Develop the Fresnel diffraction of a slit using a coordinate system fixed to the center of the slit (but source and observation point may be off axis). Show that the phase exponent \(\kappa\left(r_{s}+r_{o}\right)\) in the integrand of the Kirchhoff integral contains both a linear and a quadratic term in the aperture coordinate. (Under what conditions can the quadratic term be neglected, giving the Fraunhofer limit?) By completing the square, show that the "moving" coordinate system used in this section (Fig. 11.2.1) eliminates the linear term, i.e., results in a pure quadratic phase exponent and the conventional Fresnel integrals.

The Kirchhoff integral is used to compute the diffraction pattern produced by a given aperture. It is given by: $$ E_t = \iint_A E_s \frac{e^{i\kappa r}}{r} cos\theta \, dA $$ where \(E_t\) is the scattered field, \(E_s\) is the field at the aperture, \(r\) is the distance between source and observation point, \(\kappa\) is the wavenumber, and \(\theta\) is the angle between the normal to the aperture and the direction of observation.

To find the phase exponent, we need to express \(r_s + r_o\), which is the total distance traveled by a field between the source and observation point, in terms of the aperture coordinate. Let's denote the source coordinates by \((x_s, y_s, z_s)\), the observation point coordinates by \((x_o, y_o, z_o)\), and the aperture coordinates by \((x, y, 0)\). Then, we have: $$ r_s = \sqrt{(x - x_s)^2 + (y - y_s)^2 + z_s^2} $$ and $$ r_o = \sqrt{(x_o - x)^2 + (y_o - y)^2 + z_o^2} $$ Now, expand the square roots and add the two terms \(r_s + r_o\): $$ r_s + r_o = \sqrt{(x - x_s)^2 + (y - y_s)^2 + z_s^2} + \sqrt{(x_o - x)^2 + (y_o - y)^2 + z_o^2} $$

Upon expanding the terms inside the square roots, we can see that there are linear and quadratic terms in the aperture coordinate. Specifically, the quadratic terms are present in the form of \((x-x_s)^2\) and \((x_o-x)^2\), whereas the linear terms are seen in the form of \({(x_s - x_o)}x\) and \({(y_s - y_o)}y\).

The quadratic term can be neglected under the following conditions: 1. The observation point is far away from the aperture, i.e., \(z_o \gg (x_o - x)\) and \(z_o \gg (y_o - y)\). 2. The source is far away from the aperture, i.e., \(z_s \gg (x_s - x)\) and \(z_s \gg (y_s - y)\). In this case, the Fraunhofer limit is obtained. Under the Fraunhofer limit, the integral becomes a simple Fourier transform of the aperture function.

To eliminate the linear term in the phase exponent, we can complete the square for the sum inside the square roots. For both terms, we will get: $$ (x - x_s + x_o/2)^2 + (y - y_s + y_o/2)^2 $$ Now, let's introduce new variables \(x’ = x + (x_s - x_o)/2\) and \(y' = y + (y_s - y_o)/2\). The total distance traveled by a field between the source and observation point in the new coordinate system becomes: $$ r'_s + r'_o = \sqrt{(x' - x'_s)^2 + (y' - y'_s)^2 + z'_s^2} + \sqrt{(x_o' - x')^2 + (y_o' - y')^2 + z_o'^2} $$ The linear term in the aperture coordinate is eliminated, and the phase exponent becomes a pure quadratic term. This results in the Fresnel integrals, given by: $$ E_t = \iint_A E_s' \frac{e^{i\kappa r'}}{r'} cos\theta' \, dA' $$ In conclusion, we have found the Fresnel diffraction of a slit in a coordinate system fixed to the center of the slit. The Kirchhoff integral contains both linear and quadratic terms in the aperture coordinate, and by completing the square, the linear term is eliminated, resulting in the Fresnel integrals.