Elements of diffraction theory part II

Transcription

1 Elements of diffraction theory part II Setosla S. Iano Department of Physics, St. Kliment Ohridski Uniersity of Sofia, 5 James Bourchier Bld, 1164 Sofia, Bulgaria (Dated: February 7, 2016 We make a brief introduction to diffraction theory. We begin with the deriation of the Helmholtz- Kirchhoff integral theorem, which we use to derie the Fresnel-Kirchhoff diffraction formula. Next we derie formulas describing Fresnel s diffraction and Fraunhofer s diffraction and finally we calculate the Fraunhofer diffraction pattern from rectangular and circular apertures. Outline 1. Kirchhoff s diffraction theory 2. Fresnel s diffraction and Fraunhofer s diffraction 3. Fraunhofer s diffraction rectangular and circular apertures 4. Fraunhofer s diffraction Diffraction gratings 5. ppendix I. KIRCHHOFF S DIFFRCTION THEORY Let the point P 0 from Fig. 1 a be a light source emitting spherical waes. Consider a particular wae-front S haing radius r 0. ccording to the Huygens-Fresnel (HF principle each point of the front is a centre of a secondary spherical waelet and the wae-front S 1 at a later instant is the enelope of these waelets. Therefore the light disturbance at a point P arises from the superposition of secondary waes that can be regarded as to originate from a surface, e.g. S, situated between P 0 and P (Fig. 1 a. By disturbance we will mean a monochromatic scalar wae, e.g. one component of the E field or the B field, say E x (in absence of polarization coupling. a б P S FIG. 1: a Illustration of the Huygens-Fresnel principle. P 0 and P are the light source and the receier, respectiely, r 0 is the radius of the wae front with surface S, S 1 is the wae front at a later instant and χ is the inclination angle at some point Q. b Deriation of the Helmholtz-Kirchhoff integral (10. S and S denote the outer and the inner surfaces, respectiely, is olume, n is unit ector normal to the surface, P is arbitrary internal point. This setting generalizes the setting from Fig. 1 a. Kirchhoff realized that the Huygens-Fresnel principle can be put on a mathematical basis by using an already known integral theorem which expresses the disturbance at any point P in the field in terms of the alues of the disturbance and its deriaties at all points on an arbitrary closed surface S surrounding P (Fig. 1 b. Here we only briefly follow Kirchhoff s steps following Ref. [1]: 1. we derie Helmholtz-Kirchhoff s integral theorem (10; 2. next we use this theorem to derie an approximate, yet ery useful, Fresnel-Kirchhoff diffraction formula (15 alid for small waelengths (λ r, s; 3. we further approximate this formula for small aperture to get the Fresnel integral (16 and the Fraunhofer integral (24; 4. finally we derie analytical expressions describing Fraunhofer diffraction from rectangular (32 and circular (42 apertures. These steps are described in separate sections, which can be read independently. You can directly go oer to Sec. III.

2 2. Deriation of Helmholtz-Kirchhoff s integral theorem Summary: This theorem gies the field disturbance U(P at the obserer s point P : 1. We use the wae equation (1 to describe the field E(x, y, z, t at any point in space and time 2. We decompose E into spacial and temporal parts: U(x, y, ze iωt and write equation (3 for U(x, y, z 3. We find U(P, which depends on the alues of U(x, y, z on the wae front S (Eq. (10. In Fig. 1 a we considered two points a source (P 0 and a receier (P, and, by the Huygens-Fresnel (HF principle, we concluded that the disturbance at P was specified entirely by the disturbances at the points forming the surface S. Thus the role of P 0 was undertaken by S. Now consider Fig. 1 b, which generalizes the setting from Fig. 1 a, where now we only hae the necessary elements from Fig. 1 a S and P. In this subsection we will set the HF principle on a rigorous mathematical ground by showing that the disturbance at P is defined by and originates from the disturbance oer S. Consider the acuum wae equation for the E field (see ppendix for deriation 2 E 1 2 E c 2 = 0, (1 t2 where the field depends on position and time, i.e. E = E(x, y, z, t. Written down for any of the amplitudes E x, E y and E z, commonly denoted as V hereafter, this equation is identical, 2 V 1 2 V c 2 t = 0, as in acuum these 2 amplitudes are independent (uncoupled. Now let us assume that our wae is monochromatic with frequency ω (the generalization to non-monochromatic waes is almost triial. Then we can split V into a spacial and a temporary part, V (x, y, z, t = U(x, y, ze iωt (2 with the spatial part satisfying the Helmholtz equation for free propagation (see ppendix B: ( 2 + k 2 U = 0. (3 In fact, U(x, y, z defines the disturbance at position (x, y, z, while the time-dependent factor e iωt is uniersal for all points in space and will therefore be omitted. Let us assume that U possesses continuous first- and second-order partial deriaties within and on the surface S. If U is another function that satisfies the same continuity requirements as U, we hae by the diergence theorem (see ppendix C (U 2 U U 2 Ud = (U U S U U ds, (4 where / denotes differentiation along the inward normal to S, denoted as n (cf. Fig. 1 b. If for U we also hae then apparently the left-hand side of Eq. (4 anishes: S ( 2 + k 2 U = 0, (5 (U U U U ds = 0. (6 Suppose we take U = e iks /s, where s denotes the distance from P to the point (x, y, z. The rationale is that U describes a spherical waelet in accordance with the HF principle. We construct another integral like Eq. (6 where the integration is carried oer the surface S, which is a sphere centered at P with radius ε. We sum both integrals: S (U U U U ds + S (U U U U Next we substitute U to obtain (see ppendix B 3 ( U ( e iks ( eiks U ds = [U eiks ik 1 eiks S s s S s s s U ] ds = Ω ds = 0. (7 [U eikε ε ( ik 1 ] eikε U ε 2 dω, (8 ε ε

3 where dω denotes an element of the solid angle, dω = 4π. Ω Since the integral oer S is independent of ε, we replace the integral oer S by its limiting alue as ε 0; the first and the third terms gie no contribution in the limit (they become too small, and the only term, which suries is Ue ikε dω UdΩ = 4πU(P. (9 Hence we obtain the Helmholtz-Kirchhoff s integral theorem Ω U(P = 1 4π S Ω Helmholtz-Kirchhoff s integral theorem ( U ( e iks eiks U s s ds. (10 Thus we hae shown that the disturbance U(P is defined by the alues of U and U/ oer S. In fact, this differs from our original claim that U(P is defined by the alues of U oer S only. It may, howeer, be shown from the theory of Green s functions that the alue of U oer S is sufficient to specify U at eery internal point P. Still, formula (10 is sufficient to describe diffraction. 3 B. Deriation of Fresnel-Kirchhoff s diffraction formula Summary: This formula approximates Eq. (10 and gies the field disturbance U(P for the ery typical situation from Fig. 2 a for small waelengths: 1. We specialize the integral theorem (10 for the situation from Fig. 2 a and show that only the integral oer is not zero 2. We assume that the disturbance arising from P 0 is a spherical wae: U (i = eikr r, originating from P 0 3. We assume small waelength, λ r, s, where r and s are distances, and arrie at Eq. (15. Here we will specialize the theorem stated in Eq. (10 for the ery typical situation illustrated in Fig. 2 a. Consider a monochromatic wae from a point source P 0, propagated through an opening in a plane screen, and let P be the point at which the disturbance U(P is to be determined. a б FIG. 2: a Illustrating the deriation of the Fresnel-Kirchhoff diffraction formula. The surface S from Fig. 1 b consists of, B and C. Only the integral oer is non-zero, whereas the integrals oer B and C hae zero contribution. The assumption we make is λ r, s, r, s being the distances. b Definition of angles and directions. r and s point in the direction of increasing r and s, respectiely, n is normal to the screen. To find the disturbance U(P we take Kirchhoff s integral (10 oer a surface S formed by (see Fig. 2 a: the opening, a portion B of the nonilluminated side of the screen and a portion C of a large sphere of radius R, centered at P which, together with and B, forms a closed surface S. We will show that only the integral oer is not zero.

4 Kirchhoff s integral (10 now gies U(P = 1 [ ] ( + + U ( e iks eiks U, (11 4π B C s s where the symbol [...] represents the sum of the three integrals. Kirchhoff made the following approximation he assumed that i the disturbance U oer is the same as if the screen is missing, and ii the disturbance U on B is eerywhere zero. This approximation is written in terms of the following boundary conditions for U and U : on : U = U (i U = U on B : U = 0 where (see ppendix for U (i and B 3 for U (i / U (i = eikr, r U (i = eikr r U = 0. (i, (12a 4 (12b [ ik 1 ] cos(n, r. (13 r Here U (i represents a spherical wae originating from P 0 and (n, r is the angle between n and r (see Fig. 2 (bottom. Because of Eq. (12b, we find that the integral oer B is zero. The integral oer C is zero, as well, as we are free to choose the radius R so large that light from P 0 has not yet reached the surface C at the time of interest. Thus the disturbance on C is zero, so is the integral, too. Therefore from Eqs. (11 and (13 we hae U(P = 1 [ U (i 4π ( ] e iks eiks (i U ds = 1 s s 4π [ 1 e ikr e iks 4π r s ( ik 1 s [ ( ( e ikr e iks eiks e ikr r s s r cos(n, s eiks e ikr ( ik 1 s r r ] ds = cos(n, r ] ds. (14 Now we assume that the waelength is ery small compared to the distances r and s. Thus we can approximate ik 1 r,s with ik and finally we arrie at the Fresnel-Kirchhoff diffraction formula, alid for λ r, s U(P = i e ik(r+s [cos(n, s cos(n, r] ds. (15 2λ rs This formulation allows us to improe a bit our understanding of the Huygens-Fresnel principle: a light wae falling on the aperture propagates as if eery element ds from emitted a spherical waelet, described by e iks /s. The amplitude and the phase of this waelet depend on the distance r between ds and P, since the waelet is drien by the spherical wae, emitted from P and described by e ikr /r. The factor naturally accounts for the intensity, and the cosine factors correspond to Lambert s law of surface brightness. II. FRESNEL S DIFFRCTION ND FRUNHOFER S DIFFRCTION Summary: We derie Fresnel s and Fraunhofer s diffraction integrals. 1. We approximate Eq. (15 for a r, s (small apertures to derie Fresnel s diffraction formula (16 2. We make a stronger assumption that a 2 /λ r, s (een smaller apertures to derie Fraunhofer s diffraction formula (24 Now that we hae deried the Fresnel-Kirchhoff diffraction formula (15, we are ready to calculate diffraction. Though we can treat this problem numerically with ease, an analytic solution to this integral is highly desirable but yet unknown in the general case for arbitrary apertures and positions of P 0 and P. Therefore in this section we will seek further approximation to the formula (15. Now we assume that the linear dimensions of the opening,

5 although large compared to λ, are small compared to r and s. Combined with the preious assumption we now hae λ a r, s, where a is a characteristic size of the opening (e.g. radius, if it is circular, or the longer side for rectangular. In the following we apply this assumption to approximate each of the terms inside the integral formula (15. Due to the approximation a r, s, the factor cos(n, s cos(n, r aries only slightly inside the aperture (cf. Fig. 3. We can then safely assume that cos(n, s cos(n, s and cos(n, r cos(n, r, so that cos(n, s cos(n, r cos(n, s cos(n, r is simply a constant. On the other hand, due to the approximation λ a r, s, as the element ds explores the domain of integration, r +s will change by a huge number of waelengths, so that the factor e ik(r+s will oscillate rapidly. Finally, since a r, s, the factor 1/rs may be replaced by 1/r s (Fig. 3. s a result the integral (15 reduces to Fresnel s diffraction formula, alid for λ a r, s U(P i cos(n, s cos(n, r 2λ r s e ik(r+s ds. (16 This integral describes Fresnel s diffraction. Unfortunately, only numeric treatment of this integral is possible. In the following, based on our assumption a r, s, we will approximate e ik(r+s too, thereby deriing a simplified integral describing Fraunhofer diffraction. y 5 P 0 r r` O Q s s` x P Screen z FIG. 3: Diffraction at an aperture in a plane screen. We introduce Cartesian coordinates with origin O, where the xy-plane and the aperture plane coincide and positie z direction points into the half-space containing P. n arbitrary point Q within the aperture is specified by the coordinates (ξ, η. We introduce a Cartesian reference system with origin O in the aperture and with the x- and y-axes in the plane of the aperture and choose the positie z direction to point in the half-space that contains the point P (Fig. 3. If (x 0, y 0, z 0 and (x, y, z are the coordinates of P 0 and P, respectiely, and (ξ, η are the coordinates of an arbitrary point Q in the aperture, we hae Hence we find r 2 = (ξ, η, 0 (x 0, y 0, z 0 2 = (x 0 ξ 2 + (y 0 η 2 + z 2 0, s 2 = (ξ, η, 0 (x, y, z 2 = (x ξ 2 + (y η 2 + z 2, r 2 = (0, 0, 0 (x 0, y 0, z 0 2 = x y z 2 0, s 2 = (0, 0, 0 (x, y, z 2 = x 2 + y 2 + z 2. r 2 = r 2 2(x 0 ξ + y 0 η + ξ 2 + η 2, s 2 = s 2 2(xξ + yη + ξ 2 + η 2. We can neglect the term ξ 2 + η 2 in the general case as a r, s, while the terms x 0 ξ + y 0 η and xξ + yη depend on the positions of P 0 and P (the distances from the z axis and can hae considerable alues. Therefore we write r 2 r 2 2(x 0 ξ + y 0 η, s 2 s 2 2(xξ + yη. (18a (18b (19a (19b

6 Now we use that a r, s and expand r and s in power series, where we use the formula x 1 + x = O(x2. (20 We obtain r r x 0ξ + y 0 η r, (21a s s xξ + yη s, (21b where we hae neglected the higher orders assuming that a 2 /λ r, s. We substitute r and s into the integral (15: where Thus we derie the formula for U(P = C C = i 2λ e ik( x 0 ξ+y 0 η r + xξ+yη s dξdη, (22 cos(n, s cos(n, r r s e ik(r +s. (23 Fraunhofer s diffraction, alid for λ a r, s and a 2 /λ r, s U(p, q U(P = C e ik(pξ+qη dξdη, (24 where p = x 0 r + x s, q = y 0 r + y s. (25 The integral (24 describes Fraunhofer s diffraction. Note that p and q depend on the directions of P 0 and P measured from the origin O. In fact, p = q = 0 describes the central direction, i.e. the situation when P 0 P contains O. Such P, fulfilling p = q = 0 for fixed P 0, we will call the centre of the diffraction pattern. We expect that the brightest spot is located at this centre. Let us denote with I 0 the intensity at the centre, I 0 = U(0, 0 2. From Eq. (24 we hae U(0, 0 = C dξdη = CD, (26 where D is the area of the aperture. Hence we find a simple physical meaning for C: I0 C = D. (27 6. Fresnel s number F The integral (24 describes Fraunhofer s diffraction and was obtained by neglecting quadratic and higher order terms in ξ and η in Eq. (21, under the assumption a 2 /λ r, s. When this assumption does not hold, we hae to use the more complicated formula (16 describing the more general case of Fresnel s diffraction. To distinguish between both types of diffraction we introduce Fresnel s number F = a2 λs. (28 Fresnel s diffraction and Fraunhofer s diffraction are obsered for F 1 and F 1, respectiely. B. Remark Fourier transform The Fraunhofer integral (24 can be rewritten in the way U(p, q U(P = C M(ξ, ηe ik(pξ+qη dξdη, (29 where the integration is oer an infinite plane in the aperture plane and M(ξ, η describes the aperture: M(ξ, η = 1 if (ξ, η and 0 otherwise. Hence the diffraction image U(p, q is the Fourier transform of the aperture s shape M(ξ, η. Thus one can encode a secret message, which can be decoded with a laser beam. :

7 7 III. FRUNHOFER S DIFFRCTION SIMPLE EXMPLES In the following section we will study the Fraunhofer diffraction by rectangular and circular openings. a б FIG. 4: Rectangular aperture (a and circular aperture (b.. Rectangular aperture Consider a rectangular opening with sides 2a and 2b and centre O, as shown in Fig. 4 a. Now that we hae deried the Fraunhofer integral (24 it is straightforward to obtain the diffraction pattern: Note that a b a b U(p, q = C e ik(pξ+qη dξdη = C e ik(pξ+qη dξdη = C e ikpξ dξ e ikqη dη. (30 a b a b a a e ikpξ dξ = e ikpa e ikpa ikp = 2 sin kpa. (31 kp We obtain U(p, q = 4ab C sin kpa sin kqb kpa kqb. (32 Hence for the intensity we get I(p, q U(p, q 2 = I 0 ( sin kpa kpa 2 ( 2 sin kqb, (33 kqb where I 0 = 4ab C is the intensity at the centre of the pattern, where p = q = 0 (cf. Eq. (27. Let us inestigate the function y = (sin x/x 2, the base of our result, shown in Fig. 5 a (top. The principal maximum is for x = 0 and y = 0 for x = ±π, ±2π,.... The secondary maxima occur when x tan x = 0. s x increases, the solution to this equation asymptotically approaches x m = (2m + 1π/2, where m is integer. Can you figure out why? Thus we see that the intensity I(p, q, shown in Fig. 5 a (bottom, is zero along two sets of lines parallel to the sides of the rectangle, gien by kpa = ±mπ, kqb = ±nπ (m, n = 1, 2, 3,... (34 Homework: Find the diffraction from a single slit, which is effectiely an infinite linear source (as opposed to a point source, e.g. a luminous wire, the light from which is diffracted by an infinitely long narrow slit, parallel to the source. Let both be parallel to the y-axis. Hint: for coherent (incoherent source you must integrate the disturbance (32 (the intensity (33. Since q = y/s + y 0 /r, where y 0 /r is fixed and is defined by the position of a particular element from the wire, the integral is oer q from to.

8 8 a FIG. 5: Top left: The function y = (sin x/x 2. The zeroes occur for x = ±π, ±2π,..., while the secondary maxima asymptotically reach the alues x m = (2m + 1π/2 for integer m. Bottom left: Simulated Fraunhofer diffraction pattern from a rectangular aperture. The picture is courtesy of Hecht, Eugene, Optics, 2nd Ed, ddison Wesley, Top right: The ] 2. function y = Bottom right: Simulated Fraunhofer diffraction from a circular aperture. [ 2J1 (x x B. Circular aperture Now we study the Fraunhofer diffraction from a circular opening, shown in Fig. 4 b. gain we need to sole the integral (24, where now we use polar coordinates (ρ, θ to specify a point in the aperture: ρ cos θ = ξ, ρ sin θ = η. (35 Before we calculate the integral, let us first write the expression ik(pξ + qη in the new coordinates and do a bit of simplification: ik(pξ + qη = ikρ(p cos θ + q sin θ = ikρ ( p 2 + q 2 p p2 + q cos θ + q 2 p2 + q sin θ. (36 2 Let us introduce a new ariable w = p 2 + q 2. (37 Because we hae 1 p/w 1, 1 q/w 1 and (p/w 2 + (q/w 2 = 1, we can set p/w = cos ψ and q/w = sin ψ. From now on we will specify the obserational point P by the polar ariables (w, ψ rather than (p, q. We find ik(pξ + qη = ikρw (cos ψ cos θ + sin ψ sin θ = ikρw cos(θ ψ. (38 Note that w is the sine of the angle which the direction (p, q makes with the central direction p = q = 0; w represents the sine of the angle between a point P and the centre of the diffraction pattern (see ppendix D for proof; kind of obious for x 0 = y 0 = 0. The integral now becomes U(w, ψ = C a 2π 0 0 e ikρw cos(θ ψ ρdρdθ, (39

9 9 where we hae used the Jacobian relating Cartesian to polar coordinates: dξdη = ρdρdθ. Note that 2π 0 e i x cosα dα = 2πJ 0 (x, (40 where J 0 (x belongs to the family of the Bessel functions J n (x. Thus we obtain (we drop the ψ from U, as U has become independent of ψ U(w = 2πC which finally yields (here I used Wolfram Mathematica For the intensity we obtain a 0 J 0 (kρwρdρ, (41 U(w = Cπa 2 2J 1(kaw kaw. (42 ( 2 I(w U(w 2 2J1 (kaw = I 0. (43 kaw [ ] 2 The function y = 2J1(x x is plotted in Fig. 5 b (top. Note that the central maximum is ery strong compared to the rest and comprises almost 98% of all diffracted light. Therefore the net effect of the diffraction through a circular aperture is to spread a point source of light, such as a star, into a larger circular shape, as shown in Fig. b 5 (bottom. In this sense the first zero, located at kaw 1.22π, is a good indicator and fundamentally defines the resolution of an optical imaging system. Hence we obtain the limiting angle under which two stars appear as a single spot on the diffraction screen, w min = 1.22 λ, d = 2a. (44 d C. Extending or contracting the aperture Consider aperture, which corresponds to diffraction image U(p, q gien by the integral (24: U(p, q = C e ik(pξ+qη dξdη. (45 Now let us extend or contract the aperture in one direction by factor µ, say in ξ direction. We can easily obtain the diffraction image from the new aperture. For illustratie purposes we consider rectangular aperture, though the procedure for any aperture is absolutely the same. The original diffraction image is gien by (Eq. (30 and the new image is gien by a b U(p, q = C e ik(pξ+qη dξdη (46 a b µa b µa b ( U (p, q = C e ik(pξ+qη dξdη = µc e ik(µp ξ µ ξ a b +qη d dη = µc e ik(µpξ +qη dξ dη. (47 µa b µa b µ a b Thus we obtain U (p, q = µu(µp, q, (48 which means that the image is contracted in p-direction by a factor of µ and the intensity is increased by µ 2. See Fig. 6 for comparison of diffraction at different apertures.

10 10 FIG. 6: Top: Various apertures square, rectangular, circular and elliptic. Bottom: The corresponding diffraction images. Extending the aperture in one dimension results in contracting the diffraction image in the same dimension by the same factor. D. Multiple diffraction openings Consider Fraunhofer diffraction from a screen containing a large number of identical openings. Does the result remind you of the interference principle? Can you derie diffraction from a grating? IV. FRUNHOFER S DIFFRCTION DIFFRCTION GRTINGS p FIG. 7: Left: Diffraction grating. Plane wae falls at angle θ 1, while the iewer s angle θ 2 is ariable. Right: Intensity s p for N=5, s = λ/2π and d = 4s = 2λ/π. The diffraction image from a grating is obtained with the integral (24 U(p, q = C e ik(pξ+qη dξdη, (49 with p = sin θ 2 sin θ 1 (see Fig. 7. The aperture is a collection of N slits and the grating extends from 0 to a. Because the diffraction problem from a grating is one-dimensional, we perform the integration oer η right away. We

11 11 obtain U(p = C N nd n=1 nd s e ikpξ dξ = C ikp = ic kp N n=1 We note that 1 e ix = 2ie ix/2 sin x/2, yielding The intensity is U(p = sc e ( e ikpnd e ikp(nd s = ( 1 e ikps N 1 e ikpd (e ikpd n = ic ( kp e ikpd 1 e ikps 1 e ikpdn. (50 1 e ikpd n=0 ikp/2(d s d(n 1 sin(kps/2 I(p U(p 2 = I 0 ( sin(kps/2 kps/2 kps/2 sin(kpdn/2 sin(kpd/2. (51 2 ( 2 sin(kpdn/2 (52 sin(kpd/2 and is shown in Fig. 7. Here I 0 = I(0. Show that I 0 N 2. ( 2 From lim x 0 sin x/x = 1 we find that sin(kpdn/2 N sin(kpd/2 attains its maximal alue 1 for kpd/2 = mπ, i.e. for p = mλ/d. The enelop s zeros occur for kps/2 = mπ, i.e. for p = mλ/s.. Spectral resolution of a grating 1.0 space to first minimum FIG. 8: The limit of resolution is determined by the Rayleigh criterion as applied to the diffraction maxima, i.e., two waelengths λ r and λ b are just resoled when the maximum of one lies at the first minimum of the other. Consider two waelengths λ r and λ b, shown in Fig. 8. They can be resoled if the maximum of λ b lies at the first minimum of λ r. The maximums are located at mλ r /d and mλ b /d, respectiely, where m = 1 in the figure, and the separation is m(λ r λ b /d. The distance to the first minimum is λ r /Nd. Therefore λ r and λ b can be resoled if The resoling power is defined as R = m(λ r λ b d λ λ r λ b, so we hae λ r Nd. (53 R = mn. (54

12 12 PPENDIX : WVE EQUTION IN VCUUM Here we derie the wae equation for the E field in a region with no charges (ρ = 0 and no currents (j = 0. We use Maxwell s equations: We operate with on both sides of Eq. (1c to get E = 0, B = 0, ( E = E = B t, B = 1 c 2 E t. where we hae exchanged the time and the spacial deriaties. Note that (1a (1b (1c (1d ( B = ( B, (2 t t ( E = ( E E = 2 E, where we hae used Eq. (1a. Using the aboe identity and Eq. (1d we obtain 2 E = ( 1 E t c E t c 2 t 2. (3 (4 Hereby we arrie at the wae equation for acuum (Eq. (1. PPENDIX B: SOLUTION TO THE WVE EQUTION 1. The wae equation We consider a solution to Eq. (1, which represents a spherical wae, as we deal with such waes (cf. Fig. 1. Spherical waes originate from a point and thus hae central symmetry, i.e. V only depends on r = r = x 2 + y 2 + z 2 and t. Therefore, for conenience, we switch to spherical coordinates, where a spherical solution has the form V = V (r, t. It is explicitly gien as V (r, t = r e i(ωt kr. (B1 Though simple, we will not consider the mathematical deriation of this solution, but rather we will exercise our physical intuition to understand it. We know that the phase of a spherical wae is constant oer a sphere, and this is what the exponential factor accounts for. We also know that the intensity of the wae I(r, t (note that I V 2 must decrease as 1/r 2 with r. Hence the factor 1/r. defines intensity. non-monochromatic analogue is simply a sum (or integral oer the frequency range ω with amplitude (ω. We can separate Eq. (B1 into a position-dependent and a time-dependent factor, i.e. V (r, t = U(re iωt, where U(r = r eikr, (B2 which is Eq. (2 in spherical coordinates, when central symmetry is imposed. 2. The Helmholtz equation The Helmholtz equation (3 is deried by plugging V (r, t from Eq. (2 (or Eq. (B2 if we want to impose spherical symmetry into the wae equation (1. We only need to calculate the time deriatie: 2 t 2 V = ω2 V. (B3

13 13 Thus we get, using Eq. (B2, 2 V 1 c 2 2 V t 2 which is exactly Eq. (3 for k = ω/c. 2 (Ue iωt + ω2 c 2 (Ue iωt ( 2 U + k 2 Ue iωt = 0, (B4 3. Directional deriaties ( For the calculation of the directional deriatie e iks s in Eq. (11, we hae replaced / with / s oer S, since at each point from S the ector n points along the direction of increasing s (with s being the distance from P. Thus we hae ( e iks = ( e iks = eiks s s s s lternatiely, we could hae used the definition of a directional deriatie = n. ( ik 1. (B5 s It is ery helpful in the more general case of Eq. (13, where the deriatie is worked out as follows: U (i ( ( e ikr = n = n ˆr e ikr [ = (n ˆr eikr ik 1 ] [ = eikr ik 1 ] cos(n, r, r r r r r r r where n ˆr = cos(n, r, ˆr = r r is a unit ector parallel to r. (B6 (B7 PPENDIX C: PROOF OF EQ. (4 We will proe that First note that so that Therefore we hae ( (U 2 V V 2 Ud = U V S V U ds. (a = a + ( a (, a = (a ( a (. U 2 V = U ( V = (U V ( U ( V Now from Eq. (C4 we obtain (U 2 V V 2 Ud = ( (U V ( U ( V (V U + ( V ( Ud = ( (U V (V Ud = (U V V Ud. By using the diergence theorem d = S ds = n ds, where ds = nds, with n being a normal ector to ds, we continue Eq. (C5: (U 2 V V 2 Ud = n (U V V UdS = (Un V V n UdS = S where we hae used the definition for directional deriatie (B6. Thus we hae proed Eq. (4. s s S ( U V V U ds, (C1 (C2 (C3 (C4 (C5 (C6 (C7

14 14 PPENDIX D: POLR COORDINTES (w, ψ IN THE DIFFRCTION PTTERN Here we proe that w is approximately the sine of the angle which the direction (p, q makes with the the central direction p = q = 0, as we claimed in Sec. III B. Note that the ector corresponding to the direction (p, q is s, whereas the ector corresponding to the central direction (p = 0, q = 0 is r. The sine of the angle between the two ectors is gien by the length of their cross product sin( s, r = sin(s, r = s s r r. (D1 Recall that s = (x, y, z and r = (x 0, y 0, z 0 (cf. Fig. 2 (bottom and Eq. (17. We neglect (x/z 2, (y/z 2, (x 0 /z 0 2, (y 0 /z 0 2 and thereby obtain On the other hand, sin(s, r (x 0z xz (y 0 z yz 0 2 z 2 z0 2. (D2 w = p 2 + q 2 = (x0 r + x s 2 + ( y0 r + y s 2, (D3 where r and s are gien by Eq. (17. gain we neglect (x/z 2, (y/z 2, (x 0 /z 0 2, (y 0 /z 0 2 and thereby obtain w (x 0z xz (y 0 z yz 0 2 z 2 z0 2. (D4 Hence we proe that w sin( s, r, so that w gies the angular distance between a point P and the centre of the diffraction pattern. PPENDIX E: LIST OF SSUMPTIONS Here is a (not comprehensie list of the assumptions we made concerning the deriation of: The Helmholtz-Kirchhoff theorem (10: Monochromatic wae: In Sec. I we deried the theorem for monochromatic waes. It can be generalized relatiely easy [1]. The Fresnel-Kirchhoff s diffraction formula (15: The contour of integration S: In Secs. I and I B we assumed that i in the aperture the disturbance U is the same as if there is no screen at all; ii in the nonilluminated region B, U and its deriatie are zero. Short waelength: In Sec. I we assumed that the waelength λ is much shorter than the distances r and s, i.e. λ r, s. The Fresnel diffraction formula (16: Small aperture size: In Secs. I, I B and II we assumed that the linear dimensions of the opening, although large compared to λ, are small compared to r and s, i.e. a r, s, where a is a characteristic size of the aperture (radius for circular, diagonal for rectangle, etc.. The Fraunhofer diffraction formula (24: Een smaller aperture size: In Secs. I, I B, II for the deriation of the Fraunhofer diffraction formula (24 we assumed that a 2 /λ r, s, which is a stronger condition than a r, s. [1] M. Born and E. Wolf, Principles of Optics, 7th edition (1999.