In a single-slit experiment, the central maximum is 4 mm wide when the screen is 1.0 m from the slit. The slit width is now halved and the screen is moved to 1.5 m from the slit. The new width of the central maximum is
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → 12 mm
The width of the central maximum on the screen is \( w=2\lambda D/a \). So \( w \) is proportional to \( D \) and inversely proportional to \( a \). The wavelength does not change here.
\( w=4\ \mathrm{mm} \)
\( D=1.0\ \mathrm{m} \)
\( a'=a/2 \)
\( D'=1.5\ \mathrm{m} \)
\( \begin{aligned} \frac{w'}{w} &= \frac{D'}{D}\times\frac{a}{a'} \\ &= 1.5\times2=3 \\ w' &= 3\times4=12\ \mathrm{mm} \end{aligned} \)
Hence, the new width of the central maximum is 12 mm.
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”
\( \displaystyle \theta \approx \frac{n\lambda}{a},\quad n=\pm 1,\pm 2,\pm 3,\ldots \)