In a single-slit diffraction pattern, the first secondary maximum is at an angle of \(3.0\times10^{-3}\) rad from the centre. The angle of the second minimum is
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \(4.0\times10^{-3}\) rad
The secondary maxima are at \( (n+\frac{1}{2})\lambda/a \) and the minima at \( n\lambda/a \). The first secondary maximum has \( n=1 \), so it is at \( 1.5\lambda/a \). The second minimum is at \( 2\lambda/a \).
\( \begin{aligned} 1.5\frac{\lambda}{a} &= 3.0\times10^{-3}\ \mathrm{rad} \\ \frac{\lambda}{a} &= 2.0\times10^{-3}\ \mathrm{rad} \end{aligned} \)
\( \begin{aligned} \theta_2 &= 2\frac{\lambda}{a} \\ &= 2\times2.0\times10^{-3} \\ &= 4.0\times10^{-3}\ \mathrm{rad} \end{aligned} \)
Hence, the second minimum is at an angle of \( 4.0\times10^{-3} \) rad.
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”
\( \displaystyle \theta \approx \frac{n\lambda}{a},\quad n=\pm 1,\pm 2,\pm 3,\ldots \)