# Fresnel distance and Geometrical limit

I read about the geometrical limit of wave theory. The source from where I read had a slightly different explanation to provide than here(The more rigorous answer is too complicated for me to understand). Though I do not understand completely the easier method in the linked question either, I would like to understand what the source form where I read is trying to say:-

An Aperture of size $$a$$ illuminated by a parallel beam sends diffracted beam (the central maxima) in angular width approximately $$\lambda/a$$. Travelling a distance $$z$$, it aquires the width $$z\lambda/a$$ due to diffraction. The distance at which this width equals the size of the aperture is called the fresnel distance. $$z_F=a^2/\lambda$$. It is the distance beyond which the divergence of beam of width $$a$$ becomes significant. At distances smaller than $$z_F$$, spreading due to diffraction is smaller than the width of the beam, and at distances greater than $$z_F$$, spreading due to diffraction dominates over that due to ray optics($$a$$ is the width of the aperture)

I fail to see the meaning behind this line of argument. Is ray optics valid for distances less than the fresnel distance? Is it not that the validity holds when all objects are comfortably larger, and not smaller, than the wavelength of light? How is the "divergence" due to diffraction, and its equivalence to the aperture width, related to the geometric limit of wave theory?

## 2 Answers

To understand this explanation, you need to understand Fourier decomposition of the electromagnetic field.

In any homogeneous medium, any electromagnetic field can be thought of as a linear superposition of plane waves, all in different directions. Because they run in different directions, the phase delays they undergo in propagating from, say, your aperture to another, parallel plane are all different. Therefore the wavefront gets "scrambled" owing to these direction-dependent phase delays. This interference between the different plane wave components of the electromagnetic field is what we commonly call "diffraction". I further explain this idea, as well as draw some diagrams in this answer here as well as this one here.

So with this introduction in mind, let's look at your paragraph. For simplicity, assume only one transverse direction and one axial (in the direction of propagation) direction. Let's also assume scalar optics, i.e. that the electromagnetic field is well represented by the behaviour of one of its Cartesian components, so that we can do Fourier optics on scalar field.

So we have a uniformly lit aperture of width $a$. Its transverse profile is therefore the function ${\rm rect}(2 x/a)$ where ${\rm rect}(x) = 1;\,|x| \leq 1$ and ${\rm rect}(x) = 0;\,|x| > 1$. We take a Fourier transform to find the superposition weights of each plane wave component, because each such component has a transverse variation $\exp(i\,k_x\,x)$ where $k_x$ is the Fourier transform variable with units of reciprocal length. The Fresnel distance is, as the paragraph says, simply the axial distance needed for this spread to double the beam width. So it is a rough measure of how quickly the light spreads.

So this is how the "divergence" arises from diffraction, i.e. the interference between an optical field's plane wave components as they propagate. Also $\sin\theta = k_x/k$ where $k = 2\pi/lambda$ defines the angle that this plane wave component makes with the axial direction. We take the Fourier transform, we find that there is a spread of $k_x$ values such that the plane wave components most skewed to the axial direction make an angle without direction of roughly $\lambda/a$. So, owing to these skewed components, the field's energy spreads out.

The beam width diverges slowly at first and then, after an axial distance of several Fresnel distances, the divergence speeds up so that the propagation becomes well modelled by the cone of rays diverging from the centre of the aperture. Indeed if you plot contours of constant intensity, they are hyperbolas which begin at right angles to the aperture but bend so that their asymptotes are the cone defined by ray theory. The Fresnel distance defines how far the "knee" of the hyperbol is from the aperture.

For your question:

Is it not that the validity holds when all objects are comfortably larger, and not smaller, than the wavelength of light?

This is in general right, but it breaks down near focuses and in situations like this where we are near and aperture and if the aperture is comparable to the light wavelength. In this case you should be able to understand from the Fourier analysis the reciprocal relationship between the aperture width and the angular spread.

The distance up to which the width of central maxima is the same as the aperture size and after which central maxima becomes larger than the aperture size.