Fresnel diffraction or near-field diffraction is the diffraction pattern of an electromagnetic wave obtained close to the diffracting object (often a source or aperture). More accurately, it is the diffraction case when the Fresnel number is large and thus the Fraunhofer approximation (diffraction of parallel beams) can not be used.

The Fresnel diffraction integral

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The electric field diffraction pattern at a point (x,y,z) is given by:

$E(x,y,z)=-\{i\; \backslash over\; \backslash lambda\}\; \backslash iint\{\; E(x\text{'},y\text{'},0)\; \backslash frac\{e^\{ikr\}\}\{r\}\; \backslash cos\; \backslash vartheta\}dx\text{'}dy\text{'}$

where

$r=\backslash sqrt\{(x-x\text{'})^2+(y-y\text{'})^2+z^2\}$

and

$\backslash cos\; \backslash vartheta\; =\; \backslash frac\{z\}\{r\}$

is the cosine of the angle between z and r. Analytical solution of this integral is impossible for all but the simplest diffraction geometries. Therefore, it is usually calculated numerically.

The main problem for solving the integral is the expression of r. If we consider the Taylor series:

$\backslash sqrt\{1+x\}\; =\; 1\; +\; \backslash frac\{x\}\{2\}\; -\; \backslash frac\{x^2\}\{8\}\; +\; \backslash dots$

we can use it to express r in a different way:

$r=\backslash sqrt\{(x-x\text{'})^2+(y-y\text{'})^2+z^2\}\; =z\; \backslash sqrt\{\; 1\; +\; \backslash frac\{(x-x\text{'})^2\; +\; (y-y\text{'})^2\}\{z^2\}\; \}\; =$

$z\; \backslash left1\; +\; \backslash frac\{(x-x\text{'})^2\; +\; (y-y\text{'})^2\}\{2\; z^2\}\; -\; \backslash frac\{1\}\{8\}\; \backslash left(\; \backslash frac\{(x-x\text{'})^2\; +\; (y-y\text{'})^2\}\{z^2\}\; \backslash right)^2\; +\; \backslash dots\; \backslash right=$

$z\; +\; \backslash frac\{(x-x\text{'})^2\; +\; (y-y\text{'})^2\}\{2\; z\}\; -\; \backslash frac\{+\; (y-y\text{'})^2^2\}\{8z^3\}\; +\; \backslash dots$

Note that, if we consider all the terms of Taylor series there is no approximation.There was actually an approximation in a prior step, when assuming $e^\{i\; k\; r\}/r$ is a real wave. In fact this is not a real solution to the vector Helmholtz equation, but to the scalar one. See scalar wave 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 element is very small and can be ignored. 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. $2\; \backslash pi$:

$2\; \backslash pi\; \backslash gg\; k\; \backslash frac\{*^2\}\{8\; z^3\}$

expressing k in terms of the wavelength, i. e. $k\; =\; 2\; \backslash pi\; /\; \backslash lambda$ we get the following relationship:

$z^3\; \backslash gg\; \backslash frac\{*^2\}\{8\; \backslash lambda\}$

if this is true for any value of x, x' , y and y' , then we can ignore the third term in the Taylor expression. If this term is "small", than all the others with a greater power will be even smaller, so they can be ignore as well and we can approximate the expression only with the first two terms:

$r\; \backslash approx\; z\; +\; \backslash frac\{(x-x\text{'})^2\; +(y-y\text{'})^2\}\{2\; z\}$

This is a fairly weak condition which allows all length parameters to take comparable values, provided the aperture is small compared to the path length. Moreover, if we are interested in the behaviour of the field only in a small area close to the origin, i.e. for values of x and y much smaller than z, then we can assume $\backslash vartheta\; \backslash approx\; 0$, that means $\backslash cos\; \backslash vartheta\; \backslash approx\; 1$.

Unlike Fraunhofer diffraction, Fresnel diffraction has to account for the curvature of the wavefront, in order to correctly calculate relative phase of interfering waves.

For Fresnel diffraction the electric field at point (x,y,z) is given by:

$E(x,y,z)=-\{i\; \backslash over\; \backslash lambda\}\{e^\{ikz\}\; \backslash over\; z\}\backslash iint\; E(x\text{'},y\text{'},0)e^\{i\; \backslash lambda\; z\}\; e^\{i\; \backslash frac\{k\}\{2\; z\}\; (x^2\; +\; y^2)\}$

then the integral can be expressed in terms of a convolution:

$E(x,y,z)\; =\; E(x,y,0)\; *\; h\_z\; (x,y)$

it other words we are representing the propagation using a linear-filter modeling. That is why we might call the function h_z(x,y) the impulse response of the free space.

Another possible way is through the Fourier transform. If in the integral we express k in terms of the wavelength and we calculate $(x-x\text{'})^2\; =\; x^2\; +\; x\text{'}^2\; -2\; x\; x\text{'}$, doing the same for $(y-y\text{'})^2$, then we can express the integral in terms of the two dimensional Fourier transform. Let us use the following definition:

$\backslash mathcal\{F\}\; \backslash left\backslash \{\; g(x,y)\; \backslash right\backslash \}\; \backslash equiv\; \backslash iint\_\{-\backslash infty\}^\{\backslash infty\}\; g(x,y)\; e^\{-i\; 2\; \backslash pi\; (p\; x\; +\; q\; y)\}\; dx\; dy$

where p and q are spatial frequencies. The Fresnel integral can be expressed as:

$E(x,y,z)\; =\; \backslash frac\{e^\{i\; k\; z\}\}\{i\; \backslash lambda\; z\}\; e^\{i\; \backslash frac\{\backslash pi\}\{\backslash lambda\; z\}(x^2\; +\; y^2)\}\; \backslash mathcal\{F\}$

\left. \left\{ E(x,y,0) e^{i \frac{\pi}{\lambda z} (x^2 + y^2)} \right\} \right|_{p = \frac{x}{\lambda z}; q = \frac{y}{\lambda z}}

i.e. first multiply the field to be propagated for a complex exponential, calculate its two dimensional Fourier transform, replace (p,q) with $\backslash left(\; \backslash frac\{x\}\{\backslash lambda\; z\},\; \backslash frac\{y\}\{\backslash lambda\; z\}\; \backslash right)$ and multiply it for another factor. This expression is better than the others when the process leads to a know Fourier transform.

See also

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• Fraunhofer diffraction
• Fresnel integral
• Fresnel zone
• Fresnel number
• Augustin-Jean Fresnel

Diffraction

Notes

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