Light of wavelength 600 nm is incident normally on a slit of width 0.03 cm. The width of the central maximum on the screen placed 1 m away from the slit is
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 4 mm
In single slit diffraction the first minima lie at \( \sin\theta = \lambda/a \), so the central maximum has width
\( \displaystyle \beta_0 = \frac{2\lambda D}{a} \)
\( \lambda = 600\,\mathrm{nm} = 6\times10^{-7}\,\mathrm{m} \)
\( a = 0.03\,\mathrm{cm} = 3\times10^{-4}\,\mathrm{m} \)
\( D = 1\,\mathrm{m} \)
\( \displaystyle \beta_0 = \frac{2\times6\times10^{-7}\times1}{3\times10^{-4}} = 4\times10^{-3}\,\mathrm{m} = 4\,\mathrm{mm} \)