Page 478 - Introduction to Information Optics

8.6. Near Field Optical Storage

Substituting /(x, y, z) with its Fourier components yields

2
2
(V + k ) -0.
Notice that (x, v) is the two-dimensional space domain, and («, v) is the
two-dimensional spatial-frequency domain. Changing the order'of integration
yields

d 2 2ni(ux+vy}
2
2
(2mu} + (2niv) + ~ + k F(u,i>, z)e du dv = 0. (8.14)
Since the Fourier transform (or inverse Fourier transform) of zero is also zero.
one obtains

2
2
(2muY + (2niv) + —. + k F(u, v, z) = 0, (8. .15)
r\'7 *• I
Remember k = 2n/l; one finally gets a wave equation for F(u, v, z) as follows:
2
- ;/(M (8.16)

The solution is

Notice that

izk
2
/(x, y, z) * /(x, y, 0)e ^i-*V + t= ). (8.18)
Equation (8.17) indicates that F(w, v, z) is a propagating wave, if

1
2 2
(8.19)

However, for

(8.20)

F(w, t\ z) is no longer a propagating wave, but is an evanescent wave. Thus,