In a single-slit diffraction experiment, the width of the slit is dec…

The correct answer is Option (1) → It increases.

In single-slit diffraction the minima of intensity lie at \( \theta\approx n\lambda/a \), \( n=\pm1,\pm2,\ldots \), where \( a \) is the slit width. The central maximum spreads between the first minima on either side, at \( \pm\lambda/a \). So its angular width \( 2\lambda/a \) is inversely proportional to the slit width.

Option (1) It increases. — Correct. For the same wavelength, a smaller \( a \) moves the first minima at \( \pm\lambda/a \) farther from the centre. So the central bright band becomes wider. Light spreads out more from a narrower slit.

Option (2) It decreases. — Incorrect. The central maximum becomes narrower only when the slit is made wider, since its width \( 2\lambda/a \) falls as \( a \) grows.

Option (3) It remains unchanged. — Incorrect. With the wavelength fixed, the width \( 2\lambda/a \) changes whenever the slit width changes, so it cannot stay the same.

Option (4) It does not depend on the width of the slit. — Incorrect. The intensity first falls to zero at the angles \( \pm\lambda/a \). These angles depend on the slit width. So the width of the central maximum depends on both \( \lambda \) and \( a \).

NCERT (2026-27):
The Single Slit – “… The minima (zero intensity) are at θ ≈ nλ/a, n = ±1, ±2, ±3, ....”