Diffraction. S.M.Lea. Fall 1998

Transcription

1 Diffraction.M.Lea Fall 1998 Diffraction occurs wen EM waves approac an aperture (or an obstacle) wit dimension d > λ. We sall refer to te region containing te source of te waves as region I and te region containing te diffracted fields as region II. Te total volume comprises bot regions I and II. egion II is bounded by a surface = Te surface 2 is at infinity, wile 1 is a containing apertures, or else a set of obstacles. 1 Kircoff s metod We may write all te fields in te form ψ / e iωt and ten ψ satisfies te Helmoltz equation r 2 + k 2 ψ = 0 1

2 were k = ω/c and ψ is a component of E or B, or anoter variable describing te field, suc as a component of A. We can solve te Helmoltz equation wit te outgoing wave Green s function. ince tere are no sources in region II, te volume integral is zero. ψ ( x) = ψ ( x 0 )^n r 0 G G^n r 0 i ψ ( x 0 ) (1) were Tus G = 1 e ik ψ ( x) = 1 e 1+2 ik µ ψ ( x 0 ) ik 1+ i ^n ^ + ^n r k 0 ψ ( x 0 ) (2) Te integral reduces to an integral over 1, since ψ» 1/ at 1. We can evaluate te integral using te Kircoff approximation: 1. ψ and ψ n vanis on 1 except in te openings, and 2. in te openings, ψ and ψ are equal to te values in te incident field. n trictly speaking, tis is not a matematically sound procedure, since if bot ψ and ψ n vanis on any finite surface, ten te solution ψ must be zero. Also te solution we obtain does not yield te assumed values of ψ on 1. Yet te solution does a pretty good job of reproducing experimental results. We can partially fix tings up by using a better Green s function. If ψ is known (or approximated) on 1, we sould use a Diriclet Green s function, for wic G D ( x, x 0 ) = 0 for x 0 on 1. Ten te solution is: ψ ( x) = 1 ψ ( x 0 ) ^n G D (3) imilarly, if ψ n is known or approximated on 1, we sould use te Neumann Green s function, G N n = 0 on 1. Ten ψ ( x) = G^n ψ ( x 0 ) (4) 1 If te 1 is plane we can easily find te appropriate Green s functions using te metod of images: Ã! G D,N = 1 e ik eik 0 0 were = x x 0 and 0 = x x 00 and x 00 describes te point tat is te mirror image of x 0 in 1, and we let bot points approac te surface 1. Ten and G N n 0 = G N z 0 = 1! 2z0 e ik G D! 0 as z 0! 0 (z z 0 ) µ 1 ik as z 0! (z + µ z0 ) ik 1 e ik

3 and tus equation (3) becomes: ψ ( x) = ψ ( x 0 ) 2z e ik µ 1 1 ik = 1 ψ ( x 0 ) z e ik µ 1 1 ik (5) wit z/ = cos θ, wile equation (4) becomes: ψ ( x) = 1 e 1 ik z 0ψ ( x0 ) (6) Now to estimate te differences between tese expressions, we coose a point source in region I wit coordinate so tat te wave from te source travels in direction ^ to reac te center of a small aperture at te origin. Ten we can write e i k ψ inc = ψ 0 = ψ e ik(r0+^r0 x0 ) e ikr0 e ik^r0 x 0 0 = ψ 0 and ψ inc n = ψ e ikr0 e ik^r0 x0 0 ( ik~ ^n) were ^n = cos (π θ 0 ) = cosθ 0. Ten te 3 expressions for ψ may be written: outgoing wave Greens function: ψ ( x) = 1 e 1+2 ik µ ψ ( x 0 )ik 1 + i ^n k ^ + z 0 ψ ( x0 ) ' ψ 0 e 1+2 ikr e ikr0 e ik^r0 x0 r ik^n ^ + ( ik ^n) ' ik ψ 0 e 1+2 ikr e ikr0 [cosθ 0 + cos θ] r Diriclet Green s function: and Neumann Green s function: ψ ( x) = ik ψ 0 ψ ( x) = ik ψ 0 e 1 ikr e ik r e 1 ikr e ik r cos θ cos θ 0 i If te aperture is small compared wit te distances r and, ten te factors cosθ and cos θ 0 are essentially constant trougout te range of integration. Ten all 3 expressions give te same diffraction pattern: te overall intensity smplitude differs by small factors of order 1. Tus we can use wicever of te expressions is most convenient. 3

4 2 Fraunoffer and Fresnel diffraction Tee exponential tat appears in equation (2) may be approximated: k = k j x x 0 j = k p r x x r 0 = kr 1 + r02 x0 2^r r2 r = kr k^r x 0 k r 2 02 (^r x 0 ) + 2r Tus µ e ik = e ikr e ik^r x0 exp i k r 2 02 (^r x 0 ) 2r Te te exponent in te second term is of order: k^r x 0 = d λ were d is te dimension of te aperture, and its ratio to te first term is of order d/r.. Te exponent in te tird term is of order d 2 λr In Fraunauffer diffractionwe ignore te tird term. Tis means tat we need d 2 λr 1 and tird term second term = d r 1 o we need te observation point to be a long way from te aperture. If d/λ > 1, te tird term may become important. Tis is te regime of Fresnel diffraction. Fresnel diffraction occurs for d d λ & r À d Typically we find tat Fresnel diffraction patterns are more complex tan Fraunoffer patterns. Most of te problems in te text refer to te Fraunoffer regime. 3 Vectorial diffraction teory A more careful analysis requires tat we consider te vector nature of te fields. Tus we replace equation (2) wit te vector relation: E ( x) = E ^n 0 G G ^n 0 i E First we rewrite te integrand as: 2 E ^n 0 G ^n 0 r 0 GE 4

5 And ten we convert te surface integral of te second term to a volume integral. ecall tat te normal ^n 0 is inward: ^n 0 G~E = r 2 G~E dv 0 s V i = r r G E r r G E dv V i = ^n 0 r G E ^n 0 r G E were r G E = G r E + E rg = 0 + E rg since tere is no carge density on,and r G E = rg E + G r E = rg E+G ik B Tus we ave: E( x) = = = = 2 E ^n 0 i G ^n 0 E G + ^n 0 r 0 G E+G ik B 2E ^n 0 r 0 G ^n 0 0 E r G + r 0 G ^n 0 E E ^n 0 r 0 G + Gik^n 0 B i E ^n 0 i G ^n 0 E G + G ^n 0 E + ik^n 0 BG ^n 0 E r 0 G + ^n 0 E 0 r G + ik^n 0 BG i Now we can take G = e ik / and proceed as wit te scalar teory. We can make te same Kircoff approximations, and te same inconsistencies remain. For very large, we ave G = eikr0 e ik^ ~x ' eikr0 Now if we coose 2 to be a large emispere of radius! 1, ten^ = ^n 0, and and te integral over 2 becomes: ik 2 = ik since 2 for outgoing waves, and tus: ^n 0 E G = ik^n 0 G ^n 0 G + ^n 0 E ^n 0 G ^n 0 BG i E ^n 0 Bi G = 0 i k E = ik B ^ E = B= ^n 0 E 5

6 and so ^n 0 B = ^n 0 ^n 0 E = E on 2. Tus te integral reduces to an integral over 1. Now if our observation point P is very distant from 0 te sources on 1,ten as in te scalar teory we find G = eik ' eikr r exp i~k ~x and we expect te diffracted fields to ave a similar form E d = eikr F k, k0 r Ten te amplitude of te diffracted field may be written: F k, k0 = i ^n 0 E k + = i ^n 0 E k +k^n 0 B i i k ^n 0 E k ^n 0 E + k^n 0 B were te fields in te integral are also te diffracted fields. We know tat F must be perpendicular to k. In fact we can write te last two terms in te integrand, like te first, in te form k v for some vector v. First note tat k and from Ampere s law: giving: k ^n 0 B k ^n 0 B so te terms in te integrand are: i = k k ^n 0 B i ^n k 2 0 B k ^n 0 B = ^n 0 k B = ^n 0 ke k^n 0 B k i i = ^k k ^n 0 B ^k k ^n 0 B i = k ^n 0 E ^k k ^n 0 B ^n 0 E ~ = 1 k k k ^n 0 B i and tus F k, k0 = i k i k k ^n 0 B ^n 0 E 1 Te integrand also depends on k 0 if we use te incident fields in te apertures. 4 pecial case of plane conducting Let te conducting be te z = 0 plane, wit te incident fields propagating toward te from negative z. We can decompose te total electric field into tree parts: te incident field E 0, te field due to reflection by a solid conducting seet, E r, and te diffracted field E d. Ten E = E 0 + E r + E d = E 0 + E 1 6

7 Ten in egion II (z > 0) E 0 + E r = 0. Te fields E r + E d = E 1 are produced by currents in te conducting seet: j = j x^x+j y^y Te resulting vector potential is given by: r 2 + k 2 A = c j and consequently A z = 0. Also, since te operator on te LH is even in z, and te rigt and side is zero except for z = 0, ten A is even in z. Ten is odd in z. imilarly, is even in z. Tus we ave B y = r A E x = A x t y = A x z E x, E y, and B z are even in z and E z, B x, and B y are odd in z Te fields tat are odd in z need not be zero at z = 0 were te conducting exists, because of te surface carge density and currents in te. But in te apertures, tese odd fields must be zero. Tus in te apertures, B tangential and E normal are te incident fields alone. Now te matematical problem tat determines A is a Neumann problem, because A z = B tangential is determined (troug Ampere s law) by te currents in te. Tus we must use te Neumann Green s function: A= 1 e ik A z 0da0 A x = 1 e ik A x z 0 da0 = 1 e ik B y = 1 e ik A similar expression olds for A y, and tus B 1 = r A = r 1 ^z B 1 x da0 e ik ^z B 1 (7) ince ^z B 1 = 0 in te aperture, te integral is over te conducting material only. We can evaluate te integral on eiter side of te. For z < 0, te normal becomes ^z instead of ^z, tus confirming our expectations about te oddness of B tan. Now te source free Maxwell equations are invariant under te transformation B! E, 7

8 E! B, so we can write a similar expression for E 1 : E 1 = r 1 e ik ^z E 1 Now if we want te electric field for z < 0, we must cange te sign in front of te integral as well as te sign in front of ^z in order to obtain te correct symmetry. ince E 1z is zero in te aperture, we cannot simplify tis integral as it stands. However: E 1 = r 1 = r 1 e ik e ik ^z E1 + E 0 E 0 ^z E r 1 e ik ^z E 0 were E ~ is te total electric field, wose tangential component vanises on te conducting surface. Tus te first integral is an integral over te oles alone. Te second term r 1 e ik ^z E 0 = E 0 for z > 0 Tus E = E 1 + E 0 = E d = r 1 e ik ^z E (8) oles were te field in te integrand is te total electric field in te aperture. In practical applications we can approximate by using te incident fields as we ave indicated previously. Equation (8) is te vector myte Kircoff relation. 5 Babinet s principle Babinet s principle relates te diffraction patterns due to 2 complementary s and c. Te complementary c as apertures were is solid and vice versa. Tus te sum + c is a solid surface. 5.1 calar principle For any complete closed surface we can write a matematical identityfor any scalar function ψ : equation (1) or, using te outgoing wave Green s function, equation (2) Tus: ψ ( x) = 1 e +c+2 ik = 1 +c e ik µ ψ ( x 0 ) ik 1+ i k ψ ( x 0 )ik µ 1+ i k ^n ^ + ^n ψ ( x 0 ) ^n ^ + ^n r 0 ψ ( x 0 ) were as usual, te integral over 2 vanises. Tis result is exact. Now, using te Kircoff approximation, te integral over c (te apertures in ) gives te diffracted field ψ a in region II due to te. imilarly te integral over gives te diffracted field ψ b due 8

9 to te complementary c. Tus ψ = ψ a + ψ b Now in directions were ψ = 0,we ave ψ a = ψ b, and tus te diffraction pattern, wic depends on ψ 2, is te same. Tus for example te diffraction pattern due to a ole of radius a and due to a disk of radius a are te same except straigt aead, were te incident field is non zero. 5.2 Vector principle A more careful statement can be made for a plane conducting. First we define te two situations tat are complementary: 1. Te incident fields are E 0, B 0 and te is Te incident fields are E c = B 0, B c = E 0, and te complementary c. Ten we can use equation (8) to find te diffracted field in problem 1: E 1 = r 1 e oles ik ^z E 1 in 0 For problem 2, we use equation (7): B 2 = r 1 conducting part of c e ik ^z B 2 Tese two integrals are taken over te same surface. ince tese two expressions are matematically identical, ten E 1 = B 2 in region II. However te expression for B 2 gives te field due to te currents in te seet. Te total magnetic field in region 2 is B 2,tot = B c + B 2 = E 0 + E 1. Te intensity of te radiation is given by E pattern is te same for bot s werever E 0 is zero. A similar analysis sows tat E 2 = r 1 2 oles in c e ik or equivalently B 2. Tus te diffraction ^z E 2 = B 1 were ere again B 1 is te field due to te seet alone, and te minus sign appears so tat bot waves are outgoing. Tus B 1,tot = B 0 + B 1 = B 0 E 2 = ^k 0 E 0 E 2 wicleads to te same prediction wit regard to te diffraction pattern. Babinet s principle is used in te design of microwave antennae. A slot in a conducting plate radiates te same pattern as a metal strip. Tus slots in wave guides can serve as 9

10 effective microwave radiatiors. 10