Single Slit Diffraction

In single-slit diffraction the central bright band stretches between the first dark fringes on either side, where the minima satisfy $a\sin \theta =\lambda$, giving angular position $\theta \approx \frac{\lambda }{a}$ for small angles. The linear half-width on a screen at distance $D$ is $\frac{\lambda D}{a}$, so the full central maximum width is $W=\frac{2\lambda D}{a}$. Substituting $\lambda =500\times 10^{-9}$ m, $D=2$ m and $a=0.1\times 10^{-3}$ m gives $W=\frac{2\times 500\times 10^{-9}\times 2}{0.1\times 10^{-3}}=\frac{2\times 10^{-6}}{10^{-4}}=2\times 10^{-2}$ m $=20$ mm. The central maximum is special because it spans the gap between the first minimum on each side, whereas every secondary maximum is confined between two adjacent minima and is therefore only half as wide. This unequal spacing is a direct fingerprint of diffraction from a single aperture, in contrast to the uniform fringes of double-slit interference, and it reflects how every point of the wavefront within the slit contributes to the superposition. The value 10 mm is wrong because it gives only the half-width $\frac{\lambda D}{a}$. The value 5 mm is wrong as it omits the factor of two and doubles the slit width. The value 40 mm is wrong since it double-counts the factor of two. This NCERT result shows the central maximum is twice as wide as the secondary fringes. A check confirms a narrower slit spreads light more, consistent with the centimetre-scale band.