= Solution
For a single slit of width $d$, the far-field amplitude is proportional to $\operatorname{sinc}(\pi d\sin\theta/\lambda)$. The first zeros satisfy $d\sin\theta=\pm\lambda$. At small angles the angular width between those zeros is $2\lambda/d$, or $2L\lambda/d$ on a screen a distance $L$ away.
$$
\boxed{\text{Halving the slit width doubles the central-maximum width.}}
$$
The result assumes the wavelength, propagation distance and far-field conditions stay fixed. If the slit becomes comparable to the wavelength, the exact <sine> relation replaces the small-angle scaling.
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