In a single-slit diffraction pattern, the first minimum is 4 mm from the centre of the screen. The distance of the first secondary maximum from the centre, on the same side, is
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 6 mm
In single-slit diffraction, the minima are at angles \( n\lambda/a \). The secondary maxima lie between them, at \( (n+\frac{1}{2})\lambda/a \). On a screen at distance \( D \), each angle gives a distance \( y=D\theta \).
The first minimum gives the value of \( \lambda D/a \). The first secondary maximum has \( n=1 \).
\( \begin{aligned} y_1 &= \frac{\lambda D}{a}=4\ \mathrm{mm} \\ y_1' &= \left(1+\frac{1}{2}\right)\frac{\lambda D}{a} \\ &= 1.5\times4\ \mathrm{mm} \\ &= 6\ \mathrm{mm} \end{aligned} \)
Hence, the first secondary maximum is 6 mm from the centre.
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”
\( \displaystyle \theta \approx \frac{n\lambda}{a},\quad n=\pm 1,\pm 2,\pm 3,\ldots \)