= Solution
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>:
$$
\mathcal E(\theta)=K\int_{-w/2}^{w/2}e^{-iky\sin\theta}\,dy
=Kw\frac{\sin\beta}{\beta},\qquad
\beta=\frac{\pi w\sin\theta}{\lambda},\quad k=\frac{2\pi}{\lambda}.
$$
The <light intensity> is proportional to $|\mathcal E|^2$, so
$$
\boxed{I(\theta)=I(0)\left[\frac{\sin(\pi w\sin\theta/\lambda)}{\pi w\sin\theta/\lambda}\right]^2.}
$$
The central value is the continuous limit of the <sinc function>. The first minima satisfy $\sin\theta=\pm\lambda/w$, if $\lambda\leq w$. 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 $w^2/(\lambda D)\ll1$, 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.
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