In a single-slit experiment, the second minimum is seen 4 mm from the centre of a screen. The screen is 2 m from the slit. The angular position of the third minimum is
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \(3\times10^{-3}\) rad
A minimum at distance \( y \) on a screen at distance \( D \) is at a small angle \( \theta = y/D \). The angles of the minima go as \( \theta_n \approx n\lambda/a \). So \( \theta_3 \) is 3/2 times \( \theta_2 \), and the slit width and wavelength are not needed.
\( y_2 = 4\times10^{-3}\ \mathrm{m} \)
\( D = 2\ \mathrm{m} \)
\( \begin{aligned} \theta_2 &= \frac{y_2}{D} \\ &= \frac{4\times10^{-3}}{2} \\ &= 2\times10^{-3}\ \mathrm{rad} \end{aligned} \)
\( \begin{aligned} \theta_3 &= \frac{3}{2}\,\theta_2 \\ &= 3\times10^{-3}\ \mathrm{rad} \end{aligned} \)
Hence, the third minimum is at an angle of \( 3\times10^{-3} \) rad.
NCERT (2026-27):
The single slit – “… The minima (zero intensity) are at”
\( \displaystyle \theta \approx \frac{n\lambda}{a} \)\( ,\quad n=\pm 1,\pm 2,\pm 3,\ldots \)