Narrow slit, wide pattern

Everyday intuition says a bigger gap lets more through and therefore lets it spread further. The physics runs the other way.

$$\sin\theta = \frac{\lambda}{a} \quad \text{(first minimum, slit width } a\text{)}.$$

The angle grows as $a$ shrinks. So:

1. As the opening narrows toward the wavelength, the pattern spreads wide.
2. As it widens, the angles shrink and the shadow develops sharp edges.

Doubling the slit width therefore halves the spread, rather than doubling it. The slogan worth keeping is short: narrow slit, wide pattern.

What matters is the ratio $\lambda/a$, and judging by everyday size is what makes the two standard cases look inconsistent.

1. Sound at a doorway. Wavelengths near a metre, opening about a metre, so $\lambda/a$ is near one. The wave floods the next room, which is why you can hear round a corner.
2. Light at the same doorway. Wavelength near $500$ nm, so $\lambda/a \approx 5\times10^{-7}$. The beam runs essentially straight and casts a sharp shadow.

Same opening, opposite behaviour, because the two waves sit on opposite sides of the comparison. Calling a $1$ mm slit "small" and expecting visible spreading for any wave through it makes the same mistake in the other direction: $1$ mm is enormous compared with a light wavelength and tiny compared with a sound one.

Write the ratio before judging anything. Size on its own decides nothing.

For a single slit of width $a$,

$$a\sin\theta = m\lambda, \qquad m = 1, 2, 3, \ldots$$

gives the minima. There is no $m = 0$ available, because the centre of the pattern is bright.

Carrying over the double-slit habit, where the matching equation gives maxima, puts bright bands exactly where the screen is dark. So name what the equation locates before using it: single slit with width $a$ gives minima; double slit with separation $d$ gives maxima.

The resulting picture is worth sketching once and remembering:

1. One wide central bright band, sitting between the first minimum on each side.
2. Dark fringes at the angles the equation gives.
3. Weaker maxima roughly between the dark fringes, each much dimmer than the centre.
4. The central band is twice as wide as the ones beyond it.

The condition looks like the double-slit one and comes from the opposite reasoning, which is worth seeing once so the two stop blurring together.

Treat the slit as a row of many sources across its width. At the angle where $a\sin\theta = \lambda$, the path difference between the ray from the top edge and the one from the middle is exactly $\lambda/2$, so those two cancel. Pair every source in the top half with its partner half a slit width below, and every pair cancels. The whole slit sums to zero: a dark fringe.

That is why the equation gives minima rather than maxima, and why $m = 0$ is missing: at $\theta = 0$ every path is the same length and everything adds, giving the bright centre.

It also explains the width. The first minimum sits at $\sin\theta = \lambda/a$, so the central band spans from $-\lambda/a$ to $+\lambda/a$ in $\sin\theta$, twice the spacing of the fringes beyond it.