Light of wavelength 450 nm falls normally on a slit of width 0.3 mm.…

The correct answer is Option (3) → 0.9 mm

In a single-slit pattern, the minima are at angles \( n\lambda/a \) and the secondary maxima at \( (n+\frac{1}{2})\lambda/a \). So a secondary maximum lies about halfway between two minima. On the screen, the first minimum is at \( \lambda D/a \) and the first secondary maximum at \( 1.5\lambda D/a \).

    \( \lambda=450\ \mathrm{nm}=4.5\times10^{-7}\ \mathrm{m} \)

    \( a=0.3\ \mathrm{mm}=3\times10^{-4}\ \mathrm{m} \)

    \( D=1.2\ \mathrm{m} \)

    \( \begin{aligned} \frac{\lambda D}{a} &= \frac{4.5\times10^{-7}\times1.2}{3\times10^{-4}} \\ &= 1.8\times10^{-3}\ \mathrm{m}=1.8\ \mathrm{mm} \end{aligned} \)

    \( \begin{aligned} s &= 1.5\frac{\lambda D}{a}-\frac{\lambda D}{a} \\ &= 0.5\times1.8\ \mathrm{mm} \\ &= 0.9\ \mathrm{mm} \end{aligned} \)

Hence, the first minimum and the first secondary maximum are 0.9 mm apart.

NCERT (2026-27):
The single slit – “It is observed that the intensity has a central maximum at θ = 0 and other secondary maxima at”

    \( \displaystyle \theta \approx \left(n+\frac{1}{2}\right)\frac{\lambda}{a} \)

“which go on becoming weaker and weaker with increasing n. The minima (zero intensity) are at”

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