 Solution

For a single slit of width $d$, the far-field amplitude is proportional to $\mathrm{sinc}(\pi d\sin \theta /\lambda )$. 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.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.