Light of wavelength 600 nm falls normally on a slit of width 0.3 mm.…

The correct answer is Option (2) → 9 mm

The nth minimum is at the angle \( \theta \approx n\lambda/a \). On a screen at distance \( D \), it is a distance \( y_n = D\theta \) from the centre. This gives \( y_n = n\lambda D/a \). Here the two minima are on opposite sides, so their distances add.

    \( \lambda = 600\times10^{-9}\ \mathrm{m} \)

    \( a = 0.3\times10^{-3}\ \mathrm{m} \)

    \( D = 1.5\ \mathrm{m} \)

    \( \begin{aligned} \frac{\lambda D}{a} &= \frac{600\times10^{-9}\times 1.5}{0.3\times10^{-3}} \\ &= 3\times10^{-3}\ \mathrm{m} = 3\ \mathrm{mm} \end{aligned} \)

    \( \begin{aligned} y_2 &= 2\times 3\ \mathrm{mm} = 6\ \mathrm{mm} \\ y_1 &= 3\ \mathrm{mm} \\ y_2+y_1 &= 9\ \mathrm{mm} \end{aligned} \)

So the two dark bands are 9 mm apart.

NCERT (2026-27):
The single slit – “… The minima (zero intensity) are at”

    \( \displaystyle \theta \approx \frac{n\lambda}{a} \)\( ,\quad n=\pm 1,\pm 2,\pm 3,\ldots \)