Solution:
The width of the central maximum in single slit diffraction is given by \(w = \frac{2\lambda D}{d}\). Substituting the values: \(w = \frac{2 \times 500 \times 10^{-9} \times 2}{10^{-3}} = 2.0 \times 10^{-3} \text{ m} = 2.0 \text{ mm}\).
A beam of light of wavelength \(\lambda = 500 \text{ nm}\) from a distant source falls on a single slit 1 mm wide and the resulting diffraction pattern is observed on a screen 2 m away. The width of central maxima is
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