= Solution
Use <Fraunhofer diffraction>, appropriate to observation in the far field or in the focal plane of a <lens>, and uniform illumination across the slit. At observation angle $\theta$, points separated by transverse coordinate $u$ acquire a <phase difference> $ku\sin\theta$, where $k=2\pi/\lambda$ is the <wavenumber>. The <superposition principle> for the coherent <wave amplitudes> gives
$$
A(\theta)\propto\int_{-w/2}^{w/2}e^{iku\sin\theta}\,du=w\frac{\sin\beta}{\beta},\qquad \beta=\frac{\pi w\sin\theta}{\lambda}.
$$
The <intensity> is proportional to the squared modulus of the <wave amplitude>, so
$$
\boxed{I(\theta)=I(0)\left(\frac{\sin\beta}{\beta}\right)^2.}
$$
The limit at $\beta=0$ is one. The noncentral zeros satisfy $w\sin\theta=m\lambda$ for nonzero integers $m$ for which the angle is physically allowed. In the <Rayleigh criterion>, two mutually incoherent point images are just resolved when the central maximum of one <diffraction pattern> lies at the first minimum of the other. For <single-slit diffraction> this gives
$$
\boxed{\Delta\theta_R=\arcsin\left(\frac\lambda w\right)\simeq\frac\lambda w\quad(w\gg\lambda).}
$$
The exact statement here takes one image on the slit normal; for nearby paraxial images the <angular resolution> is $\lambda/w$. The first propagating minimum exists only for $w\geq\lambda$. The numerical factor $1.22$ belongs to a circular aperture and does not apply to this slit.
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