Light of wavelength 500 nm falls on a single slit. On a screen 1.2 m away, the first and the third minima on the same side of the centre are 4 mm apart. The width of the slit is
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 0.3 mm
Each minimum lies at \( y_n = n\lambda D/a \) from the centre. Both minima are on the same side. So their gap is \( y_3 - y_1 = 2\lambda D/a \).
\( \lambda = 500\times10^{-9}\ \mathrm{m} \)
\( D = 1.2\ \mathrm{m} \)
\( y_3 - y_1 = 4\times10^{-3}\ \mathrm{m} \)
\( \begin{aligned} a &= \frac{2\lambda D}{y_3-y_1} \\ &= \frac{2\times 500\times10^{-9}\times 1.2}{4\times10^{-3}} \\ &= 3\times10^{-4}\ \mathrm{m} \\ &= 0.3\ \mathrm{mm} \end{aligned} \)
From the above, the slit is 0.3 mm wide.
NCERT (2026-27):
The single slit – “… The minima (zero intensity) are at”
\( \displaystyle \theta \approx \frac{n\lambda}{a} \)\( ,\quad n=\pm 1,\pm 2,\pm 3,\ldots \)