A beam of light of wavelength 500 nm from a distant source falls on a single slit \( 10^{-4}\,\mathrm{m} \) wide and the resulting diffraction pattern is observed on a screen 2 m away. The width of the central bright fringe is
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \( 2 \times 10^{-2}\,\mathrm{m} \)
In single-slit diffraction the first minima lie at \( \sin\theta = \pm\lambda/a \). The central maximum spreads between them, so on a screen at distance \( D \) its width is twice the distance of the first minimum from the centre.
\( \displaystyle w = \frac{2\lambda D}{a} \)
\( \lambda = 500\,\mathrm{nm} = 5\times10^{-7}\,\mathrm{m} \)
\( a = 10^{-4}\,\mathrm{m} \)
\( D = 2\,\mathrm{m} \)
\( \displaystyle w = \frac{2 \times 5\times10^{-7} \times 2}{10^{-4}} = 2\times10^{-2}\,\mathrm{m} \)