Fresnel diffraction

The Fresnel diffraction integral

According to the Rayleigh–Sommerfeld diffraction theory, the electric-field diffraction pattern at a point (x, y, z) is given by the following solution to the Helmholtz equation:

$${\displaystyle E(x,y,z)=\frac{1}{i\lambda }{\iint }_{-\mathrm{\infty }}^{+\mathrm{\infty }}E(x^{\prime },y^{\prime },0)\frac{e^{ikr}}{r}\frac{z}{r}\left(1+\frac{i}{kr}\right)d{x}^{\prime }d{y}^{\prime },}$$

where

• ${\displaystyle E(x^{\prime },y^{\prime },0)}$ is the electric field at the aperture,
• ${\displaystyle r=\sqrt{(x-x^{\prime }{)}^2+(y-y^{\prime }{)}^2+z^2},}$
• ${\displaystyle k}$ is the wavenumber ${\displaystyle 2\pi /\lambda ,}$
• ${\displaystyle i}$ is the imaginary unit.

The analytical solution of this integral quickly becomes impractically complex for all but the simplest diffraction geometries. Therefore, it is usually calculated numerically.

The Fresnel approximation

The main problem for solving the integral is the expression of r. First, we can simplify the algebra by introducing the substitution $${\displaystyle {\rho }^2=(x-x^{\prime }{)}^2+(y-y^{\prime }{)}^2.}$$

Substituting into the expression for r, we find $${\displaystyle r=\sqrt{{\rho }^2+z^2}=z\sqrt{1+\frac{{\rho }^2}{z^2}}.}$$

Next, by the binomial expansion, $${\displaystyle \sqrt{1+u}=(1+u{)}^{\frac{1}{2}}=1+\frac{u}{2}-\frac{u^2}{8}+\cdots }$$

We can express ${\displaystyle r}$ as $${\displaystyle \begin{array}{cc}r& =z\sqrt{1+\frac{{\rho }^2}{z^2}}\\ & =z\left[1+\frac{{\rho }^2}{2z^2}-\frac{1}{8}{\left(\frac{{\rho }^2}{z^2}\right)}^2+\cdots \right]\\ & =z+\frac{{\rho }^2}{2z}-\frac{{\rho }^4}{8z^3}+\cdots \end{array}}$$

If we consider all the terms of binomial series, then there is no approximation. Let us substitute this expression in the argument of the exponential within the integral; the key to the Fresnel approximation is to assume that the third term is very small and can be ignored, and henceforth any higher orders. In order to make this possible, it has to contribute to the variation of the exponential for an almost null term. In other words, it has to be much smaller than the period of the complex exponential, i.e., ${\displaystyle 2\pi }$: $${\displaystyle k\frac{{\rho }^4}{8z^3}\ll 2\pi .}$$

Expressing k in terms of the wavelength, $${\displaystyle k=\frac{2\pi }{\lambda },}$$

we get the following relationship: $${\displaystyle \frac{{\rho }^4}{z^3\lambda }\ll 8.}$$

Multiplying both sides by ${\displaystyle z^3/{\lambda }^3,}$ we have $${\displaystyle \frac{{\rho }^4}{{\lambda }^4}\ll 8\frac{z^3}{{\lambda }^3},}$$

or, substituting the earlier expression for ${\displaystyle {\rho }^2,}$ $${\displaystyle \frac{1}{{\lambda }^4}{\left[(x-x^{\prime }{)}^2+(y-y^{\prime }{)}^2\right]}^2\ll 8\frac{z^3}{{\lambda }^3}.}$$

If this condition holds true for all values of x, x', y and y', then we can ignore the third term in the Taylor expression. Furthermore, if the third term is negligible, then all terms of higher order will be even smaller, so we can ignore them as well.

For applications involving optical wavelengths, the wavelength λ is typically many orders of magnitude smaller than the relevant physical dimensions. In particular, $${\displaystyle \lambda \ll z,}$$

and $${\displaystyle \lambda \ll \rho .}$$

Thus, as a practical matter, the required inequality will always hold true as long as $${\displaystyle \rho \ll z.}$$

We can then approximate the expression with only the first two terms: $${\displaystyle r\approx z+\frac{{\rho }^2}{2z}=z+\frac{(x-x^{\prime }{)}^2+(y-y^{\prime }{)}^2}{2z}.}$$

This equation is the Fresnel approximation, and the inequality stated above is a condition for the approximation's validity.

Fresnel diffraction

The condition for validity is fairly weak, and it allows all length parameters to take comparable values, provided the aperture is small compared to the path length. For the r in the denominator we go one step further and approximate it with only the first term, ${\displaystyle r\approx z.}$ This is valid in particular if we are interested in the behaviour of the field only in a small area close to the origin, where the values of x and y are much smaller than z. In general, Fresnel diffraction is valid if the Fresnel number is approximately 1.

For Fresnel diffraction the electric field at point ${\displaystyle (x,y,z)}$ is then given by $${\displaystyle E(x,y,z)=\frac{e^{ikz}}{i\lambda z}{\iint }_{-\mathrm{\infty }}^{+\mathrm{\infty }}E(x^{\prime },y^{\prime },0)e^{\frac{ik}{2z}\left[(x-x^{\prime }{)}^2+(y-y^{\prime }{)}^2\right]}d{x}^{\prime }d{y}^{\prime }.}$$

This is the Fresnel diffraction integral; it means that, if the Fresnel approximation is valid, the propagating field is a spherical wave, originating at the aperture and moving along z. The integral modulates the amplitude and phase of the spherical wave. Analytical solution of this expression is still only possible in rare cases. For a further simplified case, valid only for much larger distances from the diffraction source, see Fraunhofer diffraction. Unlike Fraunhofer diffraction, Fresnel diffraction accounts for the curvature of the wavefront, in order to correctly calculate the relative phase of interfering waves.

Alternative forms

Convolution

The integral can be expressed in other ways in order to calculate it using some mathematical properties. If we define the function $${\displaystyle h(x,y,z)=\frac{e^{ikz}}{i\lambda z}{e}^{i\frac{k}{2z}(x^2+y^2)},}$$

then the integral can be expressed in terms of a convolution: $${\displaystyle E(x,y,z)=E(x,y,0)\ast h(x,y,z);}$$

in other words, we are representing the propagation using a linear-filter modeling. That is why we might call the function ${\displaystyle h(x,y,z)}$ the impulse response of free-space propagation.

Fourier transform

Another possible way is through the Fourier transform. If in the integral we express k in terms of the wavelength: $${\displaystyle k=\frac{2\pi }{\lambda }}$$

and expand each component of the transverse displacement: $${\displaystyle \begin{array}{cc}{\left(x-x^{\prime }\right)}^2& =x^2+x^{\prime 2}-2x{x}^{\prime },\\ {\left(y-y^{\prime }\right)}^2& =y^2+y^{\prime 2}-2y{y}^{\prime },\end{array}}$$

then we can express the integral in terms of the two-dimensional Fourier transform. Let us use the following definition: $${\displaystyle G(p,q)=\mathcal{F}\{g(x,y)\}\equiv {\iint }_{-\mathrm{\infty }}^{\mathrm{\infty }}g(x,y)e^{-i2\pi (px+qy)}dxdy,}$$

where p and q are spatial frequencies (wavenumbers). The Fresnel integral can be expressed as $${\displaystyle \begin{array}{cc}E(x,y,z)& ={\frac{e^{ikz}}{i\lambda z}{e}^{i\frac{\pi }{\lambda z}(x^2+y^2)}\mathcal{F}\left\{E(x^{\prime },y^{\prime },0)e^{i\frac{\pi }{\lambda z}(x^{\prime 2}+y^{\prime 2})}\right\}|}_{p=\frac{x}{\lambda z},q=\frac{y}{\lambda z}}\\ & =h(x,y)\cdot G(p,q){|}_{p=\frac{x}{\lambda z},q=\frac{y}{\lambda z}}.\end{array}}$$

That is, first multiply the field to be propagated by a complex exponential, calculate its two-dimensional Fourier transform, replace ${\displaystyle (p,q)}$ with ${\displaystyle \left(\frac{x}{\lambda z},\frac{y}{\lambda z}\right)}$ and multiply it by another factor. This expression is better than the others when the process leads to a known Fourier transform, and the connection with the Fourier transform is tightened in the linear canonical transformation, discussed below.

Linear canonical transformation

From the point of view of the linear canonical transformation, Fresnel diffraction can be seen as a shear in the time–frequency domain, corresponding to how the Fourier transform is a rotation in the time–frequency domain.