Far-Field Propagation of an Off-Axis Gaussian Wave
Transcript of Far-Field Propagation of an Off-Axis Gaussian Wave
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Far-field propagation of an off-axis Gaussian wave
Xiaodong Zeng, Changhong Liang, and Yuying An
Based on a far-field asymptotic solution of the Helmholtz equation a vector theory to describe thepropagation of an off-axis Gaussian wave is developed, the accurate formulas represented in terms ofelementary functions are derived, and the propagation properties such as wave spot size and divergenceangle are discussed in detail. The applicable range of scalar theory is also presented. A relative errorcriterion of optical intensity is given by e 5 sin2 a. © 1999 Optical Society of America
OCIS codes: 050.1960, 260.2110, 260.1960.
1. Introduction
The so-called off-axis Gaussian wave means that thelight wave radiates from a small Gaussian source andhas a large divergence angle. Researchers deal witha large divergence angle light wave in many areas ofapplied optics, such as a diode laser or a stronglyfocused light wave, for which the paraxial approxi-mation no longer holds. Several authors have dis-cussed light-wave propagation beyond the paraxialapproximation.1–7 Such investigations can be di-ided into two classes: one is based on the angularpectrum representation and the other is based onhe work by Lax et al.1 Lax et al. considered a wavef large divergence angle by use of the perturbationethod, in which the fields are expanded in powers of
0yl, where v0 and l are the scaling parameters forhe wave waist and diffraction length, respectively.owever, because these methods have complicatedathematics that generate results ~represented by
n infinite series! that are not easily tractable in aractical case, it is not feasible for one to performdditional mathematical analyses. To obtain usefulesults, one must use numerical techniques.
Recently, Zeng et al.8 developed a new method fordescribing the propagation of an off-axis Gaussianwave. With this scheme the propagation problemwas regarded as a far-field problem. Then a far-fieldasymptotic solution of the Helmholtz equation wasfound, and the solution was represented in terms of
The authors are with the Department of Technical Physics, Xid-ian University, 710071 Xi’an, China. The e-mail address for X.Zeng is [email protected].
Received 16 February 1999; revised manuscript received 26 May1999.
0003-6935y99y306253-04$15.00y0© 1999 Optical Society of America
elementary functions rather than a complicated rep-resentation of an infinite series. However, thisscheme is within the framework of scalar theory, andwe assume that the optical field has a unique polar-ization direction. In fact, this approximation can beused only if the light wave propagates in the paraxialregion. We point out that the longitudinal componentof the optical field cannot be neglected when the sourcesize is very small and the light wave has a large diver-gence angle. In particular, in the off-axis region theoptical field distribution is considerably different fromthat given by scalar theory. This must be taken intoaccount for many cases in which the field distributionneeds to be described accurately, such as strongly fo-cusing and highly collimating.
Here we propose a general vector field method todescribe the far-field propagation properties of theoff-axis Gaussian wave and derive explicit expres-sions for the far-field distribution. Based on the far-field expressions we analyze the wave spot size andshape and the wave divergence angle as well as theconservation relation. We also analyze the errorsowing to the scalar approximation and discuss thecondition under which the scalar theory holds. Be-cause the expressions we obtained have a simplemathematical form, it should be possible to use themmore easily for practical work.
2. Vector Solution of the Wave Equation
Let a three-dimensional vector wave expression beE~r, t! 5 E~r!exp~2ivt!, that is, we consider onlymonochromatic radiation. According to the electro-magnetic field theory, the space-dependent part ofE~r! should satisfy the Helmholtz equation, i.e.,
¹2E~r! 1 k2E~r! 5 0 (1)
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dxnhpttspbdosiiG
w
lt
t
ns
d
6
subject to the appropriate boundary conditions.Here k 5 2pyl is the free-space wave number. We
efine a Cartesian coordinate system and let the–y plane be the boundary surface. There is a pla-ar source near the origin, and it radiates in thealf-space z . 0. ~x, y, z! is a far-field observationoint. Here far field means that the distance fromhe source to the observation point is much greaterhan the wavelength as well as the size of theource. For most propagation and diffractionroblems the far-field condition is always satisfied,ecause field measurement is usually attempted atistances significantly greater than the wavelengthr the source size. With no loss of generality weuppose that the projection of the source field vectorn the x–y plane is linearly polarized, say x polar-zed, and that the source distribution function isaussian, i.e.,
Ex~x9, y9, 0! 5 A expS2x92 1 y92
v02 D , (2)
Ey~x9, y9, 0! 5 0, (3)
where A is a constant and v0 is the spot size of thesource distribution. Equation ~2! shows that thesource pattern is circularly symmetric with respect tothe z axis. As we show below, the radiation fieldvector can be determined uniquely from the trans-verse components of the field over the boundary sur-face, i.e., Ex ~x, y, 0! and Ey~x, y, 0!. Because of thecontinuity of the electric field the radiation field hasno y component, that is, Ey~x, y, z! 5 0 in the half-space z . 0. Therefore, all we need to do is deter-mine the x and z components. In the far-field regionthe x component of the electric field can be expressedas8
Ex~x, y, z! 5 Bzr2 exp~ikr!expF2
1C2
x2 1 y2
r2 G , (4)
here
B 5iApv0
2
l, (5)
C2 54
k2v02 , (6)
r2 5 x2 1 y2 1 z2. (7)
In the far-field region the electric field vector E mustbe perpendicular to the wave vector or, equivalently,to the vector between the origin and the observationpoint, as shown in Fig. 1. Therefore, the z compo-
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nent of the electric field at any far-field point ~x, y, z!can be expressed as
Ez~x, y, z! 5 2xz
Ex~x, y, z!
5 2Bxr2 exp~ikr! expF2
1C2
x2 1 y2
r2 G . (8)
Then we obtain the electric field vector
E~x, y, z! 5 Bexp~ikr!
r2 expF21C2
x2 1 y2
r2 G F z0
2xG . (9)
Equation ~9! is the accurate vector expression of theight wave in the far-field region. Compared withhe scalar theory8 we observed that the difference
between the vector theory and the scalar theory is theexistence of the z component. It is obvious that inhe paraxial region, we have uxu ,, z and Ez can be
neglected, proof that the scalar theory is sufficientlyaccurate for practical work. The far-field z compo-
ent implies the existence of a z component near theource.
3. Propagation Properties
The vector expression in Eq. ~9! provides a completedescription of the far-field propagation properties ofthe light wave. It shows that the light wave has aperfectly spherical phase distribution with a radius ofcurvature r centered at the origin. However, theamplitude vector has a slightly complicated form.Experimentally only the field intensity can be mea-sured directly. Here we discuss the far-field inten-sity distribution.
A. Wave Spot Size
By use of Eq. ~9! we can obtain the far-field intensityistribution
I~x, y, z! 5 uExu2 1 uEzu2
5 uBu2x2 1 z2
r4 expF22C2
x2 1 y2
r2 G . (10)
Fig. 1. Geometry of the far-field components.
o
T
masw
r
Bc
Equation ~10! indicates the difference between thevector Gaussian wave and the scalar Gaussian wave.The relationship of the far-field intensity with respectto the coordinates, i.e., the propagation of the lightwave in half-space z . 0, is determined by the factor~x2 1 z2!yr4 and an analogous Gaussian factor, ratherthan the z2yr4 factor and analogous Gaussian factor.8By substituting Eq. ~10! into I~x0, y0, z! 5 I~0, 0, z!y2,we can determine the half-width of the far field, i.e.,
12z2 5
x02 1 z2
r04 expF2
2C2
x02 1 y0
2
r02 G , (11)
where r02 5 x0
2 1 y02 1 z2. From Eq. ~11! it is
bvious that for a given value of z both x0 and y0 arebounded, therefore Eq. ~11! represents a closed-planecurve. In the z constant plane Eq. ~11! is an analo-gous ellipse, which shows that the intensity distribu-tion of the vector field is considerably different fromthat of the scalar field, in which the equal-intensitycurve is a circle. From Eq. ~11! it is also obvious thatthe major axis lies in the y 5 0 plane, and the minoraxis lies in the x 5 0 plane. The major and minoraxes can be obtained as follows:
When y0 5 0, from Eq. ~11! we have
12z2 5
x02 1 z2
r04 expF2
2C2
x02
r02G .
Note that now r02 5 x0
2 1 z2, so the above expressionchanges to
12
5 F1 2x0
2
r02G expF2
2C2
x02
r02G . (12)
Based on the results reported in Ref. 8, we know thatEq. ~12! has a unique solution for x0
2yr02. Let it be s0,
i.e.,
x02
x02 1 z2 5 s0
or
x0 5 Î s0
1 2 s0z. (13)
When x0 5 0, from Eq. ~11! we have
12
5 F1 2y0
2
r02G2
expF22C2
y02
r02G . (14)
Similarly, Eq. ~14! has a unique solution t0 for y02yr0
2.hus we obtain
y0 5 Î t0
1 2 t0z. (15)
Equations ~13! and ~15! show that the major andinor axes of the wave spot are proportional to prop-
gation distance z, which is the mathematical de-cription of the rectilinear propagation of the lightave in the far-field region. When the source pa-
ameters v0 and l are given, s0 and t0 can be evalu-ated by a numerical technique.
B. Wave Divergence Angle
Since the far-field intensity pattern is an ellipse, weshould define two divergence angles in the x–z planeand the y–z plane, respectively. From Eqs. ~13! and~15! we can obtain the full angles at half-maximumintensity in the two planes:
ux 5 2 arctanSx0
z D 5 2 arctan Î s0
1 2 s0, (16)
uy 5 2 arctanSy0
z D 5 2 arctan Î t0
1 2 t0. (17)
y use of a numerical method we can evaluate theurves of ux and uy with respect to v0yl, as shown in
Fig. 2. We can see that the smaller v0yl, the largerthe wave divergence angles. However, both ux anduy have upper limits, which can be obtained directlyfrom Eqs. ~12! and ~14!:
ux max 5 2 arctan 1 5 90°, (18)
uy max 5 2 arctan@Î2 2 1#1y2 < 65.5°. (19)
The results show that the maximum divergence anglein the direction parallel to the polarization plane islarger than that predicted by scalar theory.8 In thedirection perpendicular to the polarization plane thedivergence angle cannot exceed the 65.5° limit, whichreproduces the scalar theory results. From Eqs. ~12!and ~14! we can see also that the two divergence anglescan take their maximum values, only if C3 `, i.e., thesource distribution approaches a d function.
C. Conservation Relation
The vector expression in Eq. ~9! is the far-field as-ymptotic solution of the Helmholtz equation. Froma physical point of view, the light-wave propagationshould satisfy the energy conservation relation.That is, the integral of the intensity that crosses an
Fig. 2. Variance of the divergence angles with v0yl.
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a
E
2 2 2
6
arbitrary plane z is a constant and is independent ofthe value of z, which we show below. We introduce
change of variables x 5 r cos u and y 5 r sin u.Then Eq. ~10! can be rewritten as
I~u, r, z! 5 uBu2r2 cos2 u 1 z2
~r2 1 z2!2 expF22C2
r2
r2 1 z2G . (20)
The intensity integral therefore becomes
*2`
1`
*2`
1`
I~x, y, z! dxdy 5 uBu2 *0
2p
du *0
1` r2 cos2 u 1 z2
~r2 1 z2!2
3 expF22C2
r2
r2 1 z2Grdr
5 uBu2p *0
1` r2 1 2z2
~r2 1 z2!2
3 expF22C2
r2
r2 1 z2Grdr.
(21)
When u 5 r2y~r2 1 z2!, we have
*2`
1`
*2`
1`
I~x, y, z!dxdy 5puBu2
2 *0
11 2 2 u1 2 u
3 expF22C2 uGdu. (22)
The right-hand side of Eq. ~22! indicates that theintensity integral is independent of z, i.e., the prop-agation formula in Eq. ~9! satisfies the conservationrelation.
4. Applicable Range of Scalar Theory
In many practical cases one can use the scalar theoryto describe light-wave propagation. We discuss theerrors owing to the scalar approximation and theapplicable range of the scalar theory. Because therelative errors are of utmost importance, here weconsider only the relative errors of the intensity.
Under scalar approximation the electric field con-tains only the x component Ex. Then the intensitycan be expressed as Is 5 uExu2 and subscript s denotesthe scalar field. We can obtain the relative error ofthe far-field intensity
e 5uI 2 Isu
I5
x2
x2 1 z2 . (23)
quation ~23! indicates that the relative error is in-dependent of y, i.e., independent of the coordinate inthe direction perpendicular to the polarized plane.We can see that e increases with uxu, which meansthat the scalar theory is accurate in the x 5 0 plane.
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Note that x y~x 1 z ! is just the square sine of theangle between the projection of the observation pointvector in the x–z plane and the z axis. We denotethis angle as a and Eq. ~23! becomes
e 5 sin2 a. (24)
When we choose the number d as the practical crite-rion under which scalar theory holds, from sin2 a # dwe have
a # arcsin Îd. (25)
The above results show that scalar theory is applica-ble for a light wave, provided that the energy distri-bution range ~to be considered! does not exceed 2a.For other cases vector theory should be used. Forexample, if d 5 5%, from Eq. ~25! we have 2a 5 25.8°.
5. Conclusion
Based on a far-field asymptotic solution of the vectorHelmholtz equation a general method to describe thefar-field propagation of the off-axis Gaussian wavehas been developed, and we derived explicit forms forthe far-field expressions.
The validity of the scalar theory was examined, anda practical criterion, under which the scalar theoryholds, was given. The relative errors resulting fromthe scalar approximation could be estimated by theformula e 5 sin2 a. The results indicate that, todescribe the radiation field accurately, one must takeinto account the vector properties of the field, partic-ularly in the off-axis region.
Our procedure to obtain the propagation propertiesof the off-axis light wave is considerably simpler thanthe one mentioned above. This method, which to ourknowledge is new, should be useful in applications forwhich there is a need to obtain accurate solutions for alight wave with a large divergence angle.
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