In a single-slit diffraction pattern, the angle between the third minima on the two sides of the central maximum is \(9\times10^{-3}\) rad. The slit width is 0.4 mm. The wavelength of the light is
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → 600 nm
The pattern is the same on both sides of the centre. The third minimum on each side is at \( \theta_3 \approx 3\lambda/a \) from the centre line. So the given angle is twice \( \theta_3 \).
\( a = 0.4\times10^{-3}\ \mathrm{m} \)
\( \begin{aligned} 2\theta_3 &= 9\times10^{-3}\ \mathrm{rad} \\ \theta_3 &= 4.5\times10^{-3}\ \mathrm{rad} \end{aligned} \)
\( \begin{aligned} \lambda &= \frac{a\,\theta_3}{3} \\ &= \frac{0.4\times10^{-3}\times 4.5\times10^{-3}}{3} \\ &= 6\times10^{-7}\ \mathrm{m} \\ &= 600\ \mathrm{nm} \end{aligned} \)
Hence, the wavelength of the light is 600 nm.
NCERT (2026-27):
The single slit – “… The minima (zero intensity) are at”
\( \displaystyle \theta \approx \frac{n\lambda}{a} \)\( ,\quad n=\pm 1,\pm 2,\pm 3,\ldots \)