source Io/2 located at y= h and a line source -lo/2 located at y= -h. We solve each of these problems by exploiting the appropriate symmetry, and superpose the results to find the solution to the original problem For the even-symmetric case, we begin by using(4.407) to represent the field EI (,)=-x+A For y> h this becomes E(x, y, o= 2 cos k The secondary(scattered) field consists of waves propagating in both the ty-directions +jΔ [A*(r, o)- /k, '+A(kx,oeky ]e ik dkr.(5.7) ∞+j△ The impressed field is even about y =0. Since the total field E,= E;+Es must be even in y(E is parallel to the plane y=0), the scattered field must also be even. Thus, A+=a and the total field is for y > h E 2A(kr, a)cos ky,'so lo(o)2 cos ky h -jk, e -jk, dks Now the electric field must obey the boundary condition E,=0 at y=+d. How since E is even the satisfaction of this condition at y= d automatically implie atisfaction at y=-d. So we set 2A*(, o)cos kyd-a lo(o)2 cos k,h and invoke the Fourier integral theorem to get AT(Kx, o)=op lo(o) cos kyh The total field for this case is E ∞+ For the odd-symmetric case the impressed field is El (x, y, o) 一 hksi dkx ②2001source ˜I0/2 located at y = h and a line source −˜I0/2 located at y = −h. We solve each of these problems by exploiting the appropriate symmetry, and superpose the results to find the solution to the original problem. For the even-symmetric case, we begin by using (4.407) to represent the impressed field: E˜ i z(x, y,ω) = −ωµ˜ ˜I0(ω) 2 2π ∞+  j −∞+ j e− jky |y−h| + e− jky |y+h| 2ky e− jkx x dkx . For y > h this becomes E˜ i z(x, y,ω) = −ωµ˜ ˜I0(ω) 2 2π ∞+  j −∞+ j 2 cos kyh 2ky e− jky y e− jkx x dkx . The secondary (scattered) field consists of waves propagating in both the ±y-directions: E˜ s z(x, y,ω) = 1 2π ∞+  j −∞+ j A+(kx ,ω)e− jky y + A−(kx ,ω)e jky y  e− jkx x dkx . (5.7) The impressed field is even about y = 0. Since the total field Ez = Ei z + Es z must be even in y (Ez is parallel to the plane y = 0), the scattered field must also be even. Thus, A+ = A− and the total field is for y > h E˜ z(x, y,ω) = 1 2π ∞+  j −∞+ j  2A+(kx ,ω) cos ky y − ωµ˜ ˜I0(ω) 2 2 cos kyh 2ky e− jky y  e− jkx x dkx . Nowthe electric field must obey the boundary condition E˜ z = 0 at y = ±d. However, since E˜ z is even the satisfaction of this condition at y = d automatically implies its satisfaction at y = −d. So we set 1 2π ∞+  j −∞+ j  2A+(kx ,ω) cos kyd − ωµ˜ ˜I0(ω) 2 2 cos kyh 2ky e− jkyd  e− jkx x dkx = 0 and invoke the Fourier integral theorem to get A+(kx ,ω) = ωµ˜ ˜I0(ω) 2 cos kyh 2ky e− jkyd cos kyd . The total field for this case is E˜ z(x, y,ω) = −ωµ˜ ˜I0(ω) 2 2π ∞+  j −∞+ j  e− jky |y−h| + e− jky |y+h| 2ky − − 2 cos kyh 2ky e− jkyd cos kyd cos ky y  e− jkx x dkx . For the odd-symmetric case the impressed field is E˜ i z(x, y,ω) = −ωµ˜ ˜I0(ω) 2 2π ∞+  j −∞+ j e− jky |y−h| − e− jky |y+h| 2ky e− jkx x dkx ,