Light of wavelength 600 nm is incident on a slit having a width of 0.300 mm. The screen is 2.00 m from the slit. The distance of the first minima from the central maxima is:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → ± 4 × 10−3 m
In single-slit diffraction the minima occur where \( a\sin\theta = n\lambda \) (n = 1, 2, …). The angles are small, so \( \sin\theta \approx \tan\theta = y/D \), and the first minimum lies at a distance \( \lambda D/a \) on either side of the centre.
\( \lambda = 600\,\mathrm{nm} = 6\times10^{-7}\,\mathrm{m} \)
\( a = 0.300\,\mathrm{mm} = 3\times10^{-4}\,\mathrm{m} \)
\( D = 2.00\,\mathrm{m} \)
\( \displaystyle y_1 = \frac{\lambda D}{a} \)
\( \displaystyle y_1 = \frac{6\times10^{-7}\,\mathrm{m} \times 2.00\,\mathrm{m}}{3\times10^{-4}\,\mathrm{m}} \)
\( y_1 = \pm 4\times10^{-3}\,\mathrm{m} \)
The ± sign shows the two minima placed symmetrically on either side of the central maximum.