Light of wavelength 550 nm falls on a slit of width 0.11 mm. The screen is 1.2 m away. The width of the first secondary maximum, that is, the distance between the first and second minima on one side, is
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 6 mm
The minima are at angles \( \lambda/a,\ 2\lambda/a,\ 3\lambda/a,\ldots \). Two next minima are \( \lambda/a \) apart. So each secondary maximum is \( \lambda D/a \) wide on the screen. This is half the width of the central maximum.
\( \lambda=550\ \mathrm{nm}=5.5\times10^{-7}\ \mathrm{m} \)
\( a=0.11\ \mathrm{mm}=1.1\times10^{-4}\ \mathrm{m} \)
\( D=1.2\ \mathrm{m} \)
\( \begin{aligned} w &= \frac{2\lambda D}{a}-\frac{\lambda D}{a}=\frac{\lambda D}{a} \\ &= \frac{5.5\times10^{-7}\times1.2}{1.1\times10^{-4}} \\ &= 6\times10^{-3}\ \mathrm{m}=6\ \mathrm{mm} \end{aligned} \)
Hence, the first secondary maximum is 6 mm wide.
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”
\( \displaystyle \theta \approx \frac{n\lambda}{a},\quad n=\pm 1,\pm 2,\pm 3,\ldots \)