Past exam of the physics 2 paper of the University of Oxford / 2022 / 7 / i

Take coordinate $y$ across the slit and observation angle $\theta$ from its normal vector. Under Fraunhofer diffraction, the relative optical path difference from $y$ is $y\sin \theta$. Uniform coherent illumination makes the complex amplitude proportional to the aperture integral:The light intensity is proportional to $|\mathcal{E}{|}^2$, soThe central value is the continuous limit of the sinc function. The first minima satisfy $\sin \theta =\pm \lambda /w$, if $\lambda \le w$. Infinite slit height removes spreading in the other transverse direction; this expression describes the slit-width direction.

Strictly, distance $D\gg w$ alone does not ensure Fraunhofer diffraction: one also needs $w^2/(\lambda D)\ll 1$, so that quadratic aperture-dependent wave phase is negligible, or observation in a suitable focal plane. For example, $w=1\mathrm{mm}$, $\lambda =500\mathrm{nm}$ and $D=1\mathrm{m}$ have $D/w=1000$ but $w^2/(\lambda D)=2$; the Fresnel diffraction correction is then substantial. The displayed result uses the intended far-field regime of the question and the usual scalar aperture approximation.