Q.In deriving the single slit diffraction pattern, it was stated that the intensity is zero at angles of $n\lambda/a$. Justify this by suitably dividing the slit to bring out the cancellation.

Dividing the slit into matched pairs of narrow strips shows that light from each pair cancels exactly at these particular angles, so the whole slit's contribution adds up to zero. …

Step 1: The first minimum ($n=1$) — divide the slit into 2 halves

Consider a slit of width $a$, and suppose light leaves it at angle $\theta$ satisfying $a\sin\theta = \lambda$. Divide the slit into two equal halves, each of width $a/2$. Take any point in the upper half and the corresponding point directly below it in the lower half (i.e., a pair of points separated by $a/2$ across the slit). The path difference between the wavelets leaving this pair of points, in the direction $\theta$, is:

A path difference of $\lambda/2$ means these two wavelets are exactly out of phase and cancel each other. Since every point in the upper half has such a partner in the lower half with the same $\lambda/2$ path difference, the contributions from the entire slit cancel in pairs — giving zero net intensity at $a\sin\theta = \lambda$.

Step 2: Generalising to the $n$-th minimum — divide the slit into $2n$ equal zones

Now suppose $a\sin\theta = n\lambda$ for some integer $n$. Divide the slit into $2n$ equal zones, each of width $a/(2n)$, and pair up zone $1$ with zone $2$, zone $3$ with zone $4$, and so on (this gives $n$ pairs of adjacent zones spanning the whole slit). Take corresponding points in each adjacent pair of zones — these are separated by $a/(2n)$ across the slit. The path difference between wavelets from such a corresponding pair, in direction $\theta$, is: …