References:

• J. W. Goodman, Introduction to Fourier optics Roberts and Company Publishers (2005)
• M. Born and E, Wolf Principles of optics: electromagnetic theory of propagation, interference and diffraction of light Cambridge university press (1999)
• L. Mandel and E. Wolf Optical Coherence and Quantum Optics, Cambridge University Press

Definitions

Diffraction
Equations de Maxwell
Polarization of a medium
In this courses, we will only consider dielectric media for lake of simplicity \((\rho=0,\vec M=\vec 0,\vec j=\vec 0)\).
Maxwell equation for plane wave in that case:
$$\begin{align}&i\vec k.\vec D=0\implies\vec k.\vec E=0\\ &\vec\nabla\wedge\vec B=-i\omega \vec B\end{align}$$

Diffraction

Starting from the maxwell equations just below, we find the Helmotz equation.

Fresnel

When we consider this equation in the paraxial approximation (Helmotz equation (Paraxial approximation)), we can show that the solutions of this equation are the wave describe by the Diffraction (Diffraction de Fresnel).
The Helmotz equation in paraxial approxiamtion is the following:
$$\frac{\partial A}{\partial z}={{i\frac{1}{2k_0}\nabla_{\perp}^2A}}$$
With \(A\) is the envelope of the wave.
The term \(\nabla_{\perp}\) suggests a coupling between locally closed points.
If we do a Fourier transform over \(x\) and \(y\), we can rewrite this equation as:
$$\frac{\tilde A(f_x,f_y,z)}{\partial z}=-\frac{i}{2k_0}(f_x^2+f_y^2)\tilde A$$
$$\implies \tilde A(f_x,f_y,z)=\tilde A(f_x,f_y,z=0)e^{-i\pi\lambda(f_x^2+f_y^2)z}$$
We should note that the transverse spectrum \(|\tilde A(f_x,f_y,z=0)|\) is a constant because we are in a linear system which implies that we cannot create new frequencies.
We recognize that \(\tilde A\) is a product of two function where \(z\) is a parameter? It means that \(A(x,y,z)\) is a convolution product
$$A(x,y,z)=\iint_{-\infty}^{\infty}A(x',y',z=0)F[(x-x')+(y-y')]dx'dy'\quad\text{convolution}$$
And the function \(F(x,y)\) should be the inverse Fourier transform of \(e^{-i\pi\lambda(f_x^2+f_y^2)z}\) which is : \(F(x,y)=\frac{1}{i \lambda z}e^{i\frac{\pi(x^2+y^2)}{\lambda z} }\)
Thus, we get:
$$A(x,y,z)={{\frac{1}{i\lambda}\iint_{-\infty}^{\infty}A(x',y',z=0)\frac{e^{ik\frac{\left((x-x')^2+(y-y')^2\right)} {2z} } }{z} dx'dy'}}$$
This the exact envelope of the Fresnel diffraction !

Fraunhofer

The Fraunhofer diffraction is the diffraction at \(z\to \infty\).
We already have this expression:
$$A(x,y,z)=\frac 1{i\lambda z}\iint A(x',y',0)e^{\frac{ik}{2 z}\left((x'^2+y'^2)-2(xx'+yy')+(x^2+y^2)\right)}dx'dy'$$
Now, we are looking for a threshold distance which define the distance where the Fraunhofer become a good approximation. For this, we know that we have \(x_{lim}\) et \(y_{lim}\) which are define by the initial conditions.
To neglect the quadratic terms \(x_{lim}'^2+y_{lim}'^2\), we have to find a \(z_{lim}\) such as: \(\frac{k(x_{lim}'^2+y_{lim}'^2)}{2z_{lim} }\lt \lt 2\pi\)
Thus: $$z_{lim}={{\frac{(x_{lim}'^2+y_{lim}'^2)}{2} }}$$
So, we assume that \(z\gt \gt z_{lim}\) to be in the Fraunhifer diffraction case:
$$A(x,y,z)=\frac{1}{i\lambda z}e^{\frac{ik}{2 z}\left(x^2+y^2\right)}\underbrace{\iint A(x',y',0)e^{-ik\pi\left(xx'+yy'\right)}dx'dy'}_{TF_{2D} }$$
$$|A(x,y,z)|^2={{\frac{1}{(\lambda z)^2}\left|FT\left[A(x,t,z=0)\right]\right|^2}}$$
Where we define \(k_x={{\frac{kx} z}}\) and \(k_y={{\frac{ky}{z} }}\). Moreover, we know that \(||\vec k||=k_0\) and \(k_z\approx k_0,k_x\approx k_0\alpha,k_y\approx\beta\) where \(\alpha\) and \(\beta\) are the angle between \(\vec k_0\) and the respective direction. These angles are simply \(\alpha=\frac{x}{z}\) and \(\beta=\frac{y}{z}\) when we are \(z\gt \gt z_{lim}\).

Thin lens

Thin lens