The Huygen-Fresnel principle can be stated as
$$U(P_{0})=\frac{1}{i \lambda}\int \int_{\Sigma}
U(P_{1}) \frac{e^{i k r_{01}}}{r_{01}}\cos \theta ds$$
Figure 1.
This equation describes how the light travels from one plane to another a distance \(z\) apart. All points from the aperture \(\Sigma\) contribute towards the intensity at point \(P_{0}\).
Typically, this equation is too computationally intensive to compute so we
make some approximations.
We note
$$\cos \theta = \frac{z}{r_{01}}$$
So
$$U(x,y)=\frac{z}{i \lambda}\int \int_{\Sigma} U(\xi,\eta) \frac{e^{i k r_{01}}}{r_{01}^{2}} d \xi d \eta$$
We can rewrite \(r_{01}\) as
$$r_{01}=\sqrt{z^{2}+(x-\xi)^{2}+(y-\eta)^{2}}$$
$$=z \sqrt{1+\frac{(x-\xi)^{2}}{z^{2}}+\frac{(y-\eta)^{2}}{z^{2}}}$$
To simplify this further, we note that the binomial expansion below
$$\sqrt{1+b}=1+\frac{1}{2}b-\frac{1}{8}b^{2}+\cdots$$
So
$$r_{0} \approx z [1+\frac{1}{2}[(x-\xi)^{2}+(y-\eta)^{2}]-\frac{1}{8}[(x-\xi)^{2}+(y-\eta)^{2}]^{2}]$$
The third term can be dropped, providing it contributes only a small amount
usually take as less than one radian in the \(exp\) term, i.e.,
$$\frac{z}{8}\frac{2 \pi}{\lambda} \left[(x-\xi)^{2}+(y-\eta)^{2}\right]^{2} \ll 1$$
i.e.,
$$z \gg \sqrt[3]{\frac{\pi}{4 \lambda} [(x-\xi)^{2}+(y-\eta)^{2}]}$$
The error contribution from the denominator is much smaller, so only in
this case is OK to put \(r_{0} \approx z\).
We now get
$$U(x,y)=\frac{ e^{ikz}}{i \lambda z}\int \int U(\xi,\eta)e^{\frac{ik}{2z}[(x-\xi)^{2}+(y-\eta)^{2}]} d \xi d \eta$$
If we take the \(e^{\frac{ik}{2z}(x^2+y^2)}\) outside of the integral we
have
which is the original object multiplied by a quadratic phase term and
Fourier transformed – The Fresnel Diffraction Integral.
It can also be expressed as a convolution.
with a kernel of
Either can be used, sometimes one is easier to solve than the other.