CCHROMATIC ABERRATION IN METALENSES
CRISTIAN E. GUTI ´ERREZ AND AHMAD SABRAA bstract . This paper provides a mathematical approach to study chromatic aber-ration in metalenses. It is shown that radiation of a given wavelength is refractedaccording to a generalized Snell’s law which together with the notion of enve-lope yields the existence of phase discontinuities. This is then used to establisha quantitative measure of dispersion in metalenses concluding that in the visiblespectrum it has the same order of magnitude as for standard lenses. C ontents
1. Introduction 12. Generalized Snell’s law 33. Construction of gradient fields using the notion of envelope 54. Existence of phase discontinuities focusing a collimated beam into agiven set of directions 84.1. Existence of phases when m is constant 114.2. Existence of phases focusing all rays into one point 125. Analysis of chromatic dispersion in metalenses 145.1. Dispersion of dichromatic light 145.2. Analysis of the chromatic aberration for a plane metasurface 145.3. Analysis of the chromatic aberration for a general radial function u ntroduction The shaping of the light wavefronts with standard lenses relies on gradual phasechanges accumulated along the optical path. Metalenses are ultra thin surfaceswhich use nano structures to focus light in imaging. They introduce abrupt phase
Wednesday 8 th May, 2019. a r X i v : . [ phy s i c s . c l a ss - ph ] M a y C. E. GUTI ´ERREZ AND A. SABRA shifts over the scale of the wavelength along the optical path to bend light inunusual ways. These nano structures are composed of arrays of tiny pillars,rings, and other arrangements of materials, which work together to manipulatelight waves as they pass by. The subject of metalenses is a current important areaof research, one of the nine runners-up for Science’s Breakthrough of the Year2016 [S], and is potentially useful in imaging applications. Metalenses are thinnerthan a sheet of paper and far lighter than glass, and they could revolutionizeoptical imaging devices from microscopes to virtual reality displays and cameras,including the ones in smartphones, see [S] and [LSW + ff erent angle causing the phenomenon of dispersion.The failure to focus radiation composed of several wavelengths into one pointis called chromatic aberration. It is our purpose in this paper to analyze thisphenomenon in metalenses and give a measure of the dispersion. The design ofmetalenses so that chromatic aberration is controlled is an active area of research[AKG + + + + ff erent regimes of operation, see [BGN + + Γ , φ ) where Γ is a surface in 3-d spacegiven by the graph of a C function u , and φ is a C function defined in a smallneighborhood of Γ called phase discontinuity. In this paper, we first study theexistence of metalenses refracting collimated monochromatic radiation into anarbitrary field of directions. We next analyze the behavior of those metalenseswhen the incoming radiation is polychromatic. More precisely, we are given aset Ω ⊂ R , a function u ∈ C ( Ω ) and a C field of unit directions m ( x ). Let Γ = { ( x , u ( x )) , x ∈ Ω } be the graph of u and assume it is surrounded by vacuum.Our goal is to find a C function φ defined in a neighborhood of Γ so that forevery x ∈ Ω , each vertical monochromatic ray with wavelength λ emitted from x ∈ Ω is refracted at ( x , u ( x )) into the direction m ( x ); see for example Figure 1(a).We find in Section 4 necessary and su ffi cient conditions on the field m and the HROMATIC ABERRATION IN METALENSES Wednesday 8 th May, 2019 3 function u so that the phase discontinuity φ exists in a neighborhood of ( x , u ( x ))for each x ∈ Ω , see Theorem 4.1. To do this, using first the generalized Snelllaw from Section 2 we see that the values ∇ φ ( x , u ( x )) are determined. We nextapply the method from Section 3 to extend φ to a neighborhood of ( x , u ( x )). Thisconstruction is then applied when m is a constant field, and when m points towarda point P = (0 , , p ), see Sections 4.1, and 4.2. If Γ is the graph of any smoothradial function, and m points towards P = (0 , , p ), then there exists also a phase φ in a neighborhood of (0 , u (0)) such that the metalens ( Γ , φ ) focuses all verticalrays with wavelength λ into P ; see Remark 4.4. We then study the behavior ofthese metalenses when the incoming rays have wavelength λ instead of λ . Weprove that that if Γ is a horizontal plane, Section 5.2, or more generally if Γ is thegraph of a radial function, Section 5.3, then all vertical rays with wavelength λ are focused into the points P λ ( x ) = (0 , , p λ ( x )) depending on | x | . We then estimatein both cases the chromatic aberration | P − P λ ( x ) | ; see Theorem 5.1 for the planarcase and (5.11) when u is radial.Finally, in Section 6, we analyze a similar problem with a standard lens. In thiscase, it is known that the surfaces separating two media n and n with n > n that focus vertical radiation in medium n with wavelength λ into a fixed pointin medium n are hyperboloids, see [GH09]. It is then shown that the error forchromatic aberration in metalenses for visible wavelengths is of the same orderof magnitude as for standard lenses.Finally, we mention that the phase discontinuity functions needed to designmetalenses for various refraction and reflection problems with prescribed dis-tributions of energy satisfy partial di ff erential equations of Monge-Amp`ere typewhich are derived and studied in [GP18]. These type of equations appeared alsonaturally in solving problems involving aspherical lenses, see [GH09], [AGT16],[GS16], [GS18], and references therein.2. G eneralized S nell ’ s law The refractive index of a medium corresponding to an electromagnetic wavewith frequency ω is given by n = cv , where c is the velocity of the wave invacuum and v its apparent velocity in the medium. v depends on the waveand on the material [F, Vol. 1, Chap. 31], see formula (31.19) there for n thathighlights the dependence of n on ω and on the material. See also [H, Chap. 3]and [BW59, Sec. 2.3.4]. Note that the frequency ω is a characteristic of the waveand is independent of the material traversed. The wavelength of the wave in the C. E. GUTI ´ERREZ AND A. SABRA medium is the distance between two consecutive crests and is given by λ = πω v .The wave number k = πλ is the number of waves per unit distance, λ denotesthe wavelength of a wave with the same frequency ω when traveling in vacuum,and k = πλ is the corresponding wave number. We have λ = π c ω , then(2.1) n = cv = π c λω = λ λ = kk . In [GPS17], Snell’s law has been generalized for metasurfaces using wave frontsintroducing a phase shift along the optical path. Using Fermat principle, we nextderive here this law in a form needed for our purposes. Assume Γ is a surface,separating media I and II , and given by the level set ψ ( x , y , z ) =
0, with ψ a C function. Assume a phase discontinuity φ is defined in a neighborhood of thesurface Γ . A light wave with frequency ω emitted from a point A in medium I strikes Γ at the point P ( x , y , z ) and is then refracted into the point B in medium II .Let n and n be the refractive indices of media I and II corresponding to ω , andlet k , k be the corresponding wave numbers. By Fermat’s principle of stationaryphase we have that P ( x , y , z ) is a critical point to the function k | P − A | + k | B − P | − φ ( P ) , and since P ∈ Γ , then we also have that ψ ( P ) =
0. Therefore by Lagrange multi-pliers, we have that ∇ (cid:16) k | P − A | + k | B − P | − φ ( P ) (cid:17) × ∇ ψ ( P ) = . Since ∇ ψ is parallel to the normal ν to Γ at the point P , and ∇ (cid:16) k | P − A | + k | B − P | − φ ( P ) (cid:17) = k P − A | P − A | − k B − P | B − P | − ∇ φ ( P ) , denoting the unit directions of the incident and refracted rays by x = P − A | P − A | and m = B − P | B − P | , respectively, we then get the following form of Snell’s law for themetalens ( Γ , φ )(2.2) ( k x − k m ) × ν = ∇ φ × ν. From (2.1), k i = n i k and so (2.2) can be re-written as k ( n x − n m ) × ν = ∇ φ × ν, HROMATIC ABERRATION IN METALENSES Wednesday 8 th May, 2019 5
We then obtain(2.3) ( n x − n m ) × ν = (cid:16) ∇ (cid:16) λ φ/ π (cid:17)(cid:17) × ν. This is clearly equivalent to(2.4) n x − n m = µ ν + ∇ (cid:16) λ φ/ π (cid:17) , for some µ . From [GPS17, Formula (11)], the value of µ can be calculated:(2.5) µ = (cid:16) n x − ∇ (cid:16) λ φ/ π (cid:17)(cid:17) · ν − (cid:114) n − (cid:18)(cid:12)(cid:12)(cid:12)(cid:12) n x − ∇ (cid:16) λ φ/ π (cid:17)(cid:12)(cid:12)(cid:12)(cid:12) − (cid:104)(cid:16) n x − ∇ (cid:16) λ φ/ π (cid:17)(cid:17) · ν (cid:105) (cid:19) . In order to avoid total internal reflection, the term inside the square root in (2.5)must be non-negative, that is,(2.6) x − ∇ (cid:16) λ φ/ π (cid:17) n · ν ≥ (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) x − ∇ (cid:16) λ φ/ π (cid:17) n (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) − (cid:18) n n (cid:19) .
3. C onstruction of gradient fields using the notion of envelope
Let Ω ⊂ R be an open simply connected domain, u ∈ C ( Ω ), and let Γ be thegraph of u , i.e., Γ = { ( x , u ( x )) : x ∈ Ω } ; x = ( x , x ). A field V = ( V , V , V ) : Γ → R is given on Γ such that V ( x , u ( x )) ∈ C ( Ω ). Here we answer the following question:When is it possible to have a C function φ defined in a neighborhood Γ such that ∇ φ = V on Γ ?Suppose such a φ exists, that is, ∇ φ ( x , u ( x )) = V ( x , u ( x )), x = ( x , x ) ∈ Ω ,and consider f ( x ) = φ ( x , u ( x )). Since φ = φ ( x , x , x ) is di ff erentiable and de-fined in a neighborhood of Γ , f x i ( x ) = φ x i ( x , u ( x )) + φ x ( x , u ( x )) u x i ( x ) = V i ( x , u ( x )) + V ( x , u ( x )) u x i ( x ), i = ,
2, so f satisfies the system(3.1) f x ( x ) = V ( x , u ( x )) + V ( x , u ( x )) u x ( x ) f x ( x ) = V ( x , u ( x )) + V ( x , u ( x )) u x ( x ) , for x ∈ Ω . Since V ( x , u ( x )) is C and u ∈ C , it follows that f ∈ C and so f x x ( x ) = f x x ( x ) which is equivalent to(3.2) ∂ x (cid:0) V ( x , u ( x )) + V ( x , u ( x )) u x ( x ) (cid:1) = ∂ x (cid:0) V ( x , u ( x )) + V ( x , u ( x )) u x ( x ) (cid:1) , x ∈ Ω . Therefore, if such a φ exists, then V must satisfy (3.2).Vice versa, if we are now given a field V defined on Γ and satisfying (3.2), wewant to construct φ in a neighborhood of Γ so that ∇ φ = V along Γ . Actually, wewill give a su ffi cient condition on the field V so that φ exists in a neighborhood C. E. GUTI ´ERREZ AND A. SABRA of a point ( x , u ( x )) ∈ Γ . In order to do this, we will apply the implicit functiontheorem, for which we assume V ( x , u ( x )) to be C in a neighborhood of x . Indeed,if V satisfies (3.2) in a ball B = B δ ( x ) ⊂ Ω , then the system (3.1) is solvable in B ,and let f be a solution . We then have V ( x , u ( x )) · (1 , , u x ( x )) = ∂ x f ( x )(3.3) V ( x , u ( x )) · (0 , , u x ( x )) = ∂ x f ( x ) . (3.4)We seek a function φ defined in a neighborhood of the point ( x , u ( x )) so that φ ( x , u ( x )) = f ( x ), and ∇ φ ( x , u ( x )) = V ( x , u ( x )) in a neighborhood of x .Let us then define S to be the surface in R given by the vector P ( x ) = ( x , u ( x ) , f ( x )) , x ∈ B . At each point P ( x ) consider the vector N ( x ) = ( V ( x , u ( x )) , − , and the plane Π x passing through P ( x ) with normal N ( x ), that is, Π x = { y = ( y , y , y , y ) ∈ R : (cid:0) y − P ( x ) (cid:1) · N ( x ) = } . For x ∈ B and y ∈ R define the function F ( y , x ) = (cid:0) y − P ( x ) (cid:1) · N ( x )and the map(3.5) G ( y , x ) = (cid:32) F ( y , x ) , ∂ F ∂ x ( y , x ) , ∂ F ∂ x ( y , x ) (cid:33) . Consider then the system of equations(3.6) G ( y , x ) = (0 , , . We will construct φ by solving (3.6) using the implicit function theorem. Claim 3.1.
For each x ∈ B, the vector ( P ( x ) , x ) solves the system (3.6) .Proof. Clearly, from the definition of F , F ( P ( x ) , x ) =
0. From (3.3) P x ( x ) · N ( x ) = (cid:0) , , u x ( x ) , ∂ x f ( x ) (cid:1) · ( V ( x , u ( x )) , − = . There are many solutions to (3.1). However, given a point x ∈ Ω and a number y ∈ R , thereis a unique solution f satisfying (3.1) with f ( x ) = y . HROMATIC ABERRATION IN METALENSES Wednesday 8 th May, 2019 7
Hence for each ( y , x ) ∈ R × B ∂ F ∂ x = − P x ( x ) · N ( x ) + ( y − P ( x )) · N x ( x ) = ( y − P ( x )) · N x ( x ) , and so ∂ F ∂ x ( P ( x ) , x ) = . Similarly, from (3.4) P x ( x ) · N ( x ) = ∂ F ∂ x ( P ( x ) , x ) =
0, and the claimfollows. (cid:3)
Consider the point ( P ( x ) , x ) = (cid:0) x , u ( x ) , f ( x ) , x (cid:1) . From Claim 3.1, G ( P ( x ) , x ) = (0 , , ff erentiability assumptions on u and V ( x , u ( x )), G has contin-uous first order partial derivatives in a neighborhood of ( P ( x ) , x ). If the Jacobiandeterminant(3.7) ∂ G ∂ ( y , x , x ) ( P ( x ) , x ) (cid:44) , then the implicit function theorem implies that there exist an open neighborhood O ⊆ B × R of ( x , u ( x )), an open neighborhood W ⊆ R × B of ( f ( x ) , x ), and unique C functions g , g , g : O → W solving(3.8) G ( y , y , y , g ( y , y , y ) , g ( y , y , y ) , g ( y , y , y )) = (0 , , , for all ( y , y , y ) ∈ O .We shall prove that the desired φ is φ = g . Indeed, from the continuity of u ( x )and f ( x ) in B there exists a neighborhood U ⊆ B of x such that for every x ∈ U , wehave ( x , u ( x )) ∈ O , and ( f ( x ) , x ) ∈ W . Hence, for x ∈ U we have ( P ( x ) , x ) ∈ O × W ,and from Claim 3.1 G ( P ( x ) , x ) = (0 , , g , g , g , it follows that g ( x , u ( x )) = f ( x ), g ( x , u ( x )) = x , g ( x , u ( x )) = x ,for every x = ( x , x ) ∈ U . Then, it remains to show that ∇ g ( x , u ( x )) = V ( x , u ( x ))for x ∈ U , i.e., along Γ . In fact, since for ( y , y , y ) ∈ O (3.9) F ( y , y , y , g ( y , y , y ) , g ( y , y , y ) , g ( y , y , y )) = , di ff erentiating this identity with respect to y yields(3.10) 0 = F y + F y ( g ) y + F x ( g ) y + F x ( g ) y . From (3.8), F x = F x = (cid:0) y , y , y , g ( y , y , y ) , g ( y , y , y ) , g ( y , y , y ) (cid:1) .Moreover for each ( y , x ) ∈ R × B , we have F y ( y , x ) = (1 , , , · N ( x ) = V ( x , u ( x )) F y ( y , x ) = (0 , , , · N ( x ) = − . C. E. GUTI ´ERREZ AND A. SABRA
Since g ( x , u ( x )) = x and g ( x , u ( x )) = x for x = ( x , x ) ∈ U , substituting theobtained values of F x , F x , F y , F y in (3.10) yields V ( x , u ( x )) = ( g ) y ( x , u ( x )) , x ∈ U . Di ff erentiating (3.9) with respect to y and with respect to y , and proceedingsimilarly, yields V ( x , u ( x )) = ( g ) y ( x , u ( x )) , V ( x , u ( x )) = ( g ) y ( x , u ( x )) . Hence ∇ g ( x , u ( x )) = V ( x , u ( x )) for x ∈ U .We have then proved the following theorem. Theorem 3.1.
Let x ∈ Ω , u ∈ C in a neighborhood of x , and let V ( x , u ( x )) be a fieldthat is C in a neighborhood of x satisfying (3.2) in that neighborhood. If (3.7) holds,where G is defined by (3.5) , then there exists a function φ defined in a neighborhood of ( x , u ( x )) such that V = ∇ φ in that neighborhood. For our application, let us re write condition (3.7) in simpler terms. We have ∂ G ∂ ( y , x , x ) ( P ( x ) , x ) = (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ∂ F ∂ y ∂ F ∂ x ∂ y ∂ F ∂ x ∂ y ∂ F ∂ x ∂ F ∂ x ∂ F ∂ x ∂ x ∂ F ∂ x ∂ F ∂ x ∂ x ∂ F ∂ x (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ( P ( x ) , x ) . Since F y ( y , x ) = − ∂ G ∂ ( y , x , x ) ( P ( x ) , x ) = (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) − F x F x x F x x F x F x x F x x (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ( P ( x ) , x ) = − (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) F x x F x x F x x F x x (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ( P ( x ) , x ) . From the proof of Claim 3.1, F x i ( y , x ) = ( y − P ( x )) · N x i ( x ), i = ,
2, and di ff erentiatingyields F x i x j ( y , x ) = − P x j · N x i + ( y − P ) · N x i x j . Therefore (3.7) becomes(3.11) ∂ G ∂ ( y , x , x ) ( P ( x ) , x ) = − (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) P x ( x ) · N x ( x ) P x ( x ) · N x ( x ) P x ( x ) · N x ( x ) P x ( x ) · N x ( x ) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) (cid:44) .
4. E xistence of phase discontinuities focusing a collimated beam into a givenset of directions
We are given a C ( Ω ) unit field with non negative vertical component, m ( x ) = ( m ( x ) , m ( x ) , m ( x )) , x ∈ Ω , | m ( x ) | = , m ( x ) ≥ , HROMATIC ABERRATION IN METALENSES Wednesday 8 th May, 2019 9 corresponding to the set of directions where we want the radiation to be steered.Assume the medium surrounding Γ is vacuum; Γ is the graph of a function u ∈ C ( Ω ). We consider a vertical parallel beam of monochromatic light of wavelength λ . From each x ∈ Ω a ray with direction e = (0 , ,
1) strikes Γ at the point ( x , u ( x )).The goal of this section is to apply the results from Section 3 to show existenceof a phase discontinuity φ ∈ C defined in a neighborhood of Γ so that ∇ φ istangential to Γ , and for each x ∈ Ω , the vertical ray with direction e is refracted bythe metalens ( Γ , φ ) at ( x , u ( x )) into the direction m ( x ); see Figure 1(a).Let ν : = ν ( x ) = ( −∇ u , √ + |∇ u | be the outer unit normal to Γ at each point ( x , u ( x )).Since m ≥
0, it follows that m ( x ) · ν ≥ . A tangential phase discontinuity φ means ∇ φ ( x , u ( x )) · ν = x ∈ Ω .We shall prove the following theorem. Theorem 4.1.
If a tangential phase discontinuity φ solving the problem above exists,then (4.1) ∇ φ ( x , u ( x )) = πλ ( e − m − (( e − m ) · ν ) ν ) : = V ( x , u ( x )) , and m satisfies the condition (4.2) (cid:0) m ( x ) · (1 , , u x ) (cid:1) x = (cid:0) m ( x ) · (0 , , u x ) (cid:1) x , for x ∈ Ω .Conversely, suppose m is C and satisfies (4.2) , let h ∈ C ( Ω ) satisfying (4.3) ∇ h ( x ) = (cid:0) m ( x ) · (1 , , u x ) , m ( x ) · (0 , , u x ) (cid:1) , and let V ( x , u ( x )) be given by (4.1) with u ∈ C ( Ω ) . If (4.4) det D h − e · m + ( e − m ) · ν (cid:113) + u x + u y D u (cid:44) , at some x ∈ Ω , then there exist a neighborhood W of ( x , u ( x )) and a function φ definedin W satisfying (4.5) φ ( x , u ( x )) = πλ ( u ( x ) − h ( x )) + C , for some constant C, with ∇ φ ( x , u ( x )) = V ( x , u ( x )) for x in a neighborhood of x . Inaddition, for each ( x , u ( x )) ∈ W the metalens ( Γ , φ ) refracts the ray with direction e striking Γ at ( x , u ( x )) into the direction m ( x ) . Proof.
To prove the first part of the theorem, from the Snell law (2.3) with n = n =
1, the phase must satisfy (cid:16) e − m − ( λ / π ) ∇ φ (cid:17) × ν = , at all points ( x , u ( x )). Let us calculate ∇ φ from this expression. Taking crossproduct again with ν yields ν × (cid:16)(cid:16) e − m − ( λ / π ) ∇ φ (cid:17) × ν (cid:17) = . Since φ is tangential to Γ , ∇ φ · ν = ν · ν =
1, and for all vectors a , b , c we have a × ( b × c ) = b ( a · c ) − c ( a · b ), we obtain (4.1). Hence V · (cid:0) , , u x (cid:1) = πλ (cid:0) − m · (cid:0) , , u x (cid:1) + u x (cid:1) and V · (cid:0) , , u x (cid:1) = πλ (cid:0) − m · (cid:0) , , u x (cid:1) + u x (cid:1) . Therefore, from the existence of φ ,(3.2) holds and consequently (4.2) also.To prove the converse, we apply Theorem 3.1 to V given in (4.1). Since m is C and u is C , V is C . That V satisfies condition (3.2) follows from (4.2). Let usverify (3.7). Let f = πλ ( u − h ), so f solves (3.1). From the definition of G in (3.5),the vectors P and N are P ( x ) = (cid:0) x , u ( x ) , f ( x ) (cid:1) ; N ( x ) = ( V ( x , u ( x )) , − , so to show (3.7) it is enough to show (3.11). We have P x · N x = − πλ (1 , , u x , f x ) · (cid:0) m x + [(( e − m ) · ν ) ν ] x , (cid:1) = − πλ m x · (1 , , u x ) − πλ (( e − m ) · ν ) (1 , , u x ) · ν x , since ν · (1 , , u x ) = = − πλ m x · (1 , , u x ) + πλ u x x √ + |∇ u | ( e − m ) · ν. Notice that m x · (1 , , u x ) = ( m · (1 , , u x )) x − m · (0 , , u x x ) = h x x − ( e · m ) u x x . Thus P x · N x = − πλ h x x + πλ (cid:32) ( e − m ) · ν √ + |∇ u | + e · m (cid:33) u x x , and in general P x i · N x j = − πλ h x i x j + πλ (cid:32) ( e − m ) · ν √ + |∇ u | + e · m (cid:33) u x i x j . HROMATIC ABERRATION IN METALENSES Wednesday 8 th May, 2019 11
Therefore the determinant in (3.11) is − (cid:18) πλ (cid:19) times the determinant in (4.4), andconsequently if (4.4) holds we obtain (3.7). Consequently, the existence of φ follows from Theorem 3.1.With the phase φ obtained satisfying (4.5), it remains to show that the metalens( Γ , φ ) refracts incident rays with direction e into the direction m ( x ) at the point( x , u ( x )) in a neighborhood of ( x , u ( x )). Since ∇ φ satisfies (4.1), then e − m −∇ (cid:16) λ φ/ π (cid:17) is parallel to ν and so Snell’s law (2.3) is verified and consequently e isrefracted into m . However, to avoid total internal reflection, the ray e is refractedinto the medium (vacuum) above the metasurface when e and ∇ φ satisfy condition(2.6) with n = n =
1, and x = e , that is when(4.6) (cid:104)(cid:16) e − ∇ (cid:16) λ φ/ π (cid:17)(cid:17) · ν (cid:105) ≥ (cid:12)(cid:12)(cid:12)(cid:12) e − ∇ (cid:16) λ φ/ π (cid:17)(cid:12)(cid:12)(cid:12)(cid:12) − . To verify this, from (4.1) we have (cid:12)(cid:12)(cid:12)(cid:12) e − ∇ (cid:16) λ φ/ π (cid:17)(cid:12)(cid:12)(cid:12)(cid:12) − = | m + (( e − m ) · ν ) ν | − = (( e − m ) · ν ) + e − m ) · ν ) m · ν = ( e · ν ) − ( m · ν ) ≤ ( e · ν ) , and so (4.6) follows since ∇ φ · ν = (cid:3) Existence of phases when m is constant.
We apply the converse in Theorem4.1 with m ( x ) : = m a constant vector for each x ∈ Ω and assume u ∈ C . It isclear that (4.2) holds. To calculate the determinant in (4.4), since m is constant,from (4.3) it follows that h ( x ) = m · ( x , u ( x )) + C , and D h ( x ) = ( e · m ) D u ( x ). So thedeterminant in (4.4) equals(4.7) det (cid:32)(cid:40) ( e − m ) · ν √ + |∇ u | (cid:41) D u (cid:33) = (cid:32) ( e − m ) · ν √ + |∇ u | (cid:33) det D u ( x ) . If the last expression is not zero at x = x , then (4.4) holds and therefore a phasediscontinuity φ exists in a neighborhood of ( x , u ( x )). Remark 4.2.
We remark that (4.4) is only a su ffi cient condition for the existence of φ and it is not necessary. In fact, when Γ is a plane, i.e., u ( x ) = a · x + b with a ∈ R , b ∈ R , and m is a constant field, the determinant in (4.4) equals the determinantin (4.7) which is zero. However, in this particular case we can still find a phase discontinuity φ in a neighborhood of Γ . In fact, the normal ν = ( − a , √ + | a | , andfrom (4.1) if φ exists it must verify ∇ φ ( x , u ( x )) = πλ (cid:40) e − m − (cid:34) ( e − m ) · ( − a , + | a | (cid:35) ( − a , (cid:41) : = v . Letting φ ( x , x , x ) = v · ( x , x , x ) yields the desired phase and so the metalens( Γ , φ ) refracts all vertical rays into the constant direction m .4.2. Existence of phases focusing all rays into one point.
Here we show theexistence of metalenses ( Γ , φ ) that refract all vertical rays into a fixed point P = (0 , , p ), p >
0. The set of refracted directions is(4.8) m ( x ) = ( − x , p − u ( x )) (cid:112) | x | + ( p − u ( x )) , where u ( x ) < p for all x . In this case, to show existence of phase discontinuitieswill use the following proposition. Proposition 4.3.
For each function u ∈ C , m defined by (4.8) satisfies (4.2) . Moreover,the determinant in (4.4) can be written as (4.9)1 h det (cid:32) Id + ∇ u ⊗ ∇ u − h (cid:0) x + ( u − p ) ∇ u (cid:1) ⊗ (cid:0) x + ( u − p ) ∇ u (cid:1) + (cid:0) x , u − p − h (cid:1) · ( −∇ u , + |∇ u | D u (cid:33) , where ∇ h = (cid:0) m · (cid:0) , , u x (cid:1) , m · (cid:0) , , u x (cid:1)(cid:1) . Therefore, if (4.9) is not zero at x and u ∈ C ,then from Theorem 4.1 a phase discontinuity φ exists in a neighborhood of ( x , u ( x )) .Proof. We have m ( x ) · (1 , , u x ) = − x + ( p − u ( x )) u x (cid:112) | x | + ( p − u ( x )) = − (cid:26) (cid:113) | x | + ( p − u ( x )) (cid:27) x , and similarly m ( x ) · (0 , , u x ) = − (cid:110) (cid:112) | x | + ( p − u ( x )) (cid:111) x . Hence m ( x ) satisfies (4.2).If we set(4.10) h ( x ) = − (cid:113) | x | + ( p − u ( x )) , then h satisfies (4.3).Let us calculate the matrix inside the determinant in (4.4). From the definitionof m e − m = − h (cid:0) x , u − p − h (cid:1) e · m + ( e − m ) · ν √ + |∇ u | = − p − uh − h (cid:0) x , u − p − h (cid:1) · ( −∇ u , + |∇ u | . HROMATIC ABERRATION IN METALENSES Wednesday 8 th May, 2019 13
Let us next calculate D h : ∂ x i h = − x i + u x i ( u − p ) (cid:112) | x | + ( u − p ) = x i + u x i ( u − p ) h ∂ x i x j h = h − (cid:16) δ ij + u x j u x i + ( u − p ) u x i x j (cid:17) − h − (cid:0) x i + ( u − p ) u x i (cid:1) (cid:16) x j + ( u − p ) u x j (cid:17) . Then D h = − h (cid:0) x + ( u − p ) ∇ u (cid:1) ⊗ (cid:0) x + ( u − p ) ∇ u (cid:1) + h (cid:16) Id + ∇ u ⊗ ∇ u + ( u − p ) D u (cid:17) . Combining these calculations it follows that the determinant in (4.4) equals det (cid:32) h ( Id + ∇ u ⊗ ∇ u ) − h (cid:0) x + ( u − p ) ∇ u (cid:1) ⊗ (cid:0) x + ( u − p ) ∇ u (cid:1) + h (cid:0) x , u − p − h (cid:1) · ( −∇ u , + |∇ u | D u (cid:33) , and factoring out 1 / h yields (4.9). (cid:3) Remark 4.4.
With this proposition we show existence of phases when Γ is givenby the graph of a radial function u . Let us write u ( x ) : = v ( | x | ) and we have p > u ( x )for all x . Setting r = | x | = x + x it follows that u x i = x i v (cid:48) ( r ) , u x i x j = δ ij v (cid:48) ( r ) + x i x j v (cid:48)(cid:48) ( r ) . Hence ∇ u = v (cid:48) ( r ) x , D u = v (cid:48) ( r ) Id + v (cid:48)(cid:48) ( r ) x ⊗ x . We now calculate the components of the matrix A inside the determinant in (4.9): ∇ u ⊗ ∇ u = v (cid:48) ( r )) x ⊗ x (cid:0) x + ( u − p ) ∇ u (cid:1) ⊗ (cid:0) x + ( u − p ) ∇ u (cid:1) = (cid:0) x + v ( r ) − p ) v (cid:48) ( r ) x (cid:1) ⊗ (cid:0) x + v ( r ) − p ) v (cid:48) ( r ) x (cid:1) = (cid:0) + v (cid:48) ( r )( v ( r ) − p ) (cid:1) x ⊗ x ( x , u − p − h ) · ( −∇ u , + |∇ u | = ( x , v ( r ) − p − h ) · ( − v (cid:48) ( r ) x , + v (cid:48) ( r )) r = − r v (cid:48) ( r ) + v ( r ) − p − h + v (cid:48) ( r )) r . Notice that in this case, the function h in (4.10) is radial with h = h ( r ) = − (cid:112) r + ( p − v ( r )) .Replacing in the formula for A we obtain A = Id + v (cid:48) ) x ⊗ x − h (cid:0) + v (cid:48) ( v − p ) (cid:1) x ⊗ x + − r v (cid:48) + v − p − h + v (cid:48) ) r (2 v (cid:48) Id + v (cid:48)(cid:48) x ⊗ x . ) = (cid:32) + v (cid:48) − r v (cid:48) + v − p − h + v (cid:48) ) r (cid:33) Id + (cid:32) v (cid:48) ) − h (cid:0) + v (cid:48) ( v − p ) (cid:1) + − r v (cid:48) + v − p − h + v (cid:48) ) r v (cid:48)(cid:48) (cid:33) x ⊗ x . Notice that if x =
0, then A = Id , and therefore from Proposition 4.3 there is aphase discontinuity in a neighborhood of x =
5. A nalysis of chromatic dispersion in metalenses
Dispersion of dichromatic light.
Let ( Γ , φ ) be a metasurface in R sur-rounded by vacuum. A dichromatic ray with unit direction x , superpositionof two wavelengths λ and λ , strikes Γ at a point P . Let ν ( P ) be the outer-unitnormal to Γ at P , and let m , m be the unit directions of the refracted rays at P corresponding to λ and λ . From (2.3), the vectors x − m − ∇ ( λ φ ( P ) / π ) and x − m − ∇ ( λ φ ( P ) / π ) are parallel to ν ( P ). We show that the following statementsare equivalent(1) m = m ,(2) ∇ φ ( P ) is colinear to ν ( P ),(3) m = m = x .This means that if ∇ φ ( P ) is not parallel to ν ( P ), then rays composed with di ff erentcolors will be dispersed by the metalens, i.e., refracted in di ff erent directions.And only in case ∇ φ ( P ) is parallel to ν ( P ), the incident ray propagates through themetalens without changing direction. Proof. (3) = ⇒ (1) is trivial, so we will show that (1) = ⇒ (2) = ⇒ (3) . (1) = ⇒ (2) . Since n = n , from (2.3)( x − m ) × ν ( P ) = ∇ (cid:16) λ φ ( P ) / π (cid:17) × ν ( P ) , ( x − m ) × ν ( P ) = ∇ (cid:16) λ φ ( P ) / π (cid:17) × ν ( P ) , implying λ − λ π ∇ φ ( P ) × ν ( P ) = , and since λ (cid:44) λ , (2) follows.(2) = ⇒ (3) . If ∇ φ ( P ) is parallel to ν ( P ), then from (2.3) ( x − m ) × ν ( P ) = , sothere exists µ ∈ R such that(5.1) x − m = µ ν ( P ) . Dotting (5.1) with x and m yields1 − m · x = µ x · ν ( P ) m · x − = µ m · ν ( P ) , and adding these identities, we obtain µ ( x + m ) · ν ( P ) = . Hence µ = x · ν ( P ) = m · ν ( P ) =
0, since x · ν ( P ) and m · ν ( P ) are both non negative. Dotting(5.1) with ν ( P ), if x · ν ( P ) = m · ν ( P ) = µ =
0. Therefore x = m , and similarly, x = m , concluding (3). (cid:3) Analysis of the chromatic aberration for a plane metasurface.
Let Γ be thehorizontal plane x = a in R . Suppose a phase discontinuity φ is defined in aneighborhood of Γ such that vertical rays having color with wavelength λ andstriking Γ are refracted into a point P = (0 , , p ) above Γ as in Figure 1(a). We HROMATIC ABERRATION IN METALENSES Wednesday 8 th May, 2019 15 then ask how this planar metasurface ( Γ , φ ) focuses vertical rays with wavelength λ (cid:44) λ . Assuming that all incoming rays pass through the circular aperture x + x ≤ R , we shall prove that for appropriate values of λ each refracted ray withwavelength λ intersects the z -axis at some point P λ ( x ) = (0 , , p λ ( x )) depending on x = ( x , x ). In addition, we shall prove an estimate for the distance between thepoints P λ and P in terms of p − a and the ratio λ/λ , see (5.3) and Figure 1(b). Wesummarize the results of this section in the following theorem. Theorem 5.1.
Let P = (0 , , p ) . Consider the metalens ( Γ , φ ) surrounded by vacuum,with Γ the horizontal plane x = a in R , a < p and with (5.2) φ ( x , x , x ) = πλ (cid:113) x + x + ( p − a ) + g ( x ) = πλ dist (( x , a ) , P ) + g ( x ) where g is an arbitrary C function satisfying g (cid:48) ( a ) = . Suppose vertical rays thatare superposition of two colors, with wavelengths λ and λ , strike Γ at ( x , x , a ) , withx + x < R .Then each ray splits into two rays one with wavelength λ and another with wavelength λ . Each ray with wavelength λ is refracted into the point P . And, if λλ < , or < λλ ≤ (cid:114) + ( p − a ) R , then each ray with wavelength λ is refracted into the pointP λ ( x ) = (cid:18) , , λ λ (cid:113)(cid:16) − ( λ/λ ) (cid:17) | x | + ( p − a ) (cid:19) . We also have (5.3) | P − P λ ( x ) | ≤ (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) − λ λ (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) (cid:0) p − a (cid:1) for each x = ( x , x ) with x + x ≤ R .Proof. Let us first calculate the phase discontinuity focusing rays with color λ into P . The incident ray with direction e = (0 , ,
1) emanating from x = ( x , x )with x + x ≤ R strikes the plane x = a and then bends to strike the point P .Applying the set up from Section 4 in this case we have m ( x ) = P − ( x , a ) | P − ( x , a ) | andthe normal ν = e . Since we seek φ tangential to Γ , we then have from (4.1) that(5.4) V ( x , a ) = ∇ φ ( x , a ) = πλ ( − m + ( m · e ) e ) = πλ x (cid:112) | x | + ( p − a ) , . Clearly, the function φ in (5.2) is the desired phase function. Therefore, the planarmetasurface having this phase focuses all rays with color λ into the point P . ( Γ , ϕ ) xP (a) Rays with wavelength λ focus into afixed point ( Γ , ϕ ) xP λ ( x ) P (b) Rays with wavelength λ are defo-cused F igure Γ , φ ) focuses rays with wavelength λ . First,to avoid total reflection, from (2.6), rays with color λ are refracted by the metalens( Γ , φ ) if(5.5) (cid:104)(cid:16) e − ∇ (cid:16) λ φ/ π (cid:17)(cid:17) · ν (cid:105) ≥ (cid:12)(cid:12)(cid:12)(cid:12) e − ∇ (cid:16) λ φ/ π (cid:17)(cid:12)(cid:12)(cid:12)(cid:12) − . Since ν = (0 , , = e , and ∇ φ · ν =
0, then (5.5) is equivalent to0 ≤ − (cid:12)(cid:12)(cid:12)(cid:12) e − ∇ (cid:16) λ φ/ π (cid:17)(cid:12)(cid:12)(cid:12)(cid:12) = (cid:16) − ( λ/λ ) (cid:17) | x | + ( p − a ) | x | + ( p − a ) : = ∆ ( x ) | x | + ( p − a ) . Clearly, this inequality holds for λ < λ . If λ > λ , then ∆ ≥ p − a ) ≥ (cid:16) ( λ/λ ) − (cid:17) | x | . Since | x | ≤ R , if λ > λ satisfies (cid:113) ( λ/λ ) − ≤ p − aR then ∆ ≥ | x | ≤ R . Therefore for these values of λ total reflection is avoided.Next, we need to know the directions of the refracted rays with color λ . Ac-cording to (2.4), ( Γ , φ ) refracts rays with color λ into the unit direction m (cid:48) with(5.6) m (cid:48) = e − µ (cid:48) ν − ∇ (cid:16) λ φ/ π (cid:17) , HROMATIC ABERRATION IN METALENSES Wednesday 8 th May, 2019 17 where from (2.5) µ (cid:48) = − (cid:114) − (cid:12)(cid:12)(cid:12)(cid:12) e − ∇ (cid:16) λ φ/ π (cid:17)(cid:12)(cid:12)(cid:12)(cid:12) = − (cid:112) | x | + ( p − a ) (cid:112) ∆ ( x ) . Writing (5.6) in coordinates yields m (cid:48) i = − λ φ x i / π = − λλ x i (cid:112) | x | + ( p − a ) , i = , m (cid:48) = − µ = (cid:112) | x | + ( p − a ) (cid:112) ∆ ( x ) . We now see where the line t m (cid:48) + ( x , a ) intersects the vertical line x = x =
0, thatis, we find the point on this line that is focused by the refracted ray emanating from x = ( x , x ) with direction e and color λ . So we need to find t so that t m (cid:48) + x = t m (cid:48) + x =
0. This means t = λ λ (cid:112) | x | + ( p − a ) . Hence t m (cid:48) + a = λ λ √ ∆ + a ,and the focused point on the vertical line by the vertical ray emanating from x with color λ is P λ ( x ) = (cid:18) , , λ λ √ ∆ + a (cid:19) .We now calculate the distance between P and P λ ( x ). First write t m (cid:48) + a − p = λ λ √ ∆ − (cid:0) p − a (cid:1) = (cid:18) λ λ √ ∆ − (cid:0) p − a (cid:1)(cid:19) (cid:18) λ λ √ ∆ + (cid:0) p − a (cid:1)(cid:19) λ λ √ ∆ + (cid:0) p − a (cid:1) = (cid:18) λ λ (cid:19) ∆ − (cid:0) p − a (cid:1) λ λ √ ∆ + (cid:0) p − a (cid:1) = (cid:18) λ λ (cid:19) (cid:32) − (cid:18) λλ (cid:19) (cid:33) | x | + (cid:32)(cid:18) λ λ (cid:19) − (cid:33) (cid:0) p − a (cid:1) λ λ √ ∆ + (cid:0) p − a (cid:1) = (cid:32)(cid:18) λ λ (cid:19) − (cid:33) | x | + (cid:0) p − a (cid:1) λ λ √ ∆ + (cid:0) p − a (cid:1) . We then obtain(5.7) dist ( P , P λ ( x )) = (cid:12)(cid:12)(cid:12) t m (cid:48) + a − p (cid:12)(cid:12)(cid:12) = (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:18) λ λ (cid:19) − (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) | x | + (cid:0) p − a (cid:1) λ λ √ ∆ + (cid:0) p − a (cid:1) , since the right hand side is a radial function of x , then rays with color λ emittedfrom all points in a circle are focused into the same point. Set | x | = s and considerthe function F ( s ) = s + (cid:0) p − a (cid:1) λ λ (cid:113)(cid:16) − ( λ/λ ) (cid:17) s + ( p − a ) + (cid:0) p − a (cid:1) . To estimate the error (5.7) and obtain (5.3), we will find the maximum of F ( s ) for s ∈ [0 , R ]. Supposefirst that λ < λ . In this case ∆ > F (cid:48) ( s ) > , R ] and so F is increasing. Therefore when λ < λ (cid:12)(cid:12)(cid:12) t m (cid:48) + a − p (cid:12)(cid:12)(cid:12) ≤ (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:18) λ λ (cid:19) − (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) F (0) = (cid:18) λ λ − (cid:19) (cid:0) p − a (cid:1) . On the other hand, if λ (cid:114) + ( p − a ) R > λ > λ , then it is also easy to see that F is increasing in [0 , R ] and so the error (cid:12)(cid:12)(cid:12) t m (cid:48) + a − p (cid:12)(cid:12)(cid:12) ≤ (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:18) λ λ (cid:19) − (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) F (0) = (cid:18) − λ λ (cid:19) (cid:0) p − a (cid:1) . Therefore, we conclude the estimate (5.3). (cid:3)
Analysis of the chromatic aberration for a general radial function u . In thissection, the surface Γ is given by the graph of a function u ( x ) = v ( | x | ). In viewof Remark 4.4, a phase φ exists in a neighborhood of x = Γ , φ ) refracts all vertical rays with wavelength λ into a point P = (0 , , p ). Weanalyze chromatic aberration caused by ( Γ , φ ), that is, how rays with wavelength λ (cid:44) λ are focused. We shall prove that the order of magnitude of the focusingerror in the radial case is as in the planar case (5.3).In order to do this, we first find an expression for ∇ φ ( x , u ( x )) when u ( x ) = v ( | x | ) . In this case, the set of transmitted directions for rays with wavelength λ is m ( x ) = ( − x , p − u ( x )) (cid:112) | x | + ( p − u ( x )) . And from Theorem 4.1 the gradient of φ satisfies (4.1). As in Remark 4.4,we let h ( x ) = − (cid:112) | x | + ( p − u ( x )) . Set r = | x | , so ∇ u = v (cid:48) ( r ) x , and h ( r ) = − (cid:112) r + ( p − v ( r )) . Then( e − m ) · ν = − ( x , u − p − h ) h · ( −∇ u , √ + |∇ u | = − h √ + |∇ u | (cid:0) − x · ∇ u + u − p − h (cid:1) = − h (cid:112) + v (cid:48) ( r ) r (cid:0) − v (cid:48) ( r ) r + v ( r ) − p − h (cid:1) HROMATIC ABERRATION IN METALENSES Wednesday 8 th May, 2019 19 so from (4.1) ∇ φ ( x , u ( x )) = πλ − ( x , v ( r ) − p − h ) h + h (cid:112) + v (cid:48) ( r ) r (cid:0) − v (cid:48) ( r ) r + v ( r ) − p − h (cid:1) ( − v (cid:48) ( r ) x , (cid:112) + v (cid:48) ( r ) r (5.8) = − πλ h (cid:32) + v (cid:48) ( r )( v ( r ) − p − h )1 + v (cid:48) ( r ) r (cid:33) ( x , v (cid:48) ( r ) r ) . To simplify the notation set A ( r ) = + v (cid:48) ( r )( v ( r ) − p − h ).Since the phase φ is defined in a neighborhood of x =
0, we assume that | x | = r < R for some R small.As in Section 5.2, we next find conditions on λ , λ and R so that rays withwavelength λ are not totally internally reflected. We have ∇ φ · ν =
0, then from(2.6) with n = n = x = e , to avoid total reflection we should verify that(5.9) ( e · ν ) ≥ (cid:12)(cid:12)(cid:12)(cid:12) e − ∇ (cid:16) λφ/ π (cid:17)(cid:12)(cid:12)(cid:12)(cid:12) − . Indeed, we have ( e · ν ) = + v (cid:48) ( r ) r , and from (5.8) (cid:12)(cid:12)(cid:12)(cid:12) e − ∇ (cid:16) λφ/ π (cid:17)(cid:12)(cid:12)(cid:12)(cid:12) − = − e · ∇ (cid:16) λφ/ π (cid:17) + (cid:12)(cid:12)(cid:12)(cid:12) ∇ (cid:16) λφ/ π (cid:17)(cid:12)(cid:12)(cid:12)(cid:12) = − (cid:32) − λλ h (cid:32) A ( r )1 + v (cid:48) ( r ) r (cid:33) v (cid:48) ( r ) r (cid:33) + (cid:18) λλ h (cid:19) A ( r ) + v (cid:48) ( r ) r r = λλ h (cid:32) A ( r )1 + v (cid:48) ( r ) r (cid:33) (cid:18) λλ h A ( r ) + v (cid:48) ( r ) (cid:19) r . Hence (5.9) is satisfied if and only if1 ≥ λλ h A ( r ) (cid:18) λλ h A ( r ) + v (cid:48) ( r ) (cid:19) r . We conclude that taking R small enough, above inequality holds for r ≤ R , andtotal internal reflection is avoided for rays with color λ .We next study the direction of the refracted rays with wavelength λ . From (2.4)the direction of the refracted ray with color λ is m (cid:48) = e − µ (cid:48) ( − v (cid:48) ( r ) x , (cid:112) + v (cid:48) ( r ) r − λ π ∇ φ, where from (2.5) and the above calculation µ (cid:48) = e · ν − (cid:114) − (cid:18)(cid:12)(cid:12)(cid:12)(cid:12) e − ∇ (cid:16) λ φ/ π (cid:17)(cid:12)(cid:12)(cid:12)(cid:12) − ( e · ν ) (cid:19) (5.10) = (cid:112) + v (cid:48) ( r ) r − (cid:115) − λλ h (cid:32) A ( r )1 + v (cid:48) ( r ) r (cid:33) (cid:18) λλ h A ( r ) + v (cid:48) ( r ) (cid:19) r + + v (cid:48) ( r ) r = (cid:112) + v (cid:48) ( r ) r − (cid:114) − λλ h A ( r ) (cid:18) λλ h A ( r ) + v (cid:48) ( r ) (cid:19) r . Writing m (cid:48) = (cid:16) m (cid:48) , m (cid:48) , m (cid:48) (cid:17) yields (cid:16) m (cid:48) , m (cid:48) (cid:17) = µ (cid:48) v (cid:48) ( r ) x (cid:112) + v (cid:48) ( r ) r + λλ h (cid:32) A ( r )1 + v (cid:48) ( r ) r (cid:33) x = + v (cid:48) ( r ) r v (cid:48) ( r ) − (cid:114) − λλ h A ( r ) (cid:18) λλ h A ( r ) + v (cid:48) ( r ) (cid:19) r + λλ h A ( r ) x . Recall that r is chosen small enough so that the term inside the square root in theidentity above is positive.Now we find t such that t (cid:16) m (cid:48) , m (cid:48) (cid:17) + x =
0, that is, t + v (cid:48) ( r ) r v (cid:48) ( r ) − (cid:114) − λλ h A ( r ) (cid:18) λλ h A ( r ) + v (cid:48) ( r ) (cid:19) r + λλ h A ( r ) + = t = − + v (cid:48) ( r ) r v (cid:48) ( r ) (cid:32) − (cid:114) − λλ h A ( r ) (cid:18) λλ h A ( r ) + v (cid:48) ( r ) (cid:19) r (cid:33) + λλ h A ( r ) . Next, m (cid:48) = − µ (cid:48) (cid:112) + v (cid:48) ( r ) r − λ π φ x ( x , u ( x )) = − + v (cid:48) ( r ) r − (cid:114) − λλ h A ( r ) (cid:18) λλ h A ( r ) + v (cid:48) ( r ) (cid:19) r + λλ h (cid:32) A ( r )1 + v (cid:48) ( r ) r (cid:33) v (cid:48) ( r ) r = + v (cid:48) ( r ) r v (cid:48) ( r ) r + (cid:114) − λλ h A ( r ) (cid:18) λλ h A ( r ) + v (cid:48) ( r ) (cid:19) r + λλ h A ( r )(2 v (cid:48) ( r ) r ) . The ray with color λ then focuses at the point P λ = (cid:16) , , t m (cid:48) + u ( x ) (cid:17) , and we wantto see how far is this point from P = (0 , , p ). So we then need to estimate theerror t m (cid:48) + u ( x ) − p . HROMATIC ABERRATION IN METALENSES Wednesday 8 th May, 2019 21
Taking limits when r → v ( r ) → u (0) , h ( r ) → u (0) − p , A ( r ) → , t → λ λ ( p − u (0)) , m (cid:48) → . Therefore(5.11) | t m (cid:48) + u ( x ) − p | → (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) λ λ − (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) (cid:0) p − u (0) (cid:1) , as r → omparison with chromatic aberration in standard lenses We analyze here the chromatic aberration in a standard lens sandwiched by ahorizontal plane and a hyperboloid, and compare this dispersion with the oneobtained for metasurfaces.It is known that hyperboloids having appropriate eccentricity refract verticalrays into their focus point. More precisely, for a fixed wavelength λ , we havetwo materials I , II with corresponding refractive indices n , n , respectively, and apoint Y = ( y , y ) ∈ R located in material II to be focused; let κ = n / n >
1. Let h ( x ) = y n + − κ b κ − − (cid:115)(cid:32) b κ − (cid:33) + | x − y | κ − , b > , whose graph is a sheet of a hyperboloid with upper focus Y and eccentricity κ as shown in Figure 2. Suppose the material below this hyperboloid is I and the n n Y Λ e Y ′ h ( x ) F igure material above is II . If a vertical ray with wavelength λ strikes the hyperboloidat a point ( x , h ( x )), then this ray is refracted into a direction passing through thepoint Y as shown in Figure 2. We analyze how a vertical ray having di ff erentwavelength would be refracted by the same hyperboloid. In other words, howfar is the new refracted point from the focus Y ?We assume that II is vacuum, so n = n = n = n ( λ ) forthe wavelength λ , and the focus Y = (0 , , e = (0 , , e is refracted bythe hyperboloid into a ray with unit direction Λ = κ e + δ ν where ν is the outer unit normal to h at the striking point, and δ = − κ ( e · ν ) + (cid:112) − κ + κ ( e · ν ) , see [AGT16, Section 2]; with κ = n . Suppose now that the vertical ray haswavelength λ (cid:48) so the refractive index for the material I under this wavelength hasvalue n (cid:48) = n ( λ (cid:48) ). From the Snell law such a ray is then refracted by the hyperboloidinto a direction Λ (cid:48) = κ (cid:48) e + δ (cid:48) ν with δ (cid:48) = − κ (cid:48) ( e · ν ) + (cid:112) − κ (cid:48) + κ (cid:48) ( e · ν ) , where κ (cid:48) = n (cid:48) .Consider the line through the point X = ( x , h ( x )) with direction Λ (cid:48) , and wewant to determine where this line intersects the vertical line x =
0. That is, weneed to find t such that X + t Λ (cid:48) intersects the vertical line x =
0, and see howfar this intersection point is from the original focus (0 , , ν = ( −∇ h , √ + |∇ h | , so Λ (cid:48) = n (cid:48) (0 , , + δ (cid:48) ν = (cid:32) δ (cid:48) (cid:32) − ∇ h √ + |∇ h | (cid:33) , n (cid:48) + δ (cid:48) √ + |∇ h | (cid:33) and therefore t must satisfy x + t δ (cid:48) (cid:32) − ∇ h √ + |∇ h | (cid:33) = (0 , ∇ h = − (cid:112) b + ( n − | x | x ,we get √ + |∇ h | = √ b + n | x | (cid:112) b + ( n − | x | . Therefore t = − √ b + n | x | δ (cid:48) . HROMATIC ABERRATION IN METALENSES Wednesday 8 th May, 2019 23
To calculate δ (cid:48) , we first have e · ν = (cid:112) b + ( n − | x | √ b + n | x | , hence δ (cid:48) = − n (cid:48) (cid:112) b + ( n − | x | √ b + n | x | + (cid:115) − n (cid:48) + n (cid:48) (cid:32) b + ( n − | x | b + n | x | (cid:33) = − n (cid:48) (cid:112) b + ( n − | x | + (cid:112) b + ( n − n (cid:48) ) | x | √ b + n | x | . Thus, t = − b + n | x | − n (cid:48) (cid:112) b + ( n − | x | + (cid:112) b + ( n − n (cid:48) ) | x | = − (cid:0) b + n | x | (cid:1) (cid:16) n (cid:48) (cid:112) b + ( n − | x | + (cid:112) b + ( n − n (cid:48) ) | x | (cid:17) b + ( n − n (cid:48) ) | x | − n (cid:48) ( b + ( n − | x | ) = n (cid:48) (cid:112) b + ( n − | x | + (cid:112) b + ( n − n (cid:48) ) | x | n (cid:48) − . The last component of X + t Λ (cid:48) equals E ( x ) = h ( x ) + t (cid:32) n (cid:48) + δ (cid:48) √ + |∇ h | (cid:33) = − n bn − − (cid:115)(cid:32) bn − (cid:33) + | x | n − + n (cid:48) n (cid:48) (cid:112) b + ( n − | x | + (cid:112) b + ( n − n (cid:48) ) | x | n (cid:48) − − (cid:112) b + ( n − | x | = − n bn − − n − (cid:112) b + ( n − | x | + n (cid:48) − (cid:112) b + ( n − | x | + n (cid:48) n (cid:48) − (cid:112) b + ( n − n (cid:48) ) | x | = − n bn − + (cid:18) n (cid:48) − − n − (cid:19) (cid:112) b + ( n − | x | + n (cid:48) n (cid:48) − (cid:112) b + ( n − n (cid:48) ) | x | . The ray with wavelength λ (cid:48) is refracted at X into the point (0 , , E ( x )). Let g ( r ) = − n bn − + (cid:18) n (cid:48) − − n − (cid:19) (cid:112) b + ( n − r + n (cid:48) n (cid:48) − (cid:112) b + ( n − n (cid:48) ) r , we assume r > b + (cid:0) n − n (cid:48) (cid:1) r ≥ λ (cid:48) . If n > n (cid:48) , then g is strictly increasing and we obtain E ( x ) ≥ E (0) , for all x . And if n < n (cid:48) , then g is strictly decreasing and so E ( x ) ≤ E (0) , for x satisfying b + (cid:0) n − n (cid:48) (cid:1) | x | ≥ E (0) = b (cid:18) n (cid:48) − − n − (cid:19) = b n − n (cid:48) ( n (cid:48) − n − . Let us translate the refractive indices in terms of wavelengths. To see the orderof magnitude in the error E when n and n (cid:48) are given in terms of wavelengths, weuse Cauchy approximate dispersion formula, see [BW59, Sec. 2.3.4, Formula (43)]or [JW01, Sec. 23.3] (valid only in the visible spectrum) n ( λ ) = + A + B λ . Here the terms with powers of λ bigger that four in [BW59, Sec. 2.3.4, Formula(41)] have been neglected and n is replaced by 2( n −
1) when n takes values forvarious gases; see discussion in [BW59, Sec. 2.3.4, page 100] and Table 2.6 therein.We set n = n ( λ ) and n (cid:48) = n ( λ (cid:48) ), so from Cauchy’s formula E (0) = b B λ − B λ (cid:48) (cid:18) A + B λ (cid:19) (cid:18) A + B λ (cid:48) (cid:19) = b B λ (cid:18) A + B λ (cid:19) (cid:18) A + B λ (cid:48) (cid:19) (cid:32) − (cid:18) λλ (cid:48) (cid:19) (cid:33) = C ( A , B , b , λ, λ (cid:48) ) (cid:18) + λλ (cid:48) (cid:19) (cid:18) − λλ (cid:48) (cid:19) . The order of magnitude of this error, except for a bounded multiplicative constant,and in terms of λ/λ (cid:48) , is similar to the order of magnitude in (5.3) where λ playsthe role of λ and λ the role of λ (cid:48) . Notice that in the formula above for E (0), thecoe ffi cient b can be chosen arbitrarily, in particular, if b is su ffi ciently small we cancontrol the size of the multiplicative factor in front of 1 − λλ (cid:48) . Notice also that usingmore terms in the full Cauchy dispersion formula n ( λ ) = A + B /λ + C /λ + D /λ + · · · yields the same order of magnitude in the error E (0).A cknowledgements C. E. G. was partially supported by NSF grant DMS–1600578, and A. S. was partiallysupported by Research Grant 2015 / / P / ST1 / HROMATIC ABERRATION IN METALENSES Wednesday 8 th May, 2019 25
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Nanophotonics, epartment of M athematics , T emple U niversity , P hiladelphia , PA 19122 E-mail address : [email protected] F aculty of M athematics , I nformatics , and M echanics , U niversity of W arsaw , P oland C urrent A ddress : A merican U niversity of B eirut , P. O. B ox / R iad E l -S olh / B eirut ebanon E-mail address ::