The Fresnel Diffraction Integral

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

$$U(x,y)=\frac{ e^{ikz}}{i \lambda z}e^{\frac{ik}{2z}(x^2+y^2)}\int \int [U(\xi,\eta)e^{i\frac{k}{2z}(\xi^2+\eta^2)}] e^{-i\frac{2 \pi}{\lambda z}(x \xi +y \eta)} d \xi d \eta$$

which is the orignal object multiplied by a quadratic phase term and
Fourier transformed – The Fresnel Diffraction Integral.

It can also be expressed as a convolution.

$$U(x,y)=\int \int U(\xi,\eta)h(x-\xi, y-\eta)$$

with a kernel of

$$h(x,y)= \frac{ e^{ikz}}{i \lambda z}e^{\frac{ik}{2z}(x^2+y^2)}$$

Either can be used, some times one is easier to solve than the other.

2 thoughts on “The Fresnel Diffraction Integral”

1. Thank you so much for this! My Bsc project is on holographic microscope and reading up on this has been a great start!