Light of wavelength 500 nm falls on a slit of width 0.2 mm. The screen is 2 m away. A point P on the screen is 12.5 mm from the centre of the pattern. Which of the following is nearest to P?
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → the second secondary maximum
On the screen, minima are at \( n\lambda D/a \) and secondary maxima at \( (n+\frac{1}{2})\lambda D/a \). So first find \( \lambda D/a \) and compare the distance of P with it.
\( \begin{aligned} \frac{\lambda D}{a} &= \frac{5\times10^{-7}\times2}{2\times10^{-4}} \\ &= 5\times10^{-3}\ \mathrm{m}=5\ \mathrm{mm} \end{aligned} \)
\( \begin{aligned} \frac{12.5\ \mathrm{mm}}{5\ \mathrm{mm}} &= 2.5=2+\frac{1}{2} \end{aligned} \)
Option (1) the third minimum — Incorrect. The third minimum is at \( 3\lambda D/a=15 \) mm. Single-slit minima are at whole-number multiples of \( \lambda D/a \).
Option (2) the second secondary maximum — Correct. The value \( n+\frac{1}{2}=2.5 \) gives \( n=2 \). So the second secondary maximum is at P.
Option (3) the third secondary maximum — Incorrect. The third secondary maximum is at \( 3.5\lambda D/a=17.5 \) mm, farther from P.
Option (4) the first secondary maximum — Incorrect. The first secondary maximum is at \( 1.5\lambda D/a=7.5 \) mm, much closer to 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”
\( \displaystyle \theta \approx \frac{n\lambda}{a},\quad n=\pm 1,\pm 2,\pm 3,\ldots \)