( 1 + δ ) n o ( λ ) ( 1 δ ) n e ( λ ) ] 2 tan θ i
tan [ θ aeo ( θ i , λ ) ] = n ie ( θ i , λ ) sin θ i n do ( θ i , λ ) sin [ θ do ( θ i , λ ) ] n ie ( θ i , λ ) cos θ i n do ( θ i , λ ) cos [ θ do ( θ i , λ ) ]
f aeo ( θ i , λ ) = V a λ 0 [ n ie 2 ( θ i , λ ) + n do 2 ( θ i , λ ) 2 n ie ( θ i , λ ) n do ( θ i , λ ) cos ( θ do ( θ i , λ ) θ i ) ] 1 2
n io ( θ i , λ ) = ( cos 2 θ i ( 1 δ ) 2 n o 2 ( λ ) + sin 2 θ i n o 2 ( λ ) ) 1 2
n de ( θ i , λ ) = ( cos 2 [ θ de ( θ i , λ ) ] ( 1 + δ ) 2 n o 2 ( λ ) + sin 2 [ θ de ( θ i , λ ) ] n e 2 ( λ ) ) 1 2
tan [ θ de ( θ i , λ ) ] = [ ( 1 δ ) n e ( λ ) ( 1 + δ ) n o ( λ ) ] 2 tan θ i
tan [ θ aoe ( θ i , λ ) ] = n io ( θ i , λ ) sin θ i n de ( θ i , λ ) sin [ θ de ( θ i , λ ) ] n io ( θ i , λ ) cos θ i n de ( θ i , λ ) cos [ θ de ( θ i , λ ) ]
f aoe ( θ i , λ ) = V a λ 0 [ n io 2 ( θ i , λ ) + n de 2 ( θ i , λ ) 2 n io ( θ i , λ ) n de ( θ i , λ ) cos ( θ de ( θ i , λ ) θ i ) ] 1 2
( a 2 + 1 ) 2 x 6 4 ( a 6 + a 4 + a 2 ) x 2 4 ( a 6 + a 4 ) = 0
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