4.5 The Reflected Field
81
We note that reflected field also satisfies the Helmholtz equation. Let us pass to the
coordinate system (x
, y
, z
), related to the reflected field:
kn 1 x = k 11 x
+ k 21 y
+ k 31 z
,
kn 1 y = k 12 x
+ k 22 y
+ k 32 z
,
kn 1 z = k 13 x
+ k 23 y
+ k 33 z
,
where k 11 , k 12 , k 13 , k 21 , k 22 , k 23 , k 31 , k 32 , k 33 are determined by relations (4.52)–
(4.60). In this coordinate system, the reflected field takes the form
E re f = exp(−i(k + O(ε
4
))z
)
1
(2π) 2
∞
−∞
∞
−∞
dk
∧
1x dk
∧
1y ×
× exp
−z
i
−
ε
2 y
2 k
∧
1y
2k 2 n
2
1
−
ε
2 x
2 k
∧
1x
2k 2 n
2
1
+ O(ε
3
)
×
× exp[ik
∧
1x ξ
1 (a
2
11 + a 22 a 12 + a
2
13 ) + ik
∧
1y ξ
1 (a 21 a 11 + a
2
22 + a 23 a 13 )+
+ik
∧
1x ξ
2 (a 11 a 21 + a
2
22 a 13 a 23 ) + ik
∧
1y ξ
2 (a
2
21 + a 12 a 22 + a
2
23 )]A(ξ 1 , ξ 2 , ξ 3 , k 1x , k 1y )×
×
∞
−∞
∞
−∞
dξ
∧
1 dξ
∧
2 exp(−ik
∧
1x ξ
∧
1 − ik
∧
1y ξ
∧
2 )Φ(ξ
∧
1 , ξ
∧
2 ).
(4.64)
In order to calculate the field of the reflected beam, one needs to reexpand the
amplitude A in terms of a small parameter, substitute this expansion into the formula for the reflected field, and perform the integration. In the coordinate system
(x
, y
, z
), the reflected field in the section z
= 0 related to the ray takes the form
E re f =
1
(2π) 2
∞
−∞
∞
−∞
dk
∧
1x dk
∧
1y exp[ik
∧
1x ξ
1 (a
2
11 + a 22 a 12 + a
2
13 )+
+ik
∧
1y ξ
1 (a 21 a 11 + a
2
22 + a 23 a 13 )+
+ik
∧
1x ξ
2 (a 11 a 21 + a
2
22 + a 13 a 23 ) + ik
∧
1y ξ
2 (a
2
21 + a 12 a 22 + a
2
23 )]×
×A(ξ 1 , ξ 2 , ξ 3 , k 1x , k 1y )
∞
−∞
∞
−∞
dξ
∧
1 dξ
∧
2 exp(−ik
∧
1x ξ
∧
1 − ik
∧
1y ξ
∧
2 )×
× Φ(ξ
∧
1 , ξ
∧
2 ) + O(ε
2
)
(4.65)
By expanding A in terms of a small parameter ε at z
= 0, we have