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There has been a bit of discussion about this already, however, my question arises more from the mathematical requirements to allow for Fraunhofer diffraction proposed by Born and Wolf's optics book.

First: variable definitions: Given $r'$ is the distance of the source from the aperture and $s'$ is the distance of resultant pattern, and $(\zeta, \eta)$ as the aperture coordinates, and $l_0, l$ and $m, m_0$ as direction cosines of the aperture to the source/point (these are not terribly important for the discussion).

The quadratic (Fresnel) phase can be ignored under the following condition:

$\frac{1}{2} k | (\frac{1}{r'} + \frac{1}{s'}))(\zeta^2 + \eta^2) - \frac{(l_0\zeta + m_0 \eta)^2}{r'} - \frac{(l\zeta + m\eta)^2}{s'}| << 2\pi$

There are two possibilities:

(1) The distance of the source and the resulting diffraction are observed very far away relative to the aperture size divided by the wavelength, i.e.

$|r'| >> (\zeta^2 + \eta^2)_{max}/\lambda$

$|s'| >> (\zeta^2 + \eta^2)_{max}/\lambda$

This one is very familiar and intuitive: clean, constructive interference off the small slit is successfully achieved with planar waves incident on the aperture and observed far enough away.

(2) The other possibility to satisfy this is:

$\frac{1}{r'} + \frac{1}{s'} = 0$ and the angle of the source/observation are very paraxial.

The 2nd condition is very confusing, however, it seems vital to justify the usage of a lens i.e. an ideal lens placed after the aperture will always satisfy the Fraunhofer condition. How can these distances be negative? If a point source is imaged by a lens, the image plane will reconstruct this point source. But how can this end up making $r'$ negative? From geometric optics, I know that we can interpret the image as being an additional point source, but wouldn't r' stay a positive quantity?

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If you read further in the section 4.3.3 in your "Born & Wolf" from which you have quoted the equations you will find this paragraph on page 427, ($P_0$ is the location of the source):

Conditions (33) give an estimate of the distances r' and s' for which the Fraunhofer representation may be used. Conditions (34) imply that Fraunhofer diffraction also occurs when the point of observation is situated in a plane parallel to that of the aperture, provided that both the point of observation and the source are sufficiently close to the $z$-axis. Here two cases may be distinguished: When $r'$ is negative, the wave-fronts incident upon the aperture are concave to the direction of the propagation, i.e. $P_0$ is a centre of convergence and not of divergence of the incident wave. This case is of great practical importance, as it arises in the image space of a well-corrected centred system that images a point source which is not far from the axis. A Fraunhofer pattern is then formed in the Gaussian image plane and may be considered as arising from the diffraction of the image-forming wave on the exit pupil. When $r'$ is positive, the wave-fronts are convex to the direction of propagation. The diffraction phenomena are virtual, being apparently formed on a screen through the source $P_0$. This case arises, for example, when an aperture is held in front of the eye, or the object glass of a telescope adjusted for distant vision of the light source.

The sign convention is visible and explained by Figure 8.3b. showing the angles between the aperture plane normal and the radius vectors. If you flip the source point $P_0$ to the same side as where $P$ is, then the sign of the cosine angle between the normal and the radius vector is also flipped. In short, the source having a negative distance from the aperture just means a concave wavefront is propagating along the direction of propagation.

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  • $\begingroup$ Ok that's very helpful, I see how the cosine term can justify the sign convention. But, if you a have point source --> aperture --> lens, the point source is still physically behind the screen isn't it? In that paragraph, there's a confusing jump in logic, how does a lens suddenly make the source come from the final image? The only way I can justify this that the image of the source acts as an additional source, but that doesn't seem physically sound to me $\endgroup$ Commented May 1, 2023 at 23:59
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    $\begingroup$ The point source of a concave wavefront is at a "virtual" point, it is nowhere in a real space... But imagine a plane wave hitting a concave lens whose virtual focus is $P_0$, the resulting wavefront is concave spherical centered at $P_0$. $\endgroup$
    – hyportnex
    Commented May 2, 2023 at 0:12

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