On focused fields with maximum electric field components and images of electric dipoles

R. de Bruin; H. P. Urbach; S. F. Pereira

doi:10.1364/OE.19.009157

1. Introduction

When light is focused by a lens with numerical aperture larger than 0.6, the rotation of polarization becomes important. The rotation of polarization of a diffraction limited lens of high numerical aperture is well described by the model of Ignatowsky [1, 2] and Wolf and Richards [3, 4]. When the field in the lens aperture is that of a linearly polarized plane wave, the dominant electric field component near the focal point is parallel to the direction of polarization of the incident plane wave. But, as the numerical aperture increases, the maximum value of the longitudinal component of the electric field in the focal plane becomes quite substantial, although it vanishes at the focal point itself.

In many applications such as optical recording, photolithography and microscopy, it is important to be able to shape the focused spot. A field in focus with maximum electric component in a certain direction is of interest for manipulating and probing single molecules and particles, and in materials processing [5–9]. The focused wavefront can be tailored by setting a proper amplitude, phase and polarization distributions in the pupil of the focusing lens [10,11]. Nowadays it is possible to realize almost any complex transmission function in the pupil plane, using for example liquid crystal-based devices [12–15].

The optimization of focused fields has been studied by [16–18] for rotational symmetric systems. In the first two papers it was noted that the focused linearly polarized plane wave in the theory of Ignatowsky and Richard and Wolf, is in the Debye approximation identical to the sum of fields that are truncated by the lens aperture and converge to orthogonal electric and magnetic dipoles, with a particular ratio of dipole strengths. A nice treatment is given in [19].

It was noted by several authors [16, 20–23] that when a radially polarized beam is focused, the distribution of the longitudinal component can be considerably narrower than the focused spot obtained by focusing a linearly polarized plane wave. In [24] the pupil field was determined which after focusing gives the maximum longitudinal component in the focal point for a given power incident on the lens. The optimum pupil field is radially polarized and its amplitude increases in a particular way with radial distance to the center of the pupil. Furthermore, the distribution in the focal plane of the optimized longitudinal component is narrower than the Airy spot. If a recording medium would be used which is sensitive for only the longitudinal component of the electric field, this optimized longitudinal component can yield higher resolution.

In [25] this analysis was generalized to optimum pupil fields of which the electric field amplitude in an arbitrary, not necessarily longitudinal, direction in the focal point is maximum. It was in particular found that the pupil field which gives maximum transverse amplitude in focus is similar but not identical to that of a linearly polarized plane wave.

An interesting feature of the optimum focused fields is that, although the electric field amplitude at the focal points is maximum in a required direction, the electric field is in general not linearly polarized in that direction but in another direction. Exceptions are the maximum longitudinal and maximum transverse components. In those cases the optimum fields are linearly polarized along the longitudinal and transverse directions, respectively.

For certain applications it is desirable to have maximum electric field amplitude in the focal point in a given direction under the additional constraint that the electric field is also linearly polarized in that direction. We shall therefore consider this optimization problem in this paper. A major result is that as the optimization direction is varied the set of solutions is identical to that of the problem without the constraint of parallel polarization studied in [25]. To be more precise: the electric field that has maximum amplitude in a given direction but which is not linearly polarized along that direction, is also the field that has maximum amplitude in the direction of its linear polarization.

As was already explained in [25], the optimum focused fields can be realized by focusing appropriate pupil fields. These pupil fields are linearly polarized with the direction of polarization and amplitude that are functions of the pupil coordinates. An alternative method to realize the optimum fields is by imaging an electric dipole by an optical system with unit magnification, e.g. two identical lenses. The image of an electric dipole was studied previously in [6, 17, 19] and, more recently, in [27]. We will show that when the imaging system has unit magnification, the image of an electric dipole with dipole vector p = (p_x, p_y, p_z) is, for the given power, identical to the electric field in image space that has maximum amplitude in the direction (p_x, p_y, – p_z), where the z-axis is the optical axis. It follows in particular that the image of a dipole that is parallel to the optical axis is the field for which the longitudinal component is maximum.

The paper is organized as follows. In Section 2 the optimization problem and the optimum fields that have been derived in [24] and [25] are summarized. The optimization problem is formulated in homogeneous space without considering the optical system by which the optimum fields are generated. The reason is that the results thus obtained are general and independent of any model used for the optical system. In Section 3 the results of this optimization are interpreted and the new optimization problem is studied where the electric field is required to be linearly polarized along the direction for which the amplitude of the electric field is maximum. It is then shown that when this direction is varied, the set of optimum fields is identical to the set of optimum fields that are the solutions of the first optimization problem. Finally in Section 4 two methods to realize the optimum fields are considered. The first method is by focusing an appropriate pupil field. We use here the vector diffraction theory of Ignatowksy and Wolf and Richards, which is based on the Debye approximation. The second method is by imaging an electric dipole with unit magnification.

2. The optimization problem

2.1. Formulation of the optimization problem

We first define a few quantities and introduce some notations. As was stated in the Introduction, in the optimization problem we first leave the optical system out of consideration. The optimized fields are derived from the rigorous Maxwell equations and are therefore generally valid. We only assume that the field in image space has a certain numerical aperture NA ≤ n, where n is the refractive index in image space, and that the field has a prescribed power. Later we will discuss what fields in the entrance pupil of a lens with the chosen numerical aperture produce the optimum focused fields.

The plane wave vectors in image space are limited by the numerical aperture of the imaging system. Let (x,y,z) be a Cartesian coordinate system with the z-axis parallel to the optical axis. Using the plane wave expansion, the electric field in a point r in image space, which propagates in the direction of the positive z-axis, can be written as

ℰ (r, t) = Re [E (r) e^{- i ω t}] = \frac{1}{4 π^{2}} Re [\iint_{\sqrt{k_{x}^{2} + k_{y}^{2}} \leq k_{0} N A} A (k_{x}, k_{y}) e^{i (k \cdot r - ω t)} d k_{x} d k_{y}],

where E(r) is the complex field, k = (k_x, k_y, k_z) the wave vector in the medium with refractive index n, k ₀ = |k|/n = 2π/λ ₀ with λ ₀ the wavelength in vacuum and A(k_x, k_y) are the complex plane wave amplitudes. Since the field propagates in the positive z-direction, we have

k_{z} = \sqrt{k_{0}^{2} n^{2} - k_{x}^{2} - k_{y}^{2}} > 0

. The circle

\sqrt{k_{x}^{2} + k_{y}^{2}} \leq k_{0} NA

in the (k_x, k_y)-plane, which limits the integration area, will be denoted by Ω for brevity. Since the electric field is free of divergence, we have:

A (k_{x}, k_{y}) \cdot k = 0.

The time-averaged power flow is given by the integral over a plane z = constant of the z-component of the time averaged Poynting vector [25]:

\begin{array}{l} P (A) & = & \frac{1}{2} \iint Re [E (r) \times H (r) *] \cdot \hat{z} d x d y \\ = & \frac{1}{8 π^{2} k_{0}} {(\frac{ɛ_{0}}{μ_{0}})}^{1 / 2} \iint_{Ω} | A (k_{x}, k_{y}) |^{2} k_{z} d k_{x} d k_{y} . \end{array}

Let v̂ be a unit vector and consider the projection of the electric field at the origin r = 0 along this unit vector:

ℰ (0, t) \cdot \hat{v} = Re [E (0) \cdot \hat{v} e^{- i ω t}] .

The aim is to find the field for which the amplitude of ℰ (0,t) · v̂ is maximum for a given total power. Without restricting the generality we may assume that E(0) · v is real. In fact, if it is not real, applying a time shift can always make it real. Hence we will assume that

Im [E (0) \cdot \hat{v}] = \frac{1}{4 π^{2}} \iint_{Ω} Im [A (k_{x}, k_{y}) \cdot \hat{v}] d k_{x} d k_{y} = 0.

Then the optimization problem is to find the plane wave amplitudes A(k_x, k_y) for which the component of the electric field along v̂ has maximum amplitude for a given total power P ₀:

℘_{1} (\hat{v}) : max_{A \in ℋ_{0}} E_{v} (A) \equiv E (0) \cdot \hat{v}, under the constraint P (A) \leq P_{0},

where ℋ ₀ is the space of plane wave amplitudes which have finite mean flow of power through a plane z = constant and which satisfy Eq. (5). The problem is denoted as ℘ ₁(v̂) to indicate that we are optimizing in the direction of v̂. It can be shown that optimization problem ℘ ₁(v̂) has a unique solution. The uniqueness is due to the imposed condition that E(0) ·v̂ is real. Without this condition the optimum solutions are unique except for a phase shift [25].

2.2. Optimum plane wave amplitudes and field distributions

The solution of problem ℘ ₁(v̂) can be computed by applying the Lagrange multiplier rule for inequality constraints. We remark that since we optimize a linear functional on a sphere in a Hilbert space (the sphere is the set of all fields of finite given power), the solution can alternatively be derived by using the Cauchy-Schwarz inequality. In any case, the solution can be computed in closed form and is given by [25]:

A (k_{x}, k_{y}) = {(\frac{8 π P_{0}}{n})}^{1 / 2} {(\frac{μ_{0}}{ɛ_{0}})}^{1 / 4} \frac{1}{k_{z}} \frac{1}{Γ (α_{max}, v_{z})} [\hat{v} - (\hat{v} \cdot \hat{k}) \hat{k}],

where

Γ (α_{max}, v_{z}) = {\frac{4}{3} - cos α_{max} - \frac{1}{3} {cos}^{3} α_{max} - v_{z}^{2} cos α_{max} {sin}^{2} α_{max}}^{1 / 2},

for which α _max is such that

sin α_{max} = NA / n .

It follows from Eq. (6) that the electric field vector of the plane wave with wave vector k is parallel to the projection of v̂ on the plane perpendicular to k. The electric field in an arbitrary point r in image space can be obtained by substitution of the optimum plane wave amplitudes into Eq. (1). By changing the integration variables to respective polar and azimuthal angles, 0 < α < α _max and 0 < β < 2π, given by (Fig. 1)

k_{x} = k sin α cos β,

k_{y} = k sin α sin β,

k_{z} = k cos α,

and using cylindrical coordinates ρ,φ,z for the point of observation: r = x x̂ + y ŷ + z ẑ = ρ $\hat{ρ}$ + z ẑ= ρ cos φ x̂ + ρ sin φ ŷ + z ẑ, one finds for the optimum electric field [25]:

E (ρ, φ, z) = \frac{\sqrt{2 π n P_{0}}}{λ_{0}} {(\frac{μ_{0}}{ɛ_{0}})}^{1 / 4} \frac{1}{Γ (α_{max}, v_{z})} \bar{\bar{ℳ}} (ρ, φ, z) (\begin{matrix} v_{x} \\ v_{y} \\ v_{z} \end{matrix}),

where

\begin{array}{l} \bar{\bar{ℳ}} (ρ, φ, z) & = & (\begin{matrix} g_{0}^{0, 1} (ρ, z) + g_{0}^{2, 1} (ρ, z) & 0 & 0 \\ 0 & g_{0}^{0, 1} (ρ, z) + g_{0}^{2, 1} (ρ, z) & 0 \\ 0 & 0 & 2 g_{0}^{0, 3} (ρ, z) \end{matrix}) \\ + & (\begin{matrix} g_{2}^{0, 3} (ρ, z) cos (2 φ) & g_{2}^{0, 3} (ρ, z) sin (2 φ) & - 2 i g_{1}^{1, 2} (ρ, z) cos φ \\ g_{2}^{0, 3} (ρ, z) sin (2 φ) & - g_{2}^{0, 3} (ρ, z) cos (2 φ) & - 2 i g_{1}^{1, 2} (ρ, z) sin φ \\ - 2 i g_{1}^{1, 2} (ρ, z) cos φ & - 2 i g_{1}^{1, 2} (ρ, z) sin φ & 0 \end{matrix}), \end{array}

and where

g_{l}^{ν, μ} (ρ, z) = \int_{0}^{α_{max}} e^{i k_{0} n z cos α} {cos}^{ν} α {sin}^{μ} α J_{l} (k_{0} n ρ sin α) d α .

Note that the electric field vector and the matrix in Eq. (12) are on the Cartesian basis {x̂,ŷ,ẑ}.

Fig. 1 Spherical coordinates in image space with respect to wave unit vector k̂.

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3. Interpretation of optimum fields and second optimization

3.1. The electric field in focus

We consider the optimum electric field vector at the origin. Because $g_{l}^{ν, μ} (0) = 0$ when l ≥ 1 it follows from Eq. (13) that

E (0) = [g_{0}^{0, 1} (0) + g_{0}^{2, 1} (0)] v_{x} \hat{x} + [g_{0}^{0, 1} (0) + g_{0}^{2, 1} (0)] v_{y} \hat{y} + 2 g_{0}^{0, 3} (ρ, z) v_{z} \hat{z},

with

g_{0}^{0, 1} (0) = \int_{0}^{α_{max}} sin α d α = 1 - cos α_{max},

g_{0}^{2, 1} (0) = \int_{0}^{α_{max}} {cos}^{2} α sin α d α = \frac{1}{3} - \frac{1}{3} {cos}^{3} α_{max},

g_{0}^{0, 3} (0) = \int_{0}^{α_{max}} {sin}^{3} α d α = \frac{2}{3} - cos α_{max} + \frac{1}{3} {cos}^{3} α_{max} .

It is seen that the electric vector E(0) has real components, hence the electric field at r = 0 is linearly polarized. It is furthermore parallel to the plane through v̂ and the z-axis. However, the direction of polarization is in general different from v̂, i.e. from the direction along which the electric field amplitude is maximized. We shall write for convenience:

γ_{+} \equiv g_{0}^{0, 1} (0) + g_{0}^{2, 1} (0) = \frac{4}{3} - cos α_{max} - \frac{1}{3} {cos}^{3} α_{max},

γ_{-} \equiv 2 [g_{0}^{0, 1} (0) - g_{0}^{2, 1} (0)] = \frac{4}{3} - 2 cos α_{max} + \frac{2}{3} {cos}^{3} α_{max} .

Then Eq. (7) becomes:

Γ (α_{max}, v_{z}) = {γ_{+} - (γ_{+} - γ_{-}) v_{z}^{2}}^{1 / 2},

and Eq. (15) can be written as

E (0) = γ_{+} (v_{x} \hat{x} + v_{y} \hat{y}) + γ_{-} v_{z} \hat{z} .

Let û be the unit vector of the direction of polarization, i.e.

\hat{u} = \frac{E (0)}{| E (0) |} = \frac{γ_{+} (v_{x} \hat{x} + v_{y} \hat{y}) + γ_{-} v_{z} \hat{z}}{\sqrt{γ_{+}^{2} (v_{x}^{2} + v_{y}^{2}) + γ_{-}^{2} v_{z}^{2}}} .

We conclude that the direction of polarization at the origin is only parallel to the direction of optimization when v_z = 0 (transverse polarization) and when v_x = v_y = 0 (longitudinal polarization). An exception is the case that the numerical aperture NA = n. Then α _max = π/2, so that γ ₊ = γ ₋ = 4/3 and hence û and v̂ are always identical.

Without restricting the generality we may assume that v̂ lies in the (x,z)-plane, i.e. that v_y = 0. Then û also is in the (x,z)-plane. Let ϑ_v and ϑ_u be the angle that v̂ and û, respectively, make with the z-axis. Then

v_{x} = sin ϑ_{v}, v_{z} = cos ϑ_{v},

and similarly,

u_{x} = sin ϑ_{u}, u_{z} = cos ϑ_{u},

The relationship between both angles is displayed in Fig. 2 for several values of the numerical aperture. We see that always ϑ_u ≥ ϑ_v, i.e. the angle that û makes with the positive z-axis is in general larger than that of v̂. Furthermore, ϑ_u sweeps out all values between 0 and π/2 when ϑ_v varies between 0 and π/2. The difference between the two angles decreases when the numerical aperture is increased and the angles are the same in the limit NA = n.

Fig. 2 The relationship between the optimization angle ϑ_v for which problem ℘ ₁ is solved and angle ϑ_u between the direction of the optimum electric field vector E(0) and the z-axis, for several values of the numerical aperture NA.

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3.2. Optimum linearly polarized electric field

Although the solutions of problem ℘ ₁(v̂) have at the origin maximum electric field amplitude in the direction of v̂, the electric field vector at the origin is, although linearly polarized, not parallel to v̂. In fact it is pointing in the direction of the unit vector û.

The question then arises: what are the optimum fields that have maximum amplitude in a given direction v̂ and are in addition linearly polarized along v̂? The requirement that the electric field vector ℰ(0,t) = Re[E(0)exp(–iωt)] is parallel to v̂ at all times t is equivalent to demanding that the vector E(0) is proportional to v̂, which in turn is equivalent to:

E (0) \cdot (\hat{v} \times \hat{z}) = 0 and E (0) \cdot [\hat{v} \times (\hat{v} \times \hat{z})] = 0.

(When v̂ = ẑ these conditions become: E(0) ·x̂ = 0 and E(0) · ŷ = 0). Hence, the second optimization problem that we want to consider is:

℘_{2} (\hat{v}) : max_{A \in ℋ_{0}} E_{v} (A) = E (0) \cdot \hat{v}, under the constraints P (A) \leq P_{0} and (26),

where the space ℋ ₀ consists as before of all plane wave amplitudes A (k_x, k_y) such that E(0) · v̂ is real.

The solution of problem ℘ ₂(v̂) can be determined analogously to that of problem ℘ ₁(v̂) by using the Lagrange multiplier rule or alternatively using the Cauchy-Schwarz inequality. In the latter case the constraint that the electric field should be linearly polarized along the direction of v̂, should be incorporated into the Hilbert space of fields to which the Schwarz inequality is applied. Closed formulae are again obtained and the solution is again unique due to the fact that the phase is fixed by requiring that E(0) ·v̂ is real. It is then found that when v̂ is varied, more precisely, when ϑ_v is varied in the interval 0 ≤ ϑ_v ≤ π/2, the two sets of optimum solutions of the two optimization problems are identical when the proper directions are chosen. For every v̂ there is a unit vector û such that the optimum field that is the solution of ℘ ₁(v̂) is also the solution of ℘ ₂(û), and conversely. In fact, the relationship between the unit vectors v̂ and û is given by Eq. (23) defined in the previous section.

This is easy to prove. Let $E_{1}^{\hat{v}}$ and $E_{2}^{\hat{u}}$ be the solution of problem ℘ ₁(v̂) and ℘ ₂(û), respectively, and let v̂ and û be related by Eq. (23). Then we have to show that

E_{1}^{\hat{v}} = E_{2}^{\hat{u}} .

We first remark that both

E_{1}^{\hat{v}}

and

E_{2}^{\hat{u}}

are pointing in the direction of the unit vector û and are real vectors. For

E_{1}^{\hat{v}}

this follows from Eq. (15) while for

E_{2}^{\hat{u}}

it follows from the requirements for the field in optimization problem ℘ ₂(û).

It is clear that $E_{1}^{\hat{v}}$ satisfies all constraints of problem ℘ ₂(û). Hence

0 < E_{1}^{\hat{v}} (0) \cdot \hat{u} \leq E_{2}^{\hat{u}} (0) \cdot \hat{u} .

Since

E_{1}^{\hat{v}} (0)

and

E_{2}^{\hat{u}} (0)

are both pointing in the direction of û and since by Eq. (23) v̂ · û > 0, it follows that

E_{1}^{\hat{v}} (0) \cdot \hat{v} \leq E_{2}^{\hat{u}} (0) \cdot \hat{v} .

but

E_{2}^{\hat{u}}

satisfies all constraints of problem ℘ ₁(v̂) and since

E_{1}^{\hat{v}}

is the solution, it follows that in Eq. (29) equality holds. Because the solution of problem ℘ ₁(v̂) is unique, we thus conclude that Eq. (27) is indeed true.

3.3. Explicit expressions for the solution of problem ℘₂

With the results obtained above, explicit formulae for the optimum plane wave amplitudes that are solutions of optimization problem ℘ ₂(û) can be easily obtained. First we determine the unit vector v̂ by inverting Eq. (23):

\hat{v} = \frac{γ_{-} (u_{x} \hat{x} + u_{y} \hat{y}) + γ_{+} u_{z} \hat{z}}{\sqrt{γ_{-}^{2} (u_{x}^{2} + u_{y}^{2}) + γ_{+}^{2} u_{z}^{2}}} = \frac{γ_{-} \hat{u} + (γ_{+} - γ_{-}) u_{z} \hat{z}}{\sqrt{γ_{-}^{2} (u_{x}^{2} + u_{y}^{2}) + γ_{+}^{2} u_{z}^{2}}} .

By substituting this into Eq. (6), we obtain the optimum plane wave amplitudes of problem ℘ ₂(û):

\begin{array}{l} A (k_{x}, k_{y}) & = & {(\frac{8 π P_{0}}{n})}^{1 / 2} {(\frac{μ_{0}}{ɛ_{0}})}^{1 / 4} \frac{1}{k_{z}} \frac{1}{Γ (α_{max}, v_{z})} [\hat{v} - (\hat{v} \cdot \hat{k}) \hat{k}] \\ = & {(\frac{8 π P_{0}}{n})}^{1 / 2} {(\frac{μ_{0}}{ɛ_{0}})}^{1 / 4} \frac{1}{k_{z}} \frac{1}{\tilde{Γ} (α_{max}, u_{z})} \\ \times {γ_{-} [\hat{u} - (\hat{u} \cdot \hat{k}) \hat{k}] + (γ_{+} - γ_{-}) u_{z} [\hat{z} - (\hat{z} \cdot \hat{k}) \hat{k}]}, \end{array}

where

\begin{array}{l} \tilde{Γ} (α_{max}, u_{z}) & \equiv & Γ (α_{max}, v_{z}) \sqrt{γ_{-}^{2} (u_{x}^{2} + u_{y}^{2}) + γ_{+}^{2} u_{z}^{2}} \\ = & \sqrt{γ_{+} γ_{-}} {γ_{-} - (γ_{+} - γ_{-}) u_{z}^{2}}^{1 / 2}, \end{array}

Finally, the expression for the optimum electric field in an arbitrary point r becomes with Eq. (12):

E (ρ, φ, z) = \frac{\sqrt{2 π n P_{0}}}{λ_{0}} {(\frac{μ_{0}}{ɛ_{0}})}^{1 / 4} \frac{1}{\tilde{Γ} (α_{max}, u_{z})} \bar{\bar{ℳ}} (ρ, φ, z) (\begin{matrix} γ_{-} u_{x} \\ γ_{-} u_{y} \\ γ_{+} u_{z} \end{matrix}),

where ℳ̿ (ρ, φ, z) is still given by Eq. (13).

3.4. Electric field strength in focus

Let $E_{1}^{\hat{v}}$ be the solution of optimization problem ℘ ₁(v̂) for some unit vector v̂ and let û be the unit vector in the direction of $E_{1}^{\hat{v}} (0)$ as defined by Eq. (23). Without restricting the generality, we assume again that v_y = u_y = 0. At the left-side of Fig. 3 the projections of $E_{1}^{\hat{v}} (0)$ along the direction of optimization v̂ and along û are compared as function of ϑ_v for several values of the numerical aperture and for total power P ₀ = 1W. The right-side of Fig. 3 shows a comparison between the projections of the solutions of ℘ ₁(v̂) and ℘ ₂(v̂) along v̂, i.e. $E_{1}^{\hat{v}} (0) \cdot \hat{v}$ and $E_{2}^{\hat{v}} (0) \cdot \hat{v}$ (again as functions of ϑ_v for several values of the numerical aperture and for P ₀ = 1W). We see that for ϑ_v = 0^o (longitudinal polarized optimum field) and ϑ_v = 90^o (transverse polarized optimum field), the projections are identical. As was pointed out before, this is also true for all angles 0 ≤ ϑ_v ≤ π/2 when NA = n.

Fig. 3 Left: The electric field $E_{1}^{\hat{v}} (0)$ projected along v̂ and û as functions of ϑ_v for several values of the numerical aperture. : The projections of the optimum electric fields $E_{1}^{\hat{v}} (0)$ and $E_{2}^{\hat{v}} (0)$ along the direction v̂, as function of ϑ_v for several values of the numerical aperture.

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4. Realization of the optimum fields

In this section we discuss two methods for realizing the optimum fields. In the first method an appropriate pupil distribution is focused by a lens of given numerical aperture. The optimum field is the field in image space of the lens with the focal point as point were the electric field is maximum. In the second method, an electric dipole with appropriate dipole moment is imaged by a lens system with magnification 1. The field component is then maximum at the image point of the dipole. It should be noted that both realizations of the image field are approximative because they are based on a model of the focusing and imaging of a high NA lens which are valid under certain, be it rather mild, assumptions [28].

4.1. Pupil fields that when focused yield the optimum field distributions.

We consider a diffraction limited lens with (high) numerical aperture NA and a light beam which propagates parallel to the optical axis and is focused by this lens. Let {x̂,ŷ,ẑ} be a Cartesian basis in image space such that the optical axis is parallel to ẑ and the origin coincides with the focal point. Let {x̂ _p,ŷ _p} be the coordinate system in the entrance pupil of the lens, with x̂ _p and ŷ _p parallel to x̂ and ŷ, respectively. Finally, let ρ_p and φ_p be polar coordinates in the entrance pupil, i.e.

x_{p} = ρ_{p} cos φ_{p}, y_{p} = ρ_{p} sin φ_{p} .

We have

{\hat{ρ}}_{p} = cos φ_{p} {\hat{x}}_{p} + sin φ_{p} {\hat{y}}_{p},

{\hat{φ}}_{p} = - sin φ_{p} {\hat{x}}_{p} + cos φ_{p \cdot} {\hat{y}}_{p},

so that { $\hat{ρ}$ _p, $\hat{φ}$ _p,ẑ} is a positively orientated orthonormal basis. The electric field in a point (ρ_p, φ_p) of the entrance pupil of the lens is decomposed on this basis:

E^{p} (ρ_{p}, φ_{p}) = E_{ρ}^{p} (ρ_{p}, φ_{p}) {\hat{ρ}}_{p} + E_{φ}^{p} (ρ_{p}, φ_{p}) {\hat{φ}}_{p},

where the ẑ-component of the electric field is neglected because the incident beam is predominantly propagating parallel to the optical z-axis. This system is illustrated by Fig. 4.

Fig. 4 The focusing lens with numerical aperture NA.

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To determine the focused field we use the theory of Ignatowsky [1, 2] and Richards and Wolf [3,4]. This theory is based on (i) Debye’s approximation which expresses the plane wave amplitudes at the focal plane in terms of the tangential electric field in the entrance pupil, and (ii) the Abbe sine condition, which guarantees conservation of energy. Debye’s approximation is equivalent to a sharp cut-off of the plane wave spectrum as mentioned in [29] and is valid provided that [28, 30]:

f ≫ \frac{λ_{0}}{n {sin}^{2} (α_{max})},

where f is the focal distance and sinα _max = NA/n. In the model, the pupil coordinates x_p and y_p are related to the wave vector components by (see Fig. 4):

k_{x} = - k \frac{x_{p}}{f} = - k \frac{ρ_{p}}{f} cos φ_{p}, k_{y} = - k \frac{y_{p}}{f} = - k \frac{ρ_{p}}{f} sin φ_{p} .

We introduce unit vectors perpendicular to the wave vector

\hat{i_{S}} (\hat{k}) = \frac{\hat{k} \times \hat{z}}{| \hat{k} \times \hat{z} |} = \frac{1}{\sqrt{k_{x}^{2} + k_{y}^{2}}} (\begin{matrix} k_{y} \\ - k_{x} \\ 0 \end{matrix}),

\hat{i_{P}} (\hat{k}) = \frac{\hat{i_{S}} (\hat{k}) \times \hat{k}}{| \hat{i_{S}} (\hat{k}) \times \hat{k} |} = \frac{1}{k} \frac{1}{\sqrt{k_{x}^{2} + k_{y}^{2}}} (\begin{matrix} - k_{x} k_{z} \\ - k_{y} k_{z} \\ k_{x}^{2} + k_{y}^{2} \end{matrix}) .

Note that the orthonormal basis {

\hat{i_{P}} (\hat{k})

,

\hat{i_{S}} (\hat{k})

, k̂} is positively oriented. The vector amplitude A(k_x, k_y) of the plane wave with wave vector k can be written as:

A (k_{x}, k_{y}) = A_{P} (k_{x}, k_{y}) \hat{i_{P}} (\hat{k}) + A_{S} (k_{x}, k_{y}) \hat{i_{S}} (\hat{k}) .

The radial and polar components of the pupil field are proportional to A_P and A_S, respectively [29, 31]:

E_{ρ}^{p} (ρ_{p}, φ_{p}) = \frac{\sqrt{k k_{z}}}{2 π i f} A_{P} (- k \frac{ρ_{p}}{f} cos φ_{p}, - k \frac{ρ_{p}}{f} sin φ_{p}),

E_{φ}^{p} (ρ_{p}, φ_{p}) = \frac{\sqrt{k k_{z}}}{2 π i f} A_{S} (- k \frac{ρ_{p}}{f} cos φ_{p}, - k \frac{ρ_{p}}{f} sin φ_{p}),

where the factor

\sqrt{k k_{z}}

is included to account for energy conservation and

k_{z} = k \sqrt{1 - \frac{ρ_{p}^{2}}{f^{2}}} .

By substituting the plane wave amplitudes Eq. (6) of the solution of problem ℘ ₁(v̂), we obtain the entrance pupil field that after focusing by the lens yields this optimum field in the focal region:

\begin{array}{l} E^{p} (ρ_{p}, φ_{p}) & = & - i {(\frac{2 P_{0}}{π n})}^{1 / 2} {(\frac{μ_{0}}{ɛ_{0}})}^{1 / 4} \frac{1}{f Γ (α_{max}, v_{z})} \times \\ {[\frac{{(f^{2} - ρ_{p}^{2})}^{1 / 4}}{f^{1 / 2}} (cos φ_{p} v_{x} + sin φ_{p} v_{y}) + \frac{ρ_{p}}{f^{1 / 2} {(f^{2} - ρ_{p}^{2})}^{1 / 4}} v_{z}] {\hat{ρ}}_{p} \\ + \frac{f^{1 / 2}}{{(f^{2} - ρ_{p}^{2})}^{1 / 4}} (cos φ_{p} v_{y} - sin φ_{p} v_{x}) {\hat{φ}}_{p}} . \end{array}

Similarly, the pupil field that yields the solution of optimization problem ℘ ₂(û) is:

\begin{array}{l} E^{p} (ρ_{p}, φ_{p}) & = & - i {(\frac{2 P_{0}}{π n})}^{1 / 2} {(\frac{μ_{0}}{ɛ_{0}})}^{1 / 4} \frac{1}{f \tilde{Γ} (α_{max}, u_{z})} \\ \times {[\frac{{(f^{2} - ρ_{p}^{2})}^{1 / 4}}{f^{1 / 2}} γ_{-} (cos φ_{p} u_{x} + sin φ_{p} u_{y}) + \frac{ρ_{p}}{f^{1 / 2} {(f^{2} - ρ_{p}^{2})}^{1 / 4}} γ_{+} u_{z}] {\hat{ρ}}_{p} \\ + \frac{f^{1 / 2}}{{(f^{2} - ρ_{p}^{2})}^{1 / 4}} γ_{-} (cos φ_{p} u_{y} - sin φ_{p} u_{x}) {\hat{φ}}_{p}} . \end{array}

It follows from the formulas that the optimum pupil fields are linearly polarized in all points of the pupil and that in all points they are in phase. Furthermore, for the case that v_x = v_y = u_x = u_y = 0 (longitudinal case), the pupil field is radially polarized in all points of the pupil.

In Fig. 5 snapshots are shown of the optimum pupil distributions which after focusing give the solution of ℘ ₁(v̂) (left figure) and ℘ ₂(v̂) (right figure), for unit vector v̂ with v_y = 0 and ϑ_v = 45°. The numerical aperture is NA/n = 0.75 in both cases. The direction of the arrows is that of the linear polarization and their length is proportional to the electric field amplitude. The optimum pupil field at the right of Fig. 5 is similar to a radially polarized pupil field but with center shifted in the negative x_p-direction.

Fig. 5 Left: Snapshots of the electric field in the entrance pupil, E ^p, which after focusing yields the field with maximum component along v̂ obtained by solving ℘ ₁(v̂) (left) and the optimum pupil field that is solution of ℘ ₂(v̂) (right), where v̂ is such that v_y = 0 and ϑ_v = 45°. In both cases NA/n = 0.75.

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4.2. Imaging of an electric dipole

Consider an imaging system consisting of two lenses as shown in Fig. 6. We choose a Cartesian coordinate system {x̂,ŷ,ẑ} in object space which is parallel to the coordinate system in image space as defined in Subsection 4.1 but the first focal point is at the origin of this coordinate system. Although the coordinate systems in object and image systems are denoted by the same symbols, no confusion will arise. Let there be an electric dipole with dipole vector p = (p_x, p_y, p_z) in the focal point of the first lens. The electric field in object space radiated by this dipole is [32]:

E_{d} (r) = - \frac{k^{2}}{ɛ_{0}} p G (r) - \nabla (\nabla \cdot \frac{p}{ɛ_{0}} G (r)),

where G(r) is the free space Green’s function of the scalar Helmholtz equation:

G (r) = - \frac{e^{i k r}}{4 π r} .

By Fourier transformation with respect to (x,y), the plane wave expansion of E _d in a plane of constant z ≥ 0 is:

ℱ [E_{d}] (k_{x}^{o}, k_{y}^{o}, z) = - [k^{2} p - (k^{o} \cdot p) k^{o}] \frac{e^{i k_{z}^{o} z}}{2 i ɛ_{0} k_{z}^{o}},

where superscript ‘o’ indicates a wave vector in object space. Let

A^{o} (k_{x}^{o}, k_{y}^{o})

be the plane wave amplitudes in the source plane z = 0. Then

A^{o} (k_{x}^{o}, k_{y}^{o}) = ℱ [E_{d}] (k_{x}^{o}, k_{y}^{o}, 0)

= \frac{- 1}{2 i ɛ_{0} k_{z}^{o}} [k^{2} p - (k^{o} \cdot p) k^{o}] .

The field of the dipole at the focal point is collimated by the first lens and then focused by the second lens with focal length f ₂. The pupil fields are again assumed to be related to the plane wave amplitudes by Ignatowsky’s model. After writing the plane wave amplitudes in the object plane on the basis {k̂^o,

{\hat{i_{P}}}^{o}

,

{\hat{i_{S}}}^{o}

}, the electric field in the exit pupil of the left lens and the entrance pupil of the second lens are obtained by Eqs. (43) and (44). When the two lenses are identical, i.e. f ₁ = f ₂, the (paraxial) magnification is equal to 1 and the plane wave amplitudes

A^{i} (k_{x}^{i}, k_{y}^{i})

in the focal plane of the second lens become

A^{i} (k_{x}^{i}, k_{y}^{i}) \cdot {\hat{k}}^{i} = A^{o} (k_{x}^{o}, k_{y}^{o}) \cdot {\hat{k}}^{o} = 0,

A^{i} (k_{x}^{i}, k_{y}^{i}) \cdot {\hat{i_{P}}}^{i} = A^{o} (k_{x}^{o}, k_{y}^{o}) \cdot {\hat{i_{P}}}^{o},

A^{i} (k_{x}^{i}, k_{y}^{i}) \cdot {\hat{i_{S}}}^{i} = A^{o} (k_{x}^{o}, k_{y}^{o}) \cdot {\hat{i_{S}}}^{o},

where superscript ‘i’ refers to image space and the wave vector components in image and object spaces are related by:

k_{x}^{i} = - k_{x}^{o}

,

k_{y}^{i} = - k_{y}^{o}

and

k_{z}^{i} = k_{z}^{o}

. These relationships and Eqs. (40) and (41) imply that in image space:

A^{i} (k_{x}, k_{y}) = \frac{- 1}{2 i ɛ_{0} k_{z}} [k^{2} \tilde{p} - (\tilde{p} \cdot k) k],

where p̃ = (p_x, p_y, −p_z) and where we have omitted the image space superscript from the wave vector in image space. By comparing Eqs. (56) and (6), it follows that when the dipole vector has length:

| \tilde{p} | = {(\frac{32 π ɛ_{0} P_{0}}{n})}^{1 / 2} \frac{{(ɛ_{0} μ_{0})}^{1 / 4}}{Γ (α_{max}, v_{z})},

and the unit vector v̂ is defined by

\hat{v} = (v_{x}, v_{y}, v_{z}) = (p_{x}, p_{y}, - p_{z}) / | \tilde{p} | = \tilde{p} / | \tilde{p} |,

the plane wave amplitudes of the image of the dipole and that of the solution of problem ℘ ₁ (v̂) are identical (apart from a factor −i). We conclude that when an electric dipole p is imaged with unit magnification, the field in the image point has, for the given power, maximum field amplitude in the direction of the unit vector p̃ = (p_x, p_y, −p_z)/|p̃|. Because of the relationships between the solutions of optimization problems ℘ ₁ (v̂) and ℘ ₂ (û), the image of the electric dipole is also the field that has maximum amplitude and is linearly polarized in the direction of the unit vector û given by

\hat{u} = \frac{γ_{+} (p_{x} \hat{x} + p_{y} \hat{y}) - γ_{-} p_{z} \hat{z}}{\sqrt{γ_{+}^{2} (p_{x}^{2} + p_{y}^{2}) + γ_{-}^{2} p_{z}^{2}}},

which is illustrated in Fig. 7.

Fig. 6 The two lens system that is used to image the optimum electric field in image space when a dipole source is placed in object space. The focal length f ₁ denotes the radius of entrance pupil sphere 1, focal length f ₂ that of exit pupil sphere 2.

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Fig. 7 Schematics of the imaging of a dipole using the two lens system with (paraxial) magnification equal to 1. The dipole produces the electric field at the focal point which has maximum amplitude in the direction of v̂ and which has maximum amplitude and is linearly polarized in the direction of û.

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5. Conclusion

In a previous study focused fields were determined that have maximum electric field amplitude in a particular direction in the focal point. Although the optimum electric field is linear polarized in the focal point, the direction of polarization is in general not parallel to that for which the amplitude is optimum. We therefore studied the optimization problem in which it is required that the optimum field is linear polarized in the same direction as for which the amplitude is maximum. It was shown that when the direction is varied, the set of solutions of the two optimization problems are identical, i.e for every optimization direction in the unconstrained optimization problem there is a direction for the second optimization problem such that the solutions of the two problems are the same. The relationship between solutions of the two optimization problems for the same direction of optimization were derived. The optimum fields can be realized by focusing of appropriate pupil fields. Alternatively, the fields can be obtained by imaging an electric dipole with unit magnification. We demonstrated that the optimum field for a given direction is identical to the image of an electric dipole whose vector is the mirror image of the direction of optimization with the (x,y)-plane as mirror.

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On focused fields with maximum electric field components and images of electric dipoles

Abstract

1. Introduction

2. The optimization problem

2.1. Formulation of the optimization problem

2.2. Optimum plane wave amplitudes and field distributions