Light of wavelength 500 nm falls normally on a single slit of width 0.2 mm. The width of the central maximum on a screen 2 m away is
A. 5 mm
B. 2.5 mm
C. 20 mm
D. 10 mm
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Answer: D. 10 mm
\(W=\frac{2\lambda D}{a}=\frac{2\times5\times10^{-7}\times2}{2\times10^{-4}}=10^{-2}\) m \(=10\) mm.
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· 23 Oct 2026
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