In a single-slit diffraction experiment, the first minimum is 3 mm from the centre of the screen. The slit width is now halved, and everything else is kept the same. The distance of the second minimum from the centre is now
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 12 mm
On the screen, the minima lie at distances \( y_n = n\lambda D/a \) from the centre. A higher order is farther out, and a wider slit brings every minimum closer. So halving the slit width doubles every distance.
\( \begin{aligned} y_1 &= \frac{\lambda D}{a} \\ &= 3\ \mathrm{mm} \end{aligned} \)
\( a^{\prime} = \frac{a}{2} \)
\( \begin{aligned} y_2^{\prime} &= \frac{2\lambda D}{a^{\prime}} \\ &= 4\,\frac{\lambda D}{a} \\ &= 4\times 3\ \mathrm{mm} \\ &= 12\ \mathrm{mm} \end{aligned} \)
Hence, the second minimum is now 12 mm from the centre.
NCERT (2026-27):
The single slit – “… The minima (zero intensity) are at”
\( \displaystyle \theta \approx \frac{n\lambda}{a} \)\( ,\quad n=\pm 1,\pm 2,\pm 3,\ldots \)