Step 1: Set up the geometry. Consider a slit of width \(a\) on which a plane wavefront of wavelength \(\lambda\) falls normally. Light diffracts and is observed on a distant screen (or through a converging lens). Let \(\theta\) be the angle a diffracted direction makes with the normal to the slit.
Step 2: Pair-cancellation reasoning. Split the slit into two equal halves. Rays from a point in the upper half and the matching point in the lower half differ in path by \(\tfrac{a}{2}\sin\theta\). When this equals \(\tfrac{\lambda}{2}\) every such pair cancels, giving darkness; that means \(a\sin\theta=\lambda\). Splitting into \(2n\) strips generalises this.
Step 3: General result. Complete cancellation therefore happens whenever the extreme path difference is an integral multiple of \(\lambda\):
\[\boxed{\,a\sin\theta = n\lambda,\ \ n=\pm1,\pm2,\pm3,\dots\,}\]
For small angles this reduces to \(\theta_n \approx \dfrac{n\lambda}{a}\). The integer \(n\) names the order of the dark fringe, and \(n=0\) is left out since it corresponds to the bright central band.