Light of wavelength 600 nm falls on a single slit, and the screen is 1 m from the slit. The screen is moved back to 1.5 m from the slit. The third minimum moves 3 mm farther from the centre. The width of the slit is
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → 0.3 mm
The third minimum is at the fixed angle \( \theta_3 \approx 3\lambda/a \). Its distance from the centre is \( y_3 = 3\lambda D/a \). So when the screen moves back by \( \Delta D \), this distance grows by \( 3\lambda\,\Delta D/a \).
\( \lambda = 600\times10^{-9}\ \mathrm{m} \)
\( \begin{aligned} \Delta D &= 1.5 - 1 \\ &= 0.5\ \mathrm{m} \end{aligned} \)
\( \Delta y_3 = 3\times10^{-3}\ \mathrm{m} \)
\( \begin{aligned} a &= \frac{3\lambda\,\Delta D}{\Delta y_3} \\ &= \frac{3\times 600\times10^{-9}\times 0.5}{3\times10^{-3}} \\ &= 3\times10^{-4}\ \mathrm{m} \\ &= 0.3\ \mathrm{mm} \end{aligned} \)
Hence, the width of the slit is 0.3 mm.
NCERT (2026-27):
The single slit – “… The minima (zero intensity) are at”
\( \displaystyle \theta \approx \frac{n\lambda}{a} \)\( ,\quad n=\pm 1,\pm 2,\pm 3,\ldots \)