A parallel beam of monochromatic light of wavelength 489 nm passes through a single slit of width 0.3 mm. The angular width of the central maxima is
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \( 3.26 \times 10^{-3}\,\mathrm{rad} \)
In single-slit diffraction the first minima lie at \( \theta = \pm\lambda/a \), so the central maximum spreads from \( -\lambda/a \) to \( +\lambda/a \). Its full angular width is
\( \displaystyle 2\theta = \frac{2\lambda}{a} \)
\( \lambda = 489\,\mathrm{nm} = 489\times10^{-9}\,\mathrm{m} \)
\( a = 0.3\,\mathrm{mm} = 3\times10^{-4}\,\mathrm{m} \)
\( \displaystyle 2\theta = \frac{2 \times 489\times10^{-9}}{3\times10^{-4}} = 3.26\times10^{-3}\,\mathrm{rad} \)