# Near field and far field

In school, one often describes the diffraction pattern of a grating at a large distance. This is a phenomenon in the optical far field. In this case, the incoming wave fronts can, in a good approximation, be considered to be flat. At shorter distances to the grating the diffraction phenomena are more complex. In order to correctly describe phenomena in the optical near field one also has to take into account the curvature of the wave fronts.

In order to compute the wave propagation behind an arbitrary aperture, the Kirchhoff-Fresnel diffraction integral is used. Its derivation and meaning is explained here:

Extra: Mathematical background

## Mathematical background

As the stationary (time independent) Schrödinger equation is of the same form as the Helmholtz equation of electrodynamics, we find wave phenomena in quantum physics in analogy to classical optics.

If a wave with amplitude $$\psi_{\sigma}(x,y) = \psi_0(x,y) e^{i\phi(x,y)}$$ passes a diffractive structure (e.g. an aperture or a grating) with extension $$\sigma$$ new secondary waves emanate, according to Huygens principle, from every surface element $$d\sigma$$ which can contribute to the intensity at the point $$P(x’,y’)$$:

$$d\psi_p \propto \frac{\psi_{\sigma}d\sigma}{r}e^{-ikr}$$, where
$$k = 2\pi /\lambda$$ is the wave vector and $$r$$ the distance between source point and observation point.

Integrating over the area of the diffractive structure one obtains:
$$\psi_p \propto \int{\int{ \psi_{\sigma}\frac{e^{-ikr}}{r}dx dy}}$$.

The description of the diffraction becomes simpler with increasing distance to the source since the distance $$r$$ between source point and observation point $$r = \sqrt{z_0^2 + (x-x’)^2+(y-y’)^2}$$ can be approximated with fewer terms.

## Fresnel

The Fresnel approximation is such that it takes linear and quadratic terms into consideration. It is valid as long as $$\frac{x-x’}{z}<1<\frac{\sigma^2}{\lambda z}$$.

In this case, the denominator can be approximated as $$r \simeq z_0$$ and the exponent can be written as  $$r = \sqrt{z_0^2 + (x-x’)^2+(y-y’)^2} = z_0 (1 + \frac{(x-x’)^2}{2z_0^2}+ \frac{(y-y’)^2}{2z_0^2} + ….)$$.

In the approximated diffraction integral

$$\psi_P(x’, y’, z_0) \propto \int{}\int{\psi_{\sigma}(x,y) e^{-\frac{ik}{2z_0}((x-x’)^2 + (y-y’))^2}dx dy}$$

the curvature of the wave fronts is taken into account via the quadratic terms. Due to these terms, the integral can only be solved numerically.

## Fraunhofer

For larger distances $$1 \gg \frac{\sigma^2}{\lambda z}$$, the quadratic terms can be neglected and the exponent reduces to

$$r \simeq z_0 (1 + \frac{xx’}{z_0^2}+ \frac{yy’}{z_0^2} + \frac{x’^2 + y’^2}{2z_0^2})$$ .

The diffraction integral can be reduced to

$$\psi_P(x’, y’, z_0) \propto \int{}\int{\psi(x,y) e^{-\frac{ik}{z_0}(xx’ + yy’)}dx dy}$$ .

This diffraction integral is a fourier transform of the wave field in the diffractive structure. It can often be solved analytically and can always be solved numerically in an efficient way.