M. RUDAN — KIRCHHOFF’S THEORY OF DIFFRACTION 1 Kirchhoff’s Theory of Diffraction M. Rudan
I. KIRCHHOFF’S THEORY OF DIFFRACTION where the derivatives are calculated with respect to the com- q At a point p internal to V , the solution of the Helmholtz ponents of : equation by means of the Green function, using a unit vector rn = (ξ − xn)i1 + (η − yn)i2 + (ζ − zn)i3 , ν that lies along the outward normal to S, reads ∂r ∂ 1/2 ξ − x n = (ξ − x )2 + (η − y )2 + . . . = n , ∂f ∂G ∂ξ ∂ξ n n r 4πf(p)= G − f dS − Gb dV . n ZS ∂ν ∂ν ZV whence grad rn = rn/rn. As a consequence, The above renders f(p) as a function of b, of the boundary values of and , and of the auxiliary function . exp(jkrn) exp(jkrn) 1 f ∂f/∂ν G grad = jk − grad rn . Although both f and ∂f/∂ν appear in the integral above, it rn rn rn can be demonstrated that the values of either f or ∂f/∂ν are ′ If ρn is the angle between ν and rn, then sufficient to specify f at every point within V . ′ Here, f indicates the Fourier transform with respect to time cos ρn = ν • grad rn = cos(ρn + π)= − cos ρn , of the scalar potential ϕ or of any component of the vector where ρn is the angle between the inward normal −ν and rn. potential A, and b indicates the transform of −ρ/ε0 or −µ0J , i It follows i =1, 2, 3. As a consequence, f and b depend on the angular frequency ω besides the spatial coordinates, with k2 = ω2/c2 ∂f 1 exp(jkrn) = cos ρ − jk f 0 . ∂ν n r n r in vacuum. Xn n n If a material is present within V , the constants ε0, µ0 must be The above takes a simpler form if the sources are far enough replaced with the material’s parameters ε, µ. Similarly, the from A to fulfill the relation r ≫ λ, so that k ≫ 1/r . Also, expression for the angular frequency takes the form k2 = n n it will be assumed that the positions of the points p are such ω2/u2, with u the phase velocity. n as to make the angles ρ small. This yields In the following it will be assumed that the charge density n within V is zero, so that b =0. ∂f exp(jkrn) ≃−jk f 0 = −jkf(q) . Now, consider the case in which the closed surface S is made ∂ν n r Xn n of three parts, namely, an opaque screen B, an aperture A within B, and a portion C of a large sphere of radius R In a similar manner one finds centered at p ≡ (x,y,z). ∂G exp(−jks) = ν • grad . Let q ≡ (ξ,η,ζ) be any point of A, and p0, p1, p2,..., pn . . . ∂ν s a set of points belonging to the half space on the side opposite where the derivatives are calculated with respect to the com- to p with respect to the surface A∪B, with pn ≡ (xn,yn,zn). ponents of p: Also, let rn = q − pn and s = p − q. Each point pn is assumed to be a point source from which a spherical wave of s = (x1 − ξ1)i1 + (x2 − ξ2)i2 + (x3 − ξ3)i3 , the form fn0 exp(jkγn)/γn is emitted, where fn0 is a complex ∂s ∂ 2 2 1/2 x1 − ξ1 constant and γn the distance between pn and an arbitrary point = (x1 − ξ1) + (x2 − ξ2) + . . . = , ∂x1 ∂x1 s belonging to the same half space as pn. In turn, it is k =2π/λ, with λ the wavelength. whence grad s = s/s. As a consequence, Adding up all the spherical waves at q yields the boundary exp(−jks) exp(−jks) 1 condition grad = − jk + grad s . s s s exp(jkrn) f(q)= fn0 . ′ rn If σ is the angle between ν and s, then Xn ′ Note that ν points from q towards the half space of the pn. cos σ = ν • grad s = cos(σ + π)= − cos σ , As a consequence, ν • rn < 0. The other boundary condition is found from where σ is the angle between the inward normal −ν and s. It follows ∂f exp(jkrn) ∂G 1 = ν • grad f = ν • fn0 grad , = cos σ + jk G . ∂ν r Xn n ∂ν s Also in this case one assumes that p is far enough from A to E. De Castro Advanced Research Center on Electronic Systems, and Dipar- timento di Elettronica, Informatica e Sistemistica, Universit`adi Bologna, Viale fulfill the relation s ≫ λ, so that k ≫ 1/s, to yield ∂G/∂ν ≃ Risorgimento 2, I-40136 Bologna, Italy. March 27, 2008 jk cos σ G. M. RUDAN — KIRCHHOFF’S THEORY OF DIFFRACTION 2
In order to apply the solution of the Helmholtz equation to the simplicity. The distance between the two mirrors is typically calculation of f(p) it is necessary to perform the integration much larger than the mirrors’ diameters. As a consequence, over the closed surface S made of the union of A, B, and C. one may replace 1+cos σ with 2. The field fp is given by In many cases one is interested in calculating f(p) at any point belonging to the half space opposite to that of the sources (still j fp(p)= − G(|p − q|) fq(q) dA , keeping the case where p is far enough from A to maintain λ ZA the validity of the simplifications used above). For this, one may take the limit R → ∞ and assume that the Sommerfeld where the integration is carried out over the domain A of q. asymptotic conditions hold. This makes the contribution of C However, for self-consistency fp must be the same function to the integral to vanish. as fq apart from a multiplicative complex constant. As a Next, the Kirchhoff boundary conditions are imposed. They consequence the suffixes may be dropped. Moreover, the factor consist in assuming that the contribution of B to the integral −j/λ may be incorporated into the multiplicative constant be- vanishes, and that the contribution of A is the same as if cause the wavelength here is a fixed parameter, yet unknown. B were not present. This is equivalent to assuming that In conclusion the above becomes the perturbation due to the edges of B is negligible. The theory based on these boundary condition is called Kirchhoff’s f(p)= η G(|p − q|) f(q) dA . diffraction theory. ZA In conclusion, remembering that k =2π/λ one finds The latter is an integral equation, specifically, a homogeneous j Fredholm equation of the second kind. The parameter η is the f(p)= − G(|p − q|) f(q) (1+cos σ) dA , 2λ ZA eigenvalue, while f is the corresponding eigenfunction. The solutions of the above provide the modes of the resonator. which is also called the Fresnel-Kirchhoff diffraction formula Once the solutions are found, the corresponding eigenfunctions and is the mathematical expression of the Huygens-Fresnel may be used as boundary conditions to determine the field at Principle. The Huygens coefficient is (1 + cos σ) /(2jλ). all internal points of the resonator. If the position of p is such as to make the angle σ small, the above becomes The theory depicted above is due to Fox and Li. j f(p) ≃− G(|p − q|) f(q) dA . λ Z A IV. RECIPROCITY THEOREM OF HELMHOLTZ II. BABINET’S PRINCIPLE Consider the special case where there is only one source at, Consider the dual situation in which A is the opaque screen say, p0. Letting r = |q − p0|, cos ρ = −ν • grad r, it follows and B the aperture. The dual situation will be indicated with a prime. As exp(jkr) f(q)= f(|q − p |)= f00 , + = 0 r ZA ZB′ Z∞ one finds fA(p) + fB′ (p) = f∞(p), where fA(p) is the ∂f 1 exp(jkr) p ′ p = cos ρ − jk f00 ≃−jkf(q)cos ρ . solution at in the original case, fB ( ) is the solution in ∂ν r r the dual case, and f∞(p) is the solution in the case where no obstacle is present. This result is called Babinet’s principle. In the above it has been assumed that k ≫ 1/r, whereas no An immediate consequence of the above is that, if at some assumption has been made about the position of p0. As a con- point p it is fA =0, at the same point it is fB′ (p)= f∞(p). sequence, the angle ρ is not necessarily small. Remembering Namely, where the intensity is zero in presence of one of the that s = |p − q|, cos σ = −ν • grad s one finds screens, the intensity in presence of the dual screen is the same as if no screen were present. j exp[jk(r − s)] f(p)= f00 ν • grad(r + s) dA . 2λ ZA rs III. APPLICATION TO THE FABRY-PEROT RESONATOR The Fresnel-Kirchhoff diffraction formula Consider now the case where the source is at p and the solution j is sought at p0. The same procedure used so far yields f(p)= − G(|p − q|) f(q) (1+cos σ) dA , 2λ ZA j exp[jk(s − r)] may also be exploited to improve the description of the optical f(p0)= − f00 ν • grad(r + s) dA , 2λ ZA rs resonators with respect to that given by the geometrical optics. In this case one must assume that p belongs to the first mirror where the minus sign comes from the replacement ν ←−ν. ∗ ∗ while q belongs to the second one. Clearly, in this case the As f(p0) = f (p), the real solution (f + f )/2 is invariant field distribution f(q) within the integral is not prescribed any after inverting p with p0. In other terms, a point source at more, but becomes a problem’s unknown in itself. p0 produces at p the same effect as a point source of equal Let the fields over the two mirrors be indicated with fp and fq, intensity placed at p would produce at p0. This is called the respectively. The mirrors are assumed identical for the sake of reciprocity (or reversion) theorem of Helmholtz. M. RUDAN — KIRCHHOFF’S THEORY OF DIFFRACTION 3
V. FRAUNHOFER AND FRESNEL DIFFRACTION VI. EXAMPLEOFTHE FRAUNHOFER DIFFRACTION
Still consider the case of a single point source placed at p0, By way of example consider a rectangular aperture of sides which yields 2a1 and 2a2 aligned with the ξ1 and ξ2 axes, and take the origin at the center of the rectangle. It follows j exp[jk(r − s)] f(p)= f00 ν • grad(r + s) dA , +a1 +a2 2λ ZA rs f(p)= C exp(jk1ξ1) dξ1 exp(jk2ξ2) dξ2 = Z− Z− with r = |q − p0| and s = |p − q|. As q spans the integration a1 a2 domain the difference r − s will in general change by many sin(k1a1) sin(k2a2) wavelengths, this making the exponential to oscillate rapidly. = Ae C , k1a1 k2a2 On the contrary, if the distances of p and p0 from the screen are large compared with the linear dimensions of the aperture with Ae = 4a1a2 the rectangle’s area. The intensity I is A, the terms rs and ν • grad(r + s)= −(cos ρ + cos σ) will proportional to not change appreciably. 2 2 sin(k1a1) sin(k2a2) As a consequence, choosing the origin at any point within A, I ∝ . k1a1 k2a2 one may let rs ≃ r0s0, where r0 (s0) is the distance between p0 (p) and the origin. Similarly, one may replace both cos ρ If the point source is placed at x10 = x20 =0 one has and with . In conclusion, cos σ 1 x1 x2 k1a1 = ka1 , k2a2 = ka2 . j f00 s0 s0 f(p) ≃− exp[jk(r − s)] dA . λ r0s0 ZA Over a plane x1x2 parallel to ξ1ξ2 the width of the central Assuming that A belongs to the plane ξ1, ξ2, let x1, x2, x3 be peak of the diffraction pattern along each direction x1 or x2 the coordinates of p and x10, x20, x30 those of p0. One finds is found by letting kiai = ±π, namely, 2 2 2 2 + − s0 s0 r = (x10 − ξ1) + (x20 − ξ2) + x30 = xi − xi =2π = λ . kai ai 2 2 2 = r0 − 2(x10ξ1 + x20ξ2)+ ξ1 + ξ2 , The above may be rewritten + − where ξ1, ξ2 vary while the other coordinates are kept fixed. xi − xi λ Similarly, = , s0 ai 2 2 2 2 s = (x1 − ξ1) + (x2 − ξ2) + x3 = where the left hand side is approximately the angle within
2 2 2 the xix3 plane under which the central peak is seen from any = s0 − 2(x1ξ1 + x2ξ2)+ ξ1 + ξ2 . point of the aperture A. It is worth noting that such an angle Taking the square root of the above transforms the exponent is independent on the distance r0 of the point source from the aperture. In contrast, in geometrical optics the angle due jk(r − s) within the integral into a function of ξ1, ξ2 which, in turn, may be expanded into a Taylor series. The result to parallax could be made arbitrarily small by increasing r0. obtained when the series is truncated to the first order is called On the other hand, here the angle depends on the wavelength, Fraunhofer diffraction, while that obtained when the series is which would not happen in geometrical optics. It follows that truncated to the second order is called Fresnel diffraction. The λ/ai is an approximate expression for the intrinsic divergence former approximation is applicable in most cases, and yields in the xi direction due to diffraction.
x10ξ1 + x20ξ2 x1ξ1 + x2ξ2 r ≃ r0 − , s ≃ s0 − VII. MEANINGOFTHE FRESNEL NUMBERS r0 s0 The Fresnel numbers have been defined with reference to whence, letting a prismatic domain of sides d1, d2, d3 aligned along the . x1 x10 . x2 x20 coordinate axes, 0 ≤ xi ≤ di. They are dimensionless k1 = k − , k2 = k − , s0 r0 s0 r0 quantities given by 2 . j f00 di di/d3 C = − exp[jk(r0 − s0)] , Ni = = , i =1, 2 , λ r0s0 λd3 λ/di one finds with λ = 2π/k one of the resonance wavelengths. Now, if d3 ≫ di the ratio di/d3 is the angle within the xix3 plane f(p) ≃ C exp[j(k1ξ1 + k2ξ2)] dξ1 dξ2 . under which one mirror is seen from any point of the other ZA mirror. In turn, λ/di is approximately the intrinsic divergence For the above to hold it is necessary that the remainder R of in the xi direction, due to diffraction, of a plane wave whose the expansion, transversal extension is di. The Fresnel number turns out to be the ratio between such two angles. k(r − s)= k(r0 − s0) + (k1ξ1 + k2ξ2)+ R(ξ1, ξ2) Using, e.g., di ∼ 1 mm, d3 ∼ 10 cm, λ ∼ 0.5 µm, one finds −2 −4 fulfills the relation |R| ≪ 2π. di/d3 = 10 , λ/di =5 × 10 . M. RUDAN — KIRCHHOFF’S THEORY OF DIFFRACTION 4
VIII. FAR-FIELD APPROXIMATION This result shows that the image of f(ξ1, ξ2) over the u1,u2 In the following the assumption of a single point source will plane is f itself apart from a multiplicative factor on the be disposed of. The simplified form of the Fresnel-Kirchhoff amplitude, scaling factors on the coordinates, and the inversion diffraction formula will be used, that holds if the position of due to the negative sign of the scaling factors. p is such as to make the angle σ small: In the practical cases the far-field condition is obtained by placing a lens with a focal length s0/2 at the midpoint between j f(p) ≃− G(|p − q|) f(q) dA . the ξ1, ξ2 and x1, x2 planes, and a lens with a focal length λ ZA w0/2 at the midpoint between the x1, x2 and u1,u2 planes. The above is also called far-field approximation. Letting As is well known the image on the third plane is inverted with respect to that of the first plane. Note that . j exp(−jks0) x 2 Cξ = − , 0 λ s0 2 u x s0 s 4π Cx Cξ 2 = − exp[−jk(s0 + w0)] , k w0 the same reasoning as in the single-point case yields whence for w0 = s0 the relation between the source and the x image reduces to 1 2 1 1 2 2 . f(p) ≃ Cξ f(ξ1, ξ2) exp[jk(x1ξ1 + x2ξ2)/s0] dξ1 dξ2 . f(u ,u )= f(ξ = −u , ξ = −u ) ZAξ The result above shows that, in the far-field approximation, IX. OPTICAL FILTERING over a plane x1x2 parallel to ξ1ξ2 the scalar and monochro- The possibility given by the far-field condition of obtaining matic components f are found by calculating the Fourier the spatial Fourier transform of an image lends itself to the transform of the source. Note that in this case s0 is well filtering of optical data. For instance, if a rectangular screen approximated by the distance between the two planes. The D of sides 2a1, 2a2 is placed over the x1, x2 plane, centered integration domain has been indicated with Aξ for the sake at the origin, the function f(x1, x2) is replaced with of clarity. The Fourier transform is obtained apart from a x ˜ 0 |xi|
−u1 −u2 It plays the role of a low-pass filter and smoothes out the × f(x1, x2) exp −jk x1 + x2 dx1 dx2 , image. ZAx w0 w0 with X. OPTICAL CORRELATION AND CONVOLUTION u . j exp(−jkw0) Cx = − . It has been shown that, in the far-field approximation, over λ w0 a plane x1x2 parallel to ξ1ξ2 the scalar and monochromatic If the expression of f(x1, x2) is inserted into that of f(u1,u2), components f are found by calculating the Fourier transform the integrals over x1, x2 may be separated and performed first. of the source: The following is of use: x f(x1, x2)= Cξ × ξi ui kξi kui exp jk + xi dxi ≃ 2πδ + , −x1 −x2 Z s0 w0 s0 w0 × f(ξ1, ξ2) exp −jk ξ1 + ξ2 dξ1 dξ2 . ZAξ s0 s0 where the approximation stems from the fact that the integra- x where C = exp(−jks0)/(jλs0), with s0 approximately tion domain A is finite due to the far-field constraint. Next, ξ x equal to the distance between the two planes. Now, consider integrating over ξ1, ξ2 yields a second function ϕ(ξ1, ξ2) and let 2 2 u x s0 x f(u1,u2)=4π C C × ϕ(x1, x2)= C × x ξ k2 ξ −x1 −x2 s0 s0 × ϕ(ξ1, ξ2) exp −jk ξ1 + ξ2 dξ1 dξ2 . ×f ξ1 = − u1, ξ2 = − u2 , s0 s0 w0 w0 ZAξ where f at the left hand side is calculated over the u1,u2 Defining the sum plane while f at the right hand side is calculated over the α ϕ(ξ1, ξ2)+ x δ(ξ1 − h, ξ2) , ξ1, ξ2 plane. Cξ M. RUDAN — KIRCHHOFF’S THEORY OF DIFFRACTION 5 with α and h real constants, the image of the sum over the namely, f˜− is proportional to the convolution integral of f and x1, x2 plane reads ϕ calculated at the point (−u1s0/w0 + h, −u2s0/w0). Note that ϕ(x1, x2)+ α exp(jkx1h/s0) . 2 2 s0 u x 2 jα 4π 2 αCx |Cξ | = − exp(−jkw0) . Let T (x1, x2) be the square modulus of the above: k w0λ 2 2 2 ∗ The term f˜0 is difficult to calculate analytically, unless α ≫ |ϕ| + α + αϕ exp(jkx1h/s0)+ αϕ exp(−jkx1h/s0) . |ϕ|2, in which case one simply has Finally, consider a third plane u1u2 parallel to x1x2, placed at ˜ 2 a distance w0 such that the far-field approximation applies, and f0(u1,u2) ≃ α f(u1,u2) . assume that over the x1x2 plane the function fT is present. + In any case, f˜0 does not contain the parameter h, whereas f˜ One finds − and f˜ do. The effect of such a parameter, which may be ˜ u chosen arbitrarily, is that of displacing the image along the u1 f(u1,u2)= Cx f(x1, x2) T (x1, x2)× ZAx axis. In practice, the signal δ(ξ1 − h, ξ2) is obtained by means of −u1 −u2 × exp −jk x1 + x2 dx1 dx2 . a small hole made at the point (h, 0) in the screen belonging w0 w0 to the ξ1, ξ2 plane. The signal fT over the x1, x2 plane is
The symbol f˜(u1,u2) reminds that the source of such a obtained through the following steps: x function is different from f(x1, x2). Due to the form of T , 1) The signal ϕ(ξ1, ξ2)+(α/Cξ ) δ(ξ1−h, ξ2) alone is input the above splits as from the ξ1, ξ2 plane after placing a film over the x1, x2 + − plane. f˜(u1,u2)= f˜0 + f˜ + f˜ 2) The film stores the intensity of ϕ(x1, x2) + where, letting α exp(jkx1h/s0), which is proportional to T . After the storage is completed, the transparency of the film . k ± . k Φ = (u1x1 + u2x2) , Φ =Φ ± hx1 , depends linearly on T (x1, x2). w0 s0 x 3) The signal ϕ(ξ1, ξ2) + (α/Cξ ) δ(ξ1 − h, ξ2) is removed, it is and the signal f(ξ1, ξ2) is placed over the ξ1, ξ2 plane. This provides the signal f(x1, x2) immediately before ˜ . u 2 2 f0 = Cx f |ϕ| + α exp(j Φ) dx1 dx2 , the plane and the signal Z x1, x2 f(x1, x2) T (x1, x2) Ax immediately after it. + . u ∗ + f˜ = αC f ϕ exp j Φ dx1 dx2 , In conclusion, the process depicted above provides the corre- x Z Ax lation and convolution integrals of two signals over the u1,u2 plane. Such integrals are located at different positions of the ˜− . u − f = αCx f ϕ exp j Φ dx1 dx2 . u1,u2 plane depending on the value of the parameter h. In Z Ax this manner, thay are easily separated in space and made Remembering the expressions of f(x1, x2) and ϕ(x1, x2) one distinguishable. finds
+ 2 u x 2 ∗ ′ ′ XI. APPLICATIONS OF LASERS f˜ =4π αC |C | f(ξ1, ξ2) ϕ (ξ1, ξ2) dξ1 dξ2× x ξ Z Z Aξ Aξ A. Second-harmonic generation ′ ′ u1 ξ1 − ξ1 + h u2 ξ2 − ξ2 ′ ′ Using specific properties of the laser emission, like ×δ k + k , k + k dξ1 dξ2 . w0 s0 w0 s0 monochromaticity and brilliance, it is possible to investigate Integrating over ξ′ , ξ′ yields phenomena related to the interaction of radiation with matter. 1 2 Among these, the non-linear optical effects. 2 ˜+ 2 s0 u x 2 A typical non-linear optical effect is the second-harmonic f =4π 2 αCx |Cξ | f(ξ1, ξ2)× k ZAξ generation. It is convenient to use a frequency for which the material does not provide any resonant transition. In this case, ∗ s0 s0 ×ϕ u1 + h + ξ1, u2 + ξ2 dξ1 dξ2 . the dominant phenomenon is the material polarization. In the w0 w0 linear regime one has It follows that f˜+ is proportional to the correlation integral E ∗ P = χ of f and ϕ calculated at the point (u1s0/w0 + h,u2s0/w0). Similarly, which is in fact the first-order term of a Taylor expansion. 2 Considering the scalar case one may generalize the above as ˜− 2 s0 u x 2 f =4π 2 αCx |Cξ | f(ξ1, ξ2)× 2 3 k ZAξ P = χ1 E + χ2 E + χ3 E + . . .
s0 s0 where χ = χ1. In isotropic materials it is P (−E) = −P (E), ×ϕ − u1 + h − ξ1, − u2 − ξ2 dξ1 dξ2 , w0 w0 whence χ2 = χ4 = . . . = 0. In anisotropic materials the M. RUDAN — KIRCHHOFF’S THEORY OF DIFFRACTION 6 even-order coefficient do not vanish and, if the field has the As a consequence, the solution for x ≤ 0 reduces to f = ′ form (f0/k) sin(kx)+ f0 cos(kx), which is easily seen to fulfill ′′ 2 E = E0 sin(ωt) , the equation f +k f =0 with the boundary conditions given above. Similarly, for x > d it is the second-order summand contributes the term ′ d f0 1 1 2 f = sin(kx)+ f0 cos(kx)+ b(ξ) sin[k(x − ξ)] dξ . P2 = χ2 E0 [1 − cos(ωt)] 2 k k Z0 ′ ′ to the series. Such a term containes the second harmonic. On the other hand, letting fd = f(d), fd = f (d), the solution The emitted second harmonic has the same direction as the for x > d may also be put in the form incoming radiation. f ′ f = d sin[k(x − d)] + f cos[k(x − d)] . k d B. Holography The two forms of f for x > 0 found above must coincide. Consider a one-dimensional case, and subdivide the x axis This is easily proved by calculating into three intervals x ≤ 0, 0 ≤ x ≤ d, and d ≤< x. The ′ f0 1 outer intervals are free of sources, whereas a source exists in fd = sin(kd)+ f0 cos(kd)+ Is(d) , k k 0 ≤ x ≤ d. ′ ′ Letting f be Fourier transform of the electric potential ϕ or fd = f0 cos(kd) − kf0 sin(kd)+ Ic(d) ′ of any component Ai of the magnetic potential, in the interval from the first form, and replacing fd, fd into the second form. 0 ≤ x ≤ d the Helmholtz equation for f reads In conclusion, given the function b the solution at every x ′ 2 depends only on the boundary conditions f0 and f0, as should d f 2 2π + k f = b , k = > 0 , be. dx2 λ One may notice the following: if b is replaced by another with b the transform of the field sources, namely, of −ρ/ε0 ˜ ′ function b such that Is(d), Ic(d) do not change, then fd, fd or −µ0 Ji. The boundary conditions of the above are taken as ′ ′ do not change either. Due to the second form, the solution f(0) = f0, f (0) = f0. ′′ 2 for x > d is left unchanged as well. At the same time, the The Green function G fulfills the equation G + k G = 0, solution for x < 0 is also left unchanged because it depends whence G = exp(±jkx). Extracting the second derivatives ′ on f0, f0 only. Instead, the solution for 0 Is(x)= Ic(x)=0 , x ≤ 0 .