# Huygens’ Principle

The angular spectrum approach to diffraction is one physically and therefore mathematically intuitive method to find how near field diffraction occurs. Another, and a more common introductory, approach to explaining this is Huygens’ Principle, which states: “One can consider each point on a wavefront to be a source of a secondary spherical wave. The envelopes of the wavelets gives the wavefronts at a later time”. We will explore this further below!

1. Physical Picture and Results
2. Mathematical Justification

PHYSICAL PICTURE AND RESULTS

The basic physical picture of Huygens’ principle is shown below. We can see intuitively how we may imagine a plane wave as an infinite sum of point sources. Note here that only the forward propagating wave is considered. The backward propagating wave would be eliminated via the obliquity factor (we will find out how this works mathematically in the later subsection on “Fresnel zones”).

This picture of Huygens’ principle can be used to explain reflection/refraction from boundaries as well, but for diffraction we can use this as follows. Consider that we know the scalar field in some aperture A. We can consider each point in the aperture to be a source of secondary wavelets, such that the total field at some observation point P is just the integral of all of the wavelets. Roughly speaking, we can consider that the $j^{th}$ element in the aperture contributes a field $E(r_j)$ to the total field in the aperture A. The Huygens spherical wave radiating from this $j^{th}$ element is then given as:

$$E(r_j)\frac{e^{-ik\cdotR}}{R}$$

where “R” describes the path of the wave to the observation point P. We can now sum all of these contributions in a continuum limit to get the total field at P as:

$$E_p \approx \iint_AE(x’,y’;0)\frac{e^{-ik\cdotR}}{R}dx’dy’$$

If we consider (x’,y’) to be the coordinates in the aperture, and (x,y) to be the coordinates of the point P in the observation plane, we have:

$$R = \sqrt{(x-x’)^2+(y-y’)^2+z^2}$$

Then here is the physical part that gives us a beam: we will make the paraxial/Fresnel approximation such that we obtain the distance R as:

\begin{align} R & = z\sqrt{1+\frac{(x-x’)^2}{z^2}+\frac{(y-y’)^2}{z^2}} \\ & \approx z + \frac{(x-x’)^2}{2z} + \frac{(y-y’)^2}{2z} \end{align}

Then we have:

$$E_p = E(x,y;z) \approx \frac{e^{-ikz}}{z}\iint_AE(x’,y’;0)e^{-ik[(x-x’)^2+(y-y’)^2]/2z} dx’dy’$$

MATHEMATICAL JUSTIFICATION