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Vol. 32, No. 11 / November 2015 / Journal of the Optical Society of America A

Research Article

Uniform refraction in negative refractive index materials CRISTIAN E. GUTIÉRREZ*

AND

ERIC STACHURA

Department of Mathematics, Temple University, 1805 N. Broad Street, Philadelphia, Pennsylvania 19122, USA *Corresponding author: [email protected] Received 21 July 2015; revised 18 September 2015; accepted 18 September 2015; posted 21 September 2015 (Doc. ID 246300); published 20 October 2015

We study the problem of constructing an optical surface separating two homogeneous, isotropic media, one of which has a negative refractive index. In doing so, we develop a vector form of Snell’s law, which is used to study surfaces possessing a certain uniform refraction property, in both the near- and far-field cases. In the near-field problem, unlike the case when both materials have positive refractive indices, we show that the resulting surfaces can be neither convex nor concave. © 2015 Optical Society of America OCIS codes: (160.3918) Metamaterials; (080.2720) Mathematical methods (general); (080.2740) Geometric optical design; (350.3618) Left-handed materials. http://dx.doi.org/10.1364/JOSAA.32.002110

1. INTRODUCTION Given a point O inside medium I and a point P inside medium II, the question we consider is whether we can find an interface surface Γ separating medium I and medium II that refracts all rays emanating from the point O into the point P. Suppose that medium I has an index of refraction n1 and medium II has an index of refraction n2 . In the case in which both n1 , n2 > 0, this question has been studied extensively (see, in particular, [1] and the references therein). This is the so-called near-field problem. The far-field problem is as follows: given a direction m fixed in the unit sphere in R3 , if rays of light emanate from the origin inside medium I, what is the surface Γ, interface between media I and II, that refracts all these rays into rays parallel to m? In the case in which both n1 , n2 > 0, this problem has also been studied extensively [2]. A natural question in this direction is what happens in the case when one of the media has a negative index of refraction? These media, as of now still unknown to exist naturally, constitute the so-called negative refractive index materials (NIMs) or left-handed materials. The theory behind these materials was developed by V. G. Veselago in the late 1960s [3], yet was put on hold for more than 30 years until the existence of such materials was shown by a group at the University of California at San Diego [4]. For a more detailed description, including results on near- and far-field imaging with NIMs, see [5, Chaps. 6 and 8]. Refraction laws for NIMs have been previously established, for instance, at the interface between an isotropic and anisotropic medium where negative refraction occurs [6]. This note is devoted to studying surfaces having the uniform refraction property (as discussed above) in the case when 1084-7529/15/112110-13$15/0$15.00 © 2015 Optical Society of America

κ :
n2 ∕n1 < 0, in both the far-field and near-field cases. We begin discussing in Subsection 2.A the notion of negative refraction, and in Subsection 2.B, we formulate Snell’s law in vector form for NIMs. Next, in Section 3 we find the surfaces having the uniform refraction property in the setting of NIMs. In contrast with standard materials, i.e., those having positive refractive indices, the surfaces for NIMs in the near-field case can be neither convex nor concave; this is analyzed in detail in Subsection 3.D. Finally, in Section 4, using the Snell law described in Subsection 4.B, we study the Fresnel formulas of geometric optics, which give the amount of radiation transmitted and reflected through a surface separating two homogeneous and isotropic media, one of them being a NIMs medium. We expect that the study in this paper will be useful in the design of surfaces refracting energy in a prescribed manner for NIMs as it is done for standard materials in [1,2,7].

2. GEOMETRIC OPTICS IN NEGATIVE REFRACTIVE INDEX MATERIALS A. Introduction to Negative Refractive Index Materials

The notion of negative refraction goes back to the work of Veselago in [3]. In the Maxwell system of electromagnetism, the material parameters ϵ, μ are characteristic quantities which determine the propagation of light in matter. If E
A cosr · k  ωt, then in the case of isotropic material, the dispersion relation [8, Formula (7.9)] is given by

Vol. 32, No. 11 / November 2015 / Journal of the Optical Society of America A

Research Article

jkj2



2 ω n2 ; c

with n2 being the square of the index of refraction of the material, which takes the form n
ϵμ: 2

(1)

Veselago showed that, in the case when both ϵ; μ < 0, we are forced by the Maxwell system to take the negative square root for the refractive index, i.e., pffiffiffiffiffi n
− ϵμ: (2) In particular, he observed that a slab of material with ϵ
−1

and

μ
−1

(3)

would have a refractive index n
−1 and behave like a lens. A remarkable property of this material was shown in [9]: that the focusing is perfect provided the condition (3) is exactly met. The theory behind NIMs opened the door to study the so-called metamaterials; see, e.g., [10]. We end this section by making some remarks on some important differences in geometric optics in the case of NIMs. One must generalize the classical Fermat principle of optics (see, e.g., [11, Subsection 3.3.2]) to handle materials with negative refractive indices. This was done not too long ago by Veselago in [12]. In particular, in the setting of NIMs, the optical length of a light ray propagating from one NIM to another is negative because the wave vector is opposite the direction of travel of the ray. However, in this case, it is not possible a priori to determine that the path of light is a maximum or minimum of the optical length, as this depends heavily on the geometry of the problem. Finally, the Fresnel formulas of geometric optics must also be appropriately adjusted for negative refraction; we analyze this in detail at the end. B. Snell’s Law in Vector Form for κ < 0

To explain the Snell law of refraction for media with κ < 0, we first review this for media with positive refractive indices. Suppose Γ is a surface in R3 that separates two media I and II that are homogeneous and isotropic, with refractive indices n1 and n2 , respectively. If a ray of light (assuming that light rays are monochromatic, since the refraction angle depends on the frequency of the radiation) having direction x ∈ S 2 , the unit sphere in R3 , and traveling through medium I strikes Γ at the point P, then this ray is refracted in the direction m ∈ S 2 through medium II according to the Snell law in vector form: n1 x × ν
n2 m × ν;

(4)

where ν is the unit normal to the surface to Γ at P pointing toward medium II; see [13, Subsection 4.1]. It is assumed here that x · ν ≥ 0. This has several consequences:

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Equation (4) is equivalent to n1 x − n2 m × ν
0, which means that the vector n1 x − n2 m is parallel to the normal vector ν. If we set κ
n2 ∕n1 , then x − κm
λν;

(5)

for some λ ∈ R. Notice that (5) univocally determines λ. Taking dot products with x and m in (5), we get λ
cos θ1 − κ cos θ2 , cosffi θ1
x · ν > 0, and cos θ2
m · ν
pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 1 − κ −2 1 − x · ν2 . In fact, there holds pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi (6) λ
x · ν − κ 1 − κ −2 1 − x · ν2 : It turns out that refraction behaves differently for 0 < κ < 1 and for κ > 1. 1. If 0 < κffi < 1, then for refraction to occur, we need pffiffiffiffiffiffiffiffiffiffiffi x · ν ≥ 1 − κ 2 and, in (6), we have λ > 0 and the refracted vector m is so that x · m ≥ κ. 2. If κ > 1, then refraction always occurs, and we have x · m ≥ 1∕κ, and in (6), we have λ < 0; see [2, Subsection 2.1]. We now consider the case when either medium I or medium II has a negative refractive index so that κ
nn21 < 0. (The calculation below holds, regardless of the sign of κ.) Let us take the formulation of the Snell law as in (5). This is again equivalent to x − κm
λν;

(7)

with x, m being unit vectors and ν being the unit normal to the interface pointing toward medium II. We will show that pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi (8) λ
x · ν  x · ν2 − 1 − κ 2 : In fact, taking dot products in (7), first with x, and then with m, yields 1 − κx · m
λx · ν;

and

x · m − κ
λm · ν:

This implies that 1κ − 1κ λx · ν
x · m
κ  λm · ν. We seek to get rid of the term m · ν. To do this, from (7), we have that 1 1 m
x−λν κ , so m · ν
κ x · ν − κ λ. Therefore, by substitution, we obtain the following quadratic equation in λ: λ2 − 2λx · ν  1 − κ 2 
0: Solving this equation yields pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi λ
x · ν  x · ν2 − 1 − κ 2 : If jκj ≥ 1, then x · ν2 − 1 − κ 2  ≥ 0 and the square root is real. On the other hand, pffiffiffiffiffiffiffiffiffiffiffi ffi if jκj < 1, then the square root is real only if x · ν ≥ 1 − κ 2 , which means that the incident direction x must satisfy this condition. It remains to check which sign  to take for λ. Recalling that pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi x · ν − κm · ν
λ
x · ν  x · ν2 − 1 − κ2 ;

n1 sin θ1
n2 sin θ2 ;

we see that, since κ < 0 and m · ν ≥ 0, we must take the positive square root. Hence, we conclude (8). Notice that, in contrast with the case when κ > 0, the value of λ given in (8) is always positive when κ < 0, and it can be also written in the following form: pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi λ
x · ν  jκj 1 − κ−2 1 − x · ν2 ;

where θ1 is the angle between x and ν (the angle of incidence) and θ2 is the angle between m and ν (the angle of refraction).

which is (6) when κ < 0. In the “reflection” case, κ
−1, since x · ν > 0, this formula yields

(a) the vectors x; m; ν are all on the same plane (called the plane of incidence); (b) the well-known Snell’s law in scalar form holds:

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Vol. 32, No. 11 / November 2015 / Journal of the Optical Society of America A

λ
2x · ν; and, hence, the “reflected” vector m is given by m
2x · νν − x: That is, after striking the interface Γ, the wave with direction m travels in the material with a refractive index n2 . Note that the vector formulation (7) is compatible with the Snell law for a negative refractive index [14]; see Fig. 1. Indeed, taking the cross product in (7) with the normal ν yields x × ν
κm × ν; then taking absolute values yields sin θ1
−κ sin θ2 ; where θ1 is the angle between x and ν and θ2 is the angle between m and ν. C. Physical Condition for Refraction When κ < 0

We analyze here conditions under which an electromagnetic wave is transmitted from medium I to II and there is no total internal reflection. 1. Case When −1 ≤ κ < 0

The maximum angle of refraction is θ2
π∕2 which is attained when sin θ1
−κ; that is, the critical angle is θc
arcsin−κ. If θ1 > θc , there is no refraction. Thus, θ2
arcsin− 1κ sin θ1  for 0 ≤ θ1 ≤ θc . The dot product x · m
cosθ1  θ2 . Let hθ1 
θ1  θ2
θ1  arcsin − 1κ sin θ1 . We have 1 1 h 0 θ1 
1 − cos θ1 pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ≥ 0; κ 1 − sin2 θ1 ∕κ 2 for 0 ≤ θ1 ≤ θc ; thus, h is increasing on 0; θc , and, therefore, θ1  θ2 ≤ θc  π∕2;

for 0 ≤ θ1 ≤ θc :

Then the physical constraint for refraction is x · m
cosθ1  θ2  ≥ cosθc  π∕2
− sin θc
κ: (9) 2. Case When κ < −1

In this case, 0 ≤ θ1 ≤ π∕2, and since θ2
arcsin− 1κ sin θ1 , the maximum angle of refraction is when θ1
π∕2; that is, the maximum angle of refraction is θr
arcsin−1∕κ. Now hθ1  is increasing on 0; π∕2, so θ1  θ2 ≤ π∕2  arcsin−1∕κ, and 1 x · m
cosθ1  θ2  ≥ cosπ∕2  arcsin−1∕κ
: κ (10) 3. SURFACES WITH UNIFORM REFRACTION PROPERTY

Research Article

given by rt
ρxtxt for xt ∈ S 2. According to (7), the tangent vector r 0 t to Γ satisfies r 0 t · xt − κm
0. That is, ρxt 0 xt  ρxtx 0 t · xt − κm
0, which yields ρxt1 − κm · xt
0. Therefore, b ; (11) 1 − κm · x for x ∈ S 2 and for some b ∈ R. To see the surface described by (11), we assume first that −1 < κ < 0. Since m · x ≥ −1, we have 1 − κm · x ≥ 1  κ > 0; thus, b > 0. Suppose, for simplicity, that m
e 3 , the third-coordinate vector. If y
y 0 ; y3  ∈ R3 is a point on Γ, then y
ρxx with x
y∕jyj. From (11), jyj − κy3
b; that is, jy 0 j2  y 23
κy 3  b2 , which yields jy 0 j2  1 − κ2 y 23 − 2κby 3
b2 . This equation can be written in the form   κb 2 y3 − 1−κ 2 jy 0 j2     2
1; (12) b ffi 2 b pffiffiffiffiffiffi 2 2 ρx


1−κ

1−κ

which is an ellipsoid of revolution about the y 3 axis with upper focus (0,0) and lower focus 0; 2κb∕1 − κ2 . Since jyj
κy3  b and the physical constraint for refraction (9), jyjy · e 3 ≥ κ κb is equivalent to y3 ≥ 1−κ 2 . That is, for refraction to occur, y must be in the upper half of the ellipsoid (12); we denote this semiellipsoid by Ee 3 ; b. To verify that Ee 3 ; b has the uniform refracting property, that is, that it refracts any ray emanating that (7) holds from the origin in the direction e 3 , we check  at each point. Indeed, if y ∈ Ee 3 ; b, then jyjy − κe 3 · jyjy
1 −   κe 3 · jyjy ≥ 1  κ > 0 and jyjy − κe 3 · e 3 ≥ 0; thus, jyjy − κe 3 is an outward normal to Ee 3 ; b at y. Rotating the coordinates, it is easy to see that the surface given by (11) with −1 < κ < 0 is an ellipsoid of revolution about the axis of direction m with upper focus (0,0) and lower 2κb focus 1−κ 2 m. Moreover, the semi-ellipsoid Em; b given by

b 2 ;x ∈ S ;x · m ≥ κ Em; b
ρxx : ρx
1 − κm · x (13) has the uniform refracting property: any ray emanating from the origin O is refracted in the direction m, where −1 < κ < 0. See Figs. 2 and 3 comparing refraction when κ < 0 versus κ > 0, where P
0; 2κb∕1 − κ 2  and m
e 3 .

x

θ1

n1

A. Far-Field Problem

Let m ∈ S 2 be fixed. We ask the following question: if rays of light emanate from the origin inside medium I, what is the surface Γ, interface between media I and II, that refracts all these rays into rays parallel to m? Here I has a refractive index n1 and II has a refractive index n2 such that κ
n2 ∕n1 < 0. This question can be answered using the vector form of Snell’s law. Suppose Γ is parameterized by the polar representation ρxx, where ρ > 0 and x ∈ S 2 . Consider a curve on Γ

m

θ2

θ2

n2

V Fig. 1. Snell’s law in the case of normal refraction and negative refraction (dotted arrow).

Vol. 32, No. 11 / November 2015 / Journal of the Optical Society of America A

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We claim that H m; b is the sheet with an opening in direction m of a hyperboloid of revolution of two sheets about the axis of direction m. To prove the claim, set for simplicity m
e 3 . If y
y 0 ; y 3  ∈ H e 3 ; b, then y
ρxx with jy 0 j2  x
y∕jyj. From (14), −κy3  jyj
b, and, therefore,  κb 2 2 2 0 2 2 y 3κb
2κy 3 2 b , which yields jy j − κ − 1 y 3  κ2 −1 − 
b . Thus, any point y on H e 3 ; b satisfies the κ 2 −1 equation   y3  κ2κb−1 2 jy 0 j2 
1;  b 2 −  (15) b ffi 2 pffiffiffiffiffiffi κ2 −1 2

n2

O

κ −1

n1

P

Fig. 2. Ellipsoid refracting when −1 < κ < 0.

Let us now assume that κ
−1. From (11), we have jyj  y 3
b; thus, jy 0 j2  y 23
b2 − 2by 3  y23 , and, therefore, 1 y 3
b2 − jy 0 j2 : 2b This is a paraboloid with axis e 3 and focus at O. Since κ
−1, from the physical constraint for refraction (9), we get that any ray with direction x ∈ S 2 emanating from the origin is refracted by the paraboloid. Now we turn to the case κ < −1. Due to the physical constraint of refraction (10), we must have b > 0 in (11); we determine the type of surface. Define for b > 0

b ; x ∈ S 2 ; x · m ≥ 1∕κ : H m; b
ρxx : ρx
1 − κm · x (14)

n2

which represents a hyperboloid of revolution of two sheets about the y 3 axis with foci (0,0) and 0; −2κb∕κ2 − 1. The sheets of this hyperboloid of revolution are given by sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi κb b jy 0 j2  2 y3
− 2 1   pffiffiffiffiffiffiffiffiffiffiffiffi2 : κ −1 κ −1 b∕ κ2 − 1 We decide which one satisfies the refracting property. The sheet with the minus sign in front of the square root satisfies κy3  b > 0 and, hence, has polar equation ρx
1−κeb 3 ·x ; b ≥ − κ−1 therefore, this is the sheet to consider. For a general m, by a rotation, we obtain that H m; b is the sheet with an opening in the direction opposite m of a hyperboloid of revolution of two sheets about the axis of direction m with foci (0,0) m. and −2κb κ 2 −1 Notice that the focus (0,0) is inside the region enclosed by m is outside that region. The vector H m; b and the focus −2κb κ 2 −1 y jyj − κm is an outer normal to H m; b at y, since by (14)
 
y −2κb −2b 2κ2 b m−y ≥ 2  2  κm · y − jyj − κm · 2 jyj κ −1 κ −1 κ −1


−2b 2κ 2 b  − b
b > 0: κ2 − 1 κ2 − 1

Clearly, jyjy − κm · m ≥ 1κ − κ > 0 and jyjy − κm · jyjy > 0. Therefore, H m; b satisfies the uniform refraction property. We summarize the uniform refraction property below. Theorem 3.1. Let n1 ; n2 be two indices of refraction for media I and II, respectively, and set κ
n2 ∕n1 . Assume the origin is in medium I, and Em; b, H m; b defined by (11) and (14), respectively. Then: 1. If −1 < κ < 0 and Em; b is the interface between medium I and medium II, then Em; b refracts all rays from the origin O into rays in medium II with direction m. 2. If κ < −1 and H m; b is the interface between medium I and medium II, then H m; b refracts all rays from the origin O into rays in medium II with direction m.

P

B. Near-Field Problem

n1

O

Fig. 3. Ellipsoid refracting when 0 < κ < 1.

Given a point O inside medium I and a point P inside medium II, we construct a surface Γ separating medium I and medium II such that Γ refracts all rays emanating from O into the point P. Suppose O is the origin and let X t be a curve on Γ. Recall that Snell’s law says that x − κm
λν;

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Vol. 32, No. 11 / November 2015 / Journal of the Optical Society of America A

Research Article b ≥ κjPj:

4

3 2 2 0

b  κjPj : 1  κjPj

λ


1 –2

Now suppose κjPj ≤ b ≤ jPj, and let

0

λ


–4 –1 –4

–2

0

2

4

Fig. 4. Ovals with κ
−.5, P
2; 0, and different values of −1 ≤ b ≤ 4.

where ν is the normal to the surface Γ, and we are considering the case when κ < 0. Then, via Snell’s law, we see that the tangent vector x 0 t must satisfy 
X t P − X t −κ
0: x 0 t · jX tj jP − X tj That is to say, jX tj 0  κjP − X tj 0
0: Hence, Γ is the surface jX j  κjX − Pj
b:

(16)

Compare this surface to the Cartesian ovals which occur in the near-field case when κ > 0. In this case, since the function F X 
jX j  κjX − Pj is convex and the ovals are the level sets of F , then the ovals are convex; see [1, Section 4]. However, in the case when κ < 0, the function F X  is no longer convex, and its level sets are convex sets only for a range of values in the parameter; see Fig. 4. That is, the surface (16) is, in general, not a convex surface; see Theorem 3.4. Case −1 < κ < 0. We will show that the surface (16) is nonempty if and only if b is bounded from below. In fact, this follows by the triangle inequality; since 1  κ > 0, we see that jX j  κjX − Pj ≥ κjPj for all X . Therefore, if the surface (16) is nonempty, then n2

(17)

Vice versa, if b satisfies (17), then the surface (16) is nonempty. Let X 0 be of the form X 0
λP. We will find λ > 0 so that X 0 is on the surface defined by (16). This happens if λjPj  κjλ − 1jjPj
b. Suppose b > jPj. Then we must have λ > 1. Thus, b
λjPj  κλ − 1jPj, and solving for λ gives

4

b − κjPj : 1 − κjPj

We then have 0 < λ ≤ 1, and X 0
λP is on the surface (16). (Notice that the set fX :jX j  κjX − Pj ≤ κjPj  δg, δ > 0, is contained in the ball with center zero and radius δ∕1  κ.) Case κ
−1. In this case, to have a nonempty surface, we show that b must be bounded above and below. If the surface is nonempty, then by application of the triangle inequality we get that −jPj ≤ b ≤ jPj:

(18)

Vice versa, if (18) holds, then the surface is nonempty. In fact, the point X 0
λP with λ
bjPj 2jPj belongs to the surface. Case κ < −1. We will show that the surface (16) is nonempty if and only if b is bounded above. First, notice that, by the triangle inequality and the fact that κ < −1, we have that jX j  κjX − Pj ≤ jPj for all X , and, therefore, if (16) is nonempty, then b ≤ jPj:

(19)

On the other hand, if (19) holds, then we will find X 0
λP ∈ Γ for some λ > 0. We want λjPj  κjPjjλ − 1j
b. If we let bκjPj , then λ ≥ 1 and the point λP is on the curve. λ
jPjκjPj (Notice that the set E
fX :jX j  κjX − Pj ≥ jPj − δg, δ > 0, is contained in the ball with center 0 and radius −δ∕1  κ.) C. Polar Equation of (16)

Next we will find the polar equation of the refracting surface (16) when κ < 0. We show that only a piece of this surface does the refracting job; otherwise, total internal reflection occurs. This follows from the physical conditions described in Subsection 3.C.1. Let X
ρxx with x ∈ S 2 . Then we write κjρxx − Pj
b − ρx:

P

Squaring both sides yields the following quadratic equation: 1 − κ2 ρx2  2κ 2 x · P − bρx  b2 − κ 2 jPj2
0: (20) Solving for ρ, we obtain pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi b − κ2 x · P  b − κ2 x · P2 − 1 − κ2 b2 − κ2 jPj2  ρx
: 1 − κ2 n1

Define an auxiliary quantity Δt :
b − κ 2 t2 − 1 − κ2 b2 − κ2 jPj2 ;

O

Fig. 5. Oval with κ
−.7, P
0; 2.5.

so that

Vol. 32, No. 11 / November 2015 / Journal of the Optical Society of America A

Research Article

pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi b − κ 2 x · P  Δx · P ρ x
: 1 − κ2

provided that κjPj < b. Furthermore, from the physical P−ρ xx constraint for refraction (9), in this case, with m
jP−ρ ,  xxj so x · m ≥ κ, and using Eq. (16), we must have

1. Case −1 < κ < 0

x · P ≥ b:

We begin with a proposition. Proposition 3.2 Assume −1 < κ < 0. If x ∈ S 2 ; then, Δx · P ≥ κ 2 x · P − b2 ;

(21)

with equality if and only if jx · Pj
jPj. Hence, the quantity pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi Δx · P is well defined. Proof. First notice that jx · Pj ≤ jPj comes for free since jxj
1. We have, since 0 < κ2 < 1 and x · P ≤ jPj, Δx · P − κ2 x · P − b2
−2bκ2 x · P  κ 4 x · P2  κ 2 b2 − κ4 jPj2  κ 2 jPj2 − κ2 x · P2  2κ 2 bx · P − κ2 b2 2

− κ2 x · P2

4

2

Ending with the case when −1 < κ < 0, given a point P ∈ R3 and κjPj ≤ b, a refracting oval is the set OP; b
fhx; P; bx : x ∈ S 2 ; x · P ≥ bg;

(24)

where hx; P; b

pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi b − κ 2 x · P2 − 1 − κ 2 b2 − κ 2 jPj2  :
1 − κ2 Figure 5 illustrates an example of such a refracting oval. b − κ2 x · P 

2. Case κ < −1


k x · P − κ jPj  κ jPj 4

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2

2


−κ4 jPj2 − x · P2 

 κ2 jPj2 − x · P2 
κ 2 1 − κ2 jPj2 − x · P2  ≥ 0; (22) as desired. Now, notice that in this case, to make sense of the physical problem, we must have P lying outside the oval, that is, outside the region jX j  κjX − Pj ≤ b. In addition to the requirement that the surface be nonempty, we see that the condition on b so that the ovals are meaningful is κjPj < b < jPj:

In this case, we must have O not on the oval; that is, O must lie outside the set jX j  κjX − Pj ≥ b. Hence, in combination with the requirement that the surface be nonempty, we must have κjPj < b < jPj: Notice that this is the same condition on b as in the case −1 < κ < 0. Multiplying (20) by −1 and solving for ρ yields pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi κ 2 x · P − b  b − κ2 x · P2 − κ 2 − 1κ 2 jPj2 − b2  : ρ x
κ2 − 1 To have the square root defined, we need Δx · P
b − κ 2 x · P2 − κ 2 − 1κ2 jPj2 − b2  ≥ 0:

Recalling that κjρxx − Pj
b − ρx; we see that, since κ < 0, we must have ρx > b. We now choose which sign () to take in the definition of ρ so that ρ > b. In fact, by Proposition 3.2, pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi b − κ2 x · P  Δx · P ρ x
1 − κ2 b − κ 2 x · P  jκjjb − x · Pj ≥ ≥ b; 1 − κ2 where the last inequality holds since jκj < 1. Therefore, the equality ρ x
b holds only if jx · Pj
jPj and b
x · P. In addition, b − κ 2 x · P  jκjjb − x · Pj 1
2 1−κ 1 − κ2 2 b − κ x · P  jκjb − x · P for b > x · P × b − κ2 x · P  jκjx · P − b for b < x · P

> b:

Hence, ρ x > b for all x and b such that b ≠ x · P. From (17), the oval is not degenerate only for κjPj < b. Reversing the above inequalities, one can show that ρ− x ≤ b: Thus, the polar equation of the surface Γ is given by pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi b − κ 2 x · P  Δx · P hx; P; b
ρ x
; 1 − κ2

(23)

From (19), to have a nonempty oval, we must have b ≤ jPj. To find the directions x for which ρ is well defined, we need to find the values of t for which Δt
b − κ 2 t2 − κ 2 − 1κ2 jPj2 − b2  ≥ 0: Let z
b − κ 2 t; then we want t such that z 2 ≥ κ 2 − 1κ2 jPj2 − b2 : If κ 2 jPj2 − b2 ≤ 0, then this is true for any t. On the other hand, κ2 jPj2 − b2 ≥ 0 if and only if jbj ≤ −κjPj. Thus, for b with jbj ≤ −κjPj, we have to choose t such that z
b − κ2 t satisfies (Notice that the requirement jbj ≤ −κjPj is satisfied, since we are assuming that b ≤ jPj so, in particular, jbj ≤ jPj ≤ −κjPj.) pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi jzj ≥ κ2 − 1κ 2 jPj2 − b2 : i.e., t ≤ b∕κ2 , then b − κ2 t ≥ If b − κ 2 t ≥ 0, pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2 2 2 2 κ − 1κ jPj − b ; that is, pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi b − κ2 − 1κ 2 jPj2 − b2  : t≤ κ2 i.e., t ≥ b∕κ2 , then κ2 t − b ≥ If b − κ 2 t ≤ 0, pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2 2 2 2 κ − 1κ jPj − b ; that is, pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi b  κ 2 − 1κ2 jPj2 − b2  : (25) t≥ κ2 From (10), to have refraction in this case, we need to have

Vol. 32, No. 11 / November 2015 / Journal of the Optical Society of America A

2116

x·

κ 2 x · P − b ≥ κ2 − 1ρ x: That is, we need ρ x ≤

κ2 x · P − b : κ2 − 1

Note that, since κ 2 > 1, we see that ρ x < 0 for κ 2 x · P − b < 0. Thus, we must have κ 2 x · P ≥ b if we are to have ρ x ≥ 0. From (25), we must have pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi b  κ 2 − 1κ2 jPj2 − b2  ; x·P ≥ κ2 so that Δx · P ≥ 0. We are left with choosing either ρ x or ρ− x now. However, by the physical restraint for refraction, we see that only ρ− x will do the job since pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi κ2 x · P − b − Δx · P κ2 x · P − b ρ− x
≤ ; κ2 − 1 κ2 − 1 provided that x·P ≥



P − xρ x 1 ≥ : jP − xρ xj κ

Note that this is equivalent to requiring that

b

pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi κ 2 − 1κ 2 jPj2 − b2  : κ2

Hence, for κ < −1, refraction only occurs when b ≤ jPj and the refracting piece of the oval is given by pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

b  κ 2 − 1κ2 jPj2 − b2  ; OP; b
hx; P; bx : x · P ≥ κ2 (26) with

EP; b





κ 2 x · P − b −

pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi κ2 x · P − b2 − κ 2 − 1κ2 jPj2 − b2  : κ2 − 1

A picture of the refracting piece of this oval can be obtained from Fig. 5 by reversing the roles of P and O, and changing the direction of the rays; see also the explanation at the end of the proof of Theorem 3.4.

ρx


b2 − jPj2 : 2b − x · P

To have a nonempty surface, b must satisfy (18). Since ρ ≥ 0 and the numerator in ρ is nonpositive, we must have b < x · P. The last inequality follows from the physical condition for P−ρxx . Notice that refraction (9) with κ2
2−1 and m
jP−ρxxj b −jPj ρx ≥ b; that is, 2b−x·P ≥ b, since this is equivalent to b − x · P2  jPj2 − x · P2 ≥ 0. Therefore, when κ
−1, the surface refracting the origin O into P is given by

(27)

1. If −1 < κ < 0 and Γ :
OP; b is given by (24), then Γ refracts all rays emitted from O into P. 2. If κ < −1 and Γ :
OP; b is given by (26), then Γ refracts all rays emitted from O into P. 3. If κ
−1 and Γ :
EP; b is given by (27), then Γ refracts all rays emitted from O into P. D. Analysis of the Convexity of the Surfaces

We assume κ < 0 and analyze the convexity of the surface: jX j  κjX − Pj
b: We will show the surface is convex for a range of b ’s, and neither convex nor concave for the remaining range of b ’s. (To be precise, by convexity we mean that the set that is enclosed by F X 
jX j  κjX − Pj
b is a convex set.) We begin by letting X
x 1 ; …; x n  and P
p1 ; …; pn , and analyzing the function F X 
jX j  κjX − Pj: We have the gradient DF X 


X X −P κ ; jX j jX − Pj

and x i − pi x j − pj  1 xi xj 1 − −κ  κδij ; 3 jX j jX j jX − Pj jX − Pj3

where δij is the Kronecker delta. Then, n X F x i x j X ξi ξj QX ; ξ
i;j
1




n jξj2 1 X jξj2 − x x ξ ξ  κ i j i j jX j jX j3 i;j
1 jX − Pj

−κ

3. Case κ
−1

In this case, we have jρxx − Pj
ρx − b, so squaring and solving for ρ yields

b2 − jPj2 x : x ∈ S 2; x · P > b ; 2b − x · P

with jbj ≤ jPj. The uniform refraction property is summarized below. Theorem 3.3. Let n1 ; n2 be two indices of refraction of two media, I and II, respectively, such that κ :
n2 ∕n1 < 0. Assume that O is a point inside medium I, P is a point inside medium II, and b ∈ κjPj; jPj. Then,

F x i x j
δij

hx;P;b
ρ− x

Research Article

n X 1 x − p x − p ξ ξ jX − Pj3 i;j
1 i i j j i j

jξj2 1 jξj2 2 − X · ξ  κ jX j jX j3 jX − Pj 1 X − P · ξ2 −κ jX − Pj3 1
κ jξj2 jX − Pj2 − X − P · ξ2  jX − Pj3 1  jξj2 jX j2 − X · ξ2 : jX j3 By Cauchy–Schwartz’s inequality jv · wj ≤ jvjjwj with equality if and only if v is a multiple of w. If we set ξ
X in 1 2 the quadratic form, we get QX ; X 
κ jX −Pj 3 jX j j 2 2 X − Pj − X − P · X  . Since P ≠ 0, X is not a multiple


Research Article

Vol. 32, No. 11 / November 2015 / Journal of the Optical Society of America A

of X − P if and only if X is not a multiple of P and, in this case, we get QX ; X  < 0 since κ < 0. On the other hand, if we set ξ
X − P, then QX ; X − P
jX1j3 jX − Pj2 jX j2 − X · X − P2  and, again, if X is not a multiple of P, we obtain QX ; X − P > 0. Therefore, the quadratic form QX ; ξ is indefinite, and, therefore, the function F is neither concave nor convex at each X which is not a multiple of P. Now let X have the form X
λP, 0 < λ < ∞, and see for what values of λ there holds that QλP; ξ ≥ 0 for all ξ. In other words, we study which values of λ make the Hessian F x i x j λP ≥ 0. In particular, we wish to find the range of b that gives a convex surface. The range we obtain will be proven rigorously in Theorem 3.4. First, we have 1 1 − λ2 jξj2 jPj2 − 1 − λ2 P · ξ2  QλP; ξ
κ j1 − λj3 jPj3 
1  3 3 λ2 jξj2 jPj2 − λ2 P · ξ2  λ jPj 
1 1 1 κ jξj2 jPj2 − P · ξ2  ≥ 0;
λ j1 − λj jPj3 if and only if 1 1 ϕλ :
 κ ≥ 0: λ j1 − λj Let us assume −1 < κ < 0. Then ϕλ > 0 on 0; 1∕1 − κ, ϕλ < 0 on 1∕1 − κ; 1, ϕλ < 0 on 1; 1∕1  κ, and ϕλ > 0 on 1∕1  κ; ∞. Therefore, > 0 for λ ∈ 0; 1∕1 − κ∪1∕1  κ; ∞ QλP; ξ jPj and, by the physical restriction on b, we must have κjPj < b < jPj. Thus, we see that jPj < 1 − κjPj ≤ b < jPj; which is impossible. bκjPj , we see Now let λ ∈ 1; 1∕1  κ. Then since λ
1κjPj that 1
1 since −1 < κ < 0. If λP is on the surface, which means λjPj  κλ − 1jPj
b, we have

− PjjX j2 jX j2 − X · P 1
κ 2 jX − PjjX j2  jX − Pj2 − jPj2 2  κ2 jX 4 − PjjX j2 jX j2  jX − Pj2 − jPj2 ;

2118

Vol. 32, No. 11 / November 2015 / Journal of the Optical Society of America A

and C
κjX jjX j2 − X · P2jX − Pj2  jX j2 − X · P
2κjX jjX − Pj2 jX j2 − X · P  κjX jjX j2 − X · P2
κjX jjX − Pj2 jX j2  jX − Pj2 − jPj2  1  κjX jjX j2  jX − Pj2 − jPj2 2 : 4 The numerator of QX ; −F y ; F x  then equals jX j2 jX − Pj2 κ 3 jX j  jX − Pj 1  κjX − Pj2  jX j2 − jPj2 2 jX j  κjX − Pj  κjX jjX 4 − PjjX − Pj2  jX j2 − jPj2 κjX j  jX − Pj :
H jX j; jX − Pj; κ; jPj; and we want to optimize this quantity when X is on the curve F X 
jX j  κjX − Pj
b. If X is on the curve, there holds j jX − Pj
b−jX κ ≥ 0. Hence, jX j ≥ b and jjPj − jX jj ≤ b−jX j jX − Pj
κ . Hence, b − jX j b − jX j ≤ jPj − jX j ≤ : κ κ Multiplying by κ < 0 yields −

−b  jX j ≥ κjPj − κjX j ≥ b − jX j; so 1 − κjX j ≥ b − κjPj; and 1  κjX j ≥ b  κjPj: Then the last two inequalities, together with jX j ≥ b, imply that X must satisfy

b − κjPj b  κjPj ; ;b : jX j ≥ max 1−κ 1κ j In addition, from jX − Pj
b−jX we get that κ , b−jX j jX j  jPj ≥ jX − Pj
κ , which implies jX j ≤ b−κjPj 1κ since −1 < κ < 0. (This means that the ball centered at zero with radius b−κjPj 1κ contains the oval. In fact, this is the smallest ball centered at zero that contains the oval because the point X
b−κjPj satisfies the equation F X 
b [notice λP with λ
− 1κjPj −b−jPj that λ ≤ 0 and λ − 1
1κjPj < 0 since −b ≤ −κjPj < jPj

j since −1 < κ < 0].) Substituting jX − Pj
b−jX κ in the expression for H above, we obtain the following expression in terms of κ; b; jX j and jPj:

bb2 − κ2 jPj2 2  6bκ 2 − 1b − κjPjb  κjPjjX j2 4κ3 2 2 2 4κ − 1b κ − 2  κ 2 jPj2 jX j3 − 3bκ 2 − 12 jX j4  4κ3 :
GjX j; b; κ; with jX j ≥ b. Thus, the problem of finding the b’s for which QX ; −F y ; F x  ≥ 0 for all X reduces to minimizing a polynomial of one variable jX j over the range

Research Article

bκjPj b−κjPj maxfb−κjPj 1−κ ; 1κ ; bg ≤ jX j ≤ 1κ . Notice that the value of the maximum is given by M
b−κjPj 1−κ . (Here we use that b ≤ jPj, because the point P must be outside the oval to have a problem physically meaningful. This means that the radius of the largest ball centered at zero and contained in the oval has radius b−κjPj 1−κ .) The derivative of G with respect to the first variable t
jX j equals

∂t Gt; b; κ


3−1  κ2 b − ttb2 − κ 2 jPj2  b−1  κ 2 t : k3

Therefore, if b ≠ 0, then ∂t Gt; b; κ
0 when t
0; b, or 2 κ 2 jPj2 t b :
−bb−1κ 2  . If b
0, then ∂t Gt; 0; κ
0 when t
0.

Furthermore, the coefficient of t 4 in the original function 3b 2 2 Gt; b; k equals Lb :
− 4κ 3 1 − κ  . To optimize Gt; b; κ, we proceed by analyzing the following cases: Case κjPj ≤ b < 0. Since κ < 0, then the coefficient Lb < 0. We also have t b > 0. We write κ 2 − 12 tt − bt − t b : ∂t Gt; b; κ
−3b κ3 For t b < t < ∞, we have ∂t Gt; b; κ < 0 and, for 0 < t < t b, we have ∂t Gt; b; κ > 0. We optimize the function G on the interval 
b − κjPj b − κjPj ; ⊂ 0; t b : I
1−κ 1κ We have that ∂t Gt; b; κ > 0 for t ∈ I and, thus, G is strictly increasing on I. Hence, we obtain 
b − κjPj max Gt; b; κ
G ; b; κ t∈I 1κ


b  jPj2 b − κjPj2 b  jPj − κjPj ; 1  k2

and


b − κjPj ; b; κ min Gt; b; κ
G t∈I 1−κ

b − jPj2 b − κjPj2 −b  1  κjPj : −1  k2 Note that this minimum is positive since −1 < κ < 0 and the maximum is positive as well since b ≥ κjPj. Hence, the maximum and minimum have the same sign and, as such, the curve is convex for this range of b. Case 0 < b ≤ 1  κjPj. By calculation, we have that Lb > 0. Let us analyze the sign of t b . Since b > 0 and −1 < κ < 0, we have t b > 0 if and only if b > −κjPj, and t b < 0 if and only if b < −κjPj. Let us compare −κjPj with 1  κjPj. We have −κjPj ≤ 1  κjPj if and only if κ ≤ −1∕2, and −κjPj > 1  κjPj if and only if κ > −1∕2. Therefore, if κ > −1∕2, then t b < 0. On the other hand, if κ ≤ −1∕2, then b ∈ 0; −κjPj∪−κjPj; 1  κjPj. If b ∈ 0; −κjPj, then t b < 0. If b ∈ −κjPj; 1  κjPj, then t b > 0. Notice also that t b < b since b < 1  κjPj. If κ > −1∕2, then t b < 0 and ∂t G < 0 on 0; b, while ∂t G > 0 on b; ∞. Since b−κjPj 1−κ > b, we have that the interval I ⊂ b; ∞ and


Vol. 32, No. 11 / November 2015 / Journal of the Optical Society of America A

Research Article
min Gt; b; κ
G t∈I

b − κjPj ; b; κ 1−κ



b − jPj2 b − κjPj2 −b  1  κjPj :
−1  k2 If κ ≤ −1∕2, then 0 < t b < b; thus, ∂t G > 0 on 0; t b , ∂t G < 0 on t b ; b, and ∂t G > 0 on b; ∞. Once again, I ⊂ b; ∞, and we obtain 
b − κjPj ; b; κ min Gt; b; κ
G t∈I 1−κ

4. FRESNEL FORMULAS FOR NIMS To obtain the Fresnel formulas for NIMs materials, we briefly review the calculations leading to the Fresnel formulas for standard materials; see [11, Subsection 1.5.2]. The electric field is denoted by E, and the magnetic field is denoted by H. These are three-dimensional vector fields, E
Er; t and H
Hr; t, where r represents a point in threedimensional space r
x; y; z. The way in which E and H interact is described by Maxwell’s equations [11] and [16, Subsection 4.8]:

b − jPj2 b − κjPj2 −b  1  κjPj :
−1  k2

∇×E
−

Note that this minimum is positive since b < 0. Furthermore, we have 
b − κjPj ; b; κ ; max Gt; b; κ
G t∈I 1κ which is positive because b ≥ κjPj. Hence, the maximum and minimum have the same sign and, for this range of b ’s, the curve is also convex. Case 1  κjPj < b < jPj. b−κjPj In this case, we have that t b < b < b−κjPj 1−κ ; 1κ so that the b−κjPj b−κjPj interval I
 1−κ ; 1κ  ⊂ b; ∞. Furthermore, we have that t b > 0 provided that −1 < κ < −1∕2 and −κjPj < b < jPj or, if −1∕2 < κ < 0 and 1  κjPj < b < jPj. Moreover, we have that t b < 0 if −1 < κ < −1∕2 and 1  κjPj < b < −κjPj. In addition, on the interval I, we have that ∂t Gt; b; κ > 0 so that 
b − κjPj ; bκ min Gt; b; κ
G t∈I 1−κ


b − jPj2 b − κjPj2 −b  1  κjPj ; −1  κ2

which is negative since b > 1  κjPj. Furthermore, we have that 
b − κjPj ; b; κ max Gt; b; κ
G t∈I 1κ


b  jPj2 b − κjPj2 b  1 − κjPj ; 1  κ2

which is positive since −1 < κ < 0. Thus, the minimum and maximum of G on the interval have opposite signs, which implies that the curve is neither convex nor concave for this range of b. In dimension three, the above result still holds true because the oval is radially symmetric with respect to the axis OP. Finally, suppose that κ < −1. We have X ∈ fX :jX j  κjX − Pj
bg :
O if and only if X − P ∈ fZ :jZ j  1κ jZ  Pj
bκg :
O 0 . The ovals O and O 0 have the same curvature and, by the −1 < κ < 0 case, O 0 is convex provided 1κ jPj < bκ < 1  1κ jPj, and neither convex nor concave provided 1  1κjPj < bκ < jPj. Multiplying both inequalities by κ, we obtain the desired range of b.

2119

∇×H


μ ∂H ; c ∂t

(28)

2π ϵ ∂E σE  ; c c ∂t

(29)

∇ · ϵE
4πρ;

(30)

∇ · μH
0;

(31)

c being the speed of light in vacuum. We consider plane wave solutions to the Maxwell equations (28)–(31), with ρ
0 and σ
0, and having components of the form  

r·s  δ
a cosωt − k · r  δ; a cos ω t − v with k
ωv s and a, δ being real numbers, and s being a unit vector. If Er; t
Ek · r − ωt and Hr; t
H k · r − ωt solve the Maxwell equations (28)–(30), one can show that H


c k × E; μω

and;

E
−

c k × H: ϵω

(32)

We assume that the incident vector s
si with ˆ si
sin θi ˆi  cos θi k: That is, si lives on the x‐z plane and, thus, the direction of propagation is perpendicular to the y axis. In addition, the boundary between the two media is the x‐y plane, ν denotes the normal vector to the boundary at the point P; that is, ν is on the z axis, and θi is the angle between the normal vector ν and the incident direction si (as usual, ˆi; ˆj; kˆ denote the unit coordinate vectors). Let I ∥ and I ⊥ denote the parallel and perpendicular components, respectively, of the incident field. The electric field corresponding to this incident field is 

r · si Ei r; t
−I ∥ cos θi ; I ⊥ ; I ∥ sin θi  cos ω t − v1 

i r·s ; (33)
Ei0 cos ω t − v1 with c v 1
pffiffiffiffiffiffiffiffiffi : ϵ1 μ1 From (32), the magnetic field is then

Vol. 32, No. 11 / November 2015 / Journal of the Optical Society of America A

2120

rffiffiffiffiffi ϵ1 i s × Ei  μ1 

rffiffiffiffiffi ϵ1 r · si −I cos θi ; −I ∥ ; I ⊥ sin θi  cos ω t −
μ1 ⊥ v1

 i r·s :
Hi0 cos ω t − v1

z 2 cos θi − z 1 cos θt I z 2 cos θi  z 1 cos θt ∥ z cos θi − z 2 cos θt R⊥
1 I : z 1 cos θi  z 2 cos θt ⊥

Hi r; t


Let us now introduce st , the direction of propagation of the transmitted wave, and let θt be the angle between the normal ν and st . Similarly, sr is the direction of propagation of the reflected wave, and θr is the angle between the normal ν and sr . ˆ We have that sr
sin θr ˆi  cos θr kˆ
sin θi ˆi − cos θi k. Then the corresponding electric and magnetic fields corresponding to transmission are

 r · st t E r; t
−T ∥ cos θt ; T ⊥ ; T ∥ sin θt  cos ω t − v2 

t r·s ;
Et0 cos ω t − v2 and



rffiffiffiffiffi ϵ2 r · st −T ⊥ cos θt ;−T ∥ ;T ⊥ sin θt cos ω t − H r;t
μ2 v2 

t r·s ;
Ht0 cos ω t − v2 t

with c v 2
pffiffiffiffiffiffiffiffiffi : ϵ2 μ2 There are similar formulas for the fields Er r; t; Hr r; t corresponding to reflection. Since the tangential components of the electric and magnetic fields are continuous across the boundary (see, e.g., [11, Subsection 1.1.3]), we obtain the equations I ⊥  R⊥
T ⊥;

cos θi I ∥ − R ∥ 
cos θt T ∥ ;

and I R T∥ q∥ffiffiffiffi  q∥ffiffiffiffi
qffiffiffi ffi; ϵ1 μ1

0

ϵ1 μ1

ϵ2 μ2

1

R C T BI cos θi @q⊥ffiffiffiffi − q⊥ffiffiffiffiA
cos θt q⊥ffiffiffiffi : ϵ1 μ1

ϵ1 μ1

ϵ2 μ2

The wave impedance of the medium is defined by rffiffiffi μ ; z
ϵ qμffiffiffi so setting z j
ϵjj , j
1; 2, and solving the last two sets of linear equations, yields 2z 1 cos θi I z 2 cos θi  z 1 cos θt ∥ 2z 1 cos θi T⊥
I ; z 1 cos θi  z 2 cos θt ⊥

Research Article

R∥


(35)

These are the Fresnel equations expressing the amplitudes of the reflected and transmitted waves in terms of the amplitude of the incident wave. We now apply this calculation to deal with NIMs. We will replace si by x and st by m, and we also set pffiffiffiffiffiffiffiffiffi ϵ2 μ2 κ
− pffiffiffiffiffiffiffiffiffi : ϵ1 μ1 Recall ν is the normal to the interface. We have cos θi
x · ν and cos θt
m · ν. Suppose medium I has ϵ1 > 0 and μ1 > 0; and medium II has ϵ2 < 0 and μ2 < 0. In other words, medium I is “right-handed” and medium II is “left-handed.” This means that, in medium II, the refracted ray m stays on the same side of the incident ray x. From the Snell law formulated in (7) (notice that κ < 0), we have x − κm
λν, so the Fresnel equations (34) and (35) take the form 2z 1 x · ν 2z 1 x · x − κm I∥
I ; z 2 x  z 1 m · ν z 2 x  z 1 m · x − κm ∥ 2z 1 x · ν 2z 1 x · x − κm T⊥
I⊥
I ; z 1 x  z 2 m · ν z 1 x  z 2 m · x − κm ⊥ z x − z 1 m · ν z x − z 1 m · x − κm I
2 I ; R∥
2 z 2 x  z 1 m · ν ∥ z 2 x  z 1 m · x − κm ∥ z x − z 2 m · ν z x − z 2 m · x − κm R⊥
1 I
1 I : (36) z 1 x  z 2 m · ν ⊥ z 1 x  z 2 m · x − κm ⊥ T∥


Notice that the denominators of the perpendicular components are the same and, likewise, for the parallel components. As an example, suppose that ϵ1 , μ1 are both positive, and ϵ2
−ϵ1 and μ2
−μ1 . This is the so-called mirror like material. Then z 1
z 2 and κ
−1; thus, R ∥
R ⊥
0 for all incident rays. This means that all the energy is transmitted and nothing is reflected internally. Notice also that, if κ ≠ −1, then there is a reflected wave; that is, R ⊥ and R ∥ may be different from zero. The wave impedance, like the index of refraction, is a unique characteristic of the medium in consideration. However, unlike the refractive index n, the wave impedance z remains positive for negative values of ϵi , μi . In addition, compare Eq. (35) with the table in [14, p. 765] giving R ⊥ . There the Fresnel formula for r ⊥ in the exact Fresnel formula column contains a misprint. We believe that our calculations above are correct and consistent with what is expected in the case of negative refraction. A. Brewster Angle

This is the case when R ∥
0; this means when z 2 cos θi − z 1 cos θt
0;

T∥


(34)

which, together with the Snell law (sin θi
κ sin θt ), yields
2 z2 1 cos2 θi  2 sin2 θi
1
cos2 θi  sin2 θi : z1 κ

Vol. 32, No. 11 / November 2015 / Journal of the Optical Society of America A

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2121




Thus,

 2 tan θi
2

z2 z1

−1

1 − κ12





 μ2 ϵ1 μ2 − ϵ2 μ1 ; μ1 ϵ2 μ2 − ϵ1 μ1

if 1∕κ 2 ≠ 1, and the Brewster angle is θi with sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
 μ2 ϵ1 μ2 − ϵ2 μ1 : tan θi
μ1 ϵ2 μ2 − ϵ1 μ1 Let us compare this value of θi with the case when κ > 0. As above, the Brewster angle occurs when R ∥
0. In this situation, the reflected and transmitted rays are orthogonal to each other and, via Snell’s law, it follows that tan θi
n2 ∕n1 ; which is consistent with what we would expect. For related results, see [17]. B. Fresnel Coefficients for NIMs

To calculate these coefficients we follow the calculations [11 from, Subsection 1.5.3]. The Poynting vector is given by S
4πc E × H, where c is the speed of light in free space. From (32), we get that
 rffiffiffi c c c ϵ k×E
E × s × E; S
E× 4π μω 4π μ ffi where k
ωv s, s is a unit vector, and v
pcffiffiffi ϵμ. Using the form of the incident wave (33), the amount of energy J i of the incident wave Ei flowing through a unit area of the boundary per second is then rffiffiffiffiffi c ϵ1 i 2 i i J
jS j cos θi
jE j cos θi : 4π μ1 0 Similarly, the amount of energy in the reflected and transmitted waves (also given in the previous section) leaving a unit area of the boundary per second is given by rffiffiffiffiffi c ϵ1 r 2 J r
jSr j cos θi
jE j cos θi ; 4π μ1 0 rffiffiffiffiffi c ϵ2 t 2 jE j cos θt : J t
jSt j cos θt
4π μ2 0 The reflection and transmission coefficients are defined by
r 2 Jr jE0 j ; and R
i
J jEi0 j
 rffiffiffiffiffiffiffiffiffi Jt ϵ2 μ1 cos θt jEt0 j 2 : T
i
J ϵ1 μ2 cos θi jEi0 j By conservation of energy or by direct verification, we have R  T
1. In the case when no polarization is assumed, we have from Fresnel’s Eq. (36) that jEr0 j2
R 2∥  R 2⊥  z 2 x − z 1 m · x − κm 2 2
I z 2 x  z 1 m · x − κm ∥  z 1 x − z 2 m · x − κm 2 2  I ; z 1 x  z 2 m · x − κm ⊥ and, thus,

 jEr0 j 2 R 2∥  R 2⊥ R

2 I ∥  I 2⊥ jEi0 j  z 2 x − z 1 m · x − κm 2 I 2∥
z 2 x  z 1 m · x − κm I 2∥  I 2⊥  z 1 x − z 2 m · x − κm 2 I 2⊥  z 1 x  z 2 m · x − κm I 2∥  I 2⊥ 2
32 ffi qffiffiffiffi qffiffiffiffiffiffiffi
qffiffiffiffi qffiffiffiffiffiffi μ22 μ2 ϵ2 μ2 μ1 x · m − − − 6 7 ϵ2 ϵ1 ϵ1 μ 1 ϵ21 I 2∥ 6 7
6
qffiffiffiffi qffiffiffiffiffiffiffi
qffiffiffiffi qffiffiffiffiffiffiffi 7 4 5 I 2∥  I 2⊥ μ22 μ2 ϵ2 μ2 μ1  ϵ2  ϵ1  ϵ1 μ1 x · m ϵ2 1

32 2
ffi
qffiffiffiffi qffiffiffiffiffiffiffi qffiffiffiffi qffiffiffiffiffiffi μ22 μ1 μ2 ϵ2 μ2 x ·m7 6 ϵ1 − ϵ1 μ1 − ϵ2 − ϵ21 I 2⊥ 7 6  6
qffiffiffiffi qffiffiffiffiffiffiffi
qffiffiffiffi qffiffiffiffiffiffiffi ; 7 2 5 I ∥  I 2⊥ 4 μ22 μ1 μ2 ϵ2 μ 2 x·m ϵ1  ϵ1 μ1 − ϵ2 − ϵ2 1

which is a function only of x · m. We assume in these formulas that the product ϵi μi > 0, i
1; 2. In principle, the coefficients I ∥ and I ⊥ might depend on the direction x; in other words, for each direction x we would have a wave that changes its amplitude with the direction of propagation. The energy of the incident wave would be f x
jEi0 j2
I ∥ x2  I ⊥ x2 . Notice that, if the incidence is normal, that is, x
m, then −z 1 2  , which shows that even for radiation normal R
zz12z 2 to the interface we may lose energy by reflection. The Fresnel coefficients R and T being written in terms of x and m as above are useful for studying some refraction problems in geometric optics. In fact, when κ > 0, it was shown in [7] that it is possible to construct a surface interface between media I and II in such a way that both the incident radiation and transmitted energy are prescribed, and that loss of energy due to internal reflection across the interface is taken into account.

5. CONCLUSION We have given a vector formulation of the Snell law for waves passing between two homogeneous and isotropic materials when one of them has a negative refractive index, that is, is left-handed. This formulation was used to find surfaces having the uniform refraction property in both the far-field and nearfield cases. In the near-field case, in contrast with the case when both materials are standard, these surfaces can be neither convex nor concave and can wrap around the target. A quantitative analysis in terms of the parameters defining the surfaces has been carried out. We have used the vector formulation of Snell’s law to find expressions for the Fresnel formulas and coefficients for NIMs. We expect these formulas to be useful in the design of surfaces separating materials, one of them a NIM, that refract radiation with prescribed amounts of energy given in advance. This will be done in future work and requires more sophisticated mathematical tools. Funding. National Science Foundation (NSF), Division of Mathematical Sciences (1201401).

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Research Article

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