A parallel beam of light of wavelength 500 nm is incident normally on a slit of width d. If the distance between slit and screen is 0.6 m and distance of second order minimum from the centre of the screen is 6.0 mm, what is the width of the slit?
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → 0.1 mm
In single-slit diffraction the minima occur where \( d\sin\theta = n\lambda \). For small angles \( \sin\theta \approx \tan\theta = y/D \), so the \( n \)th minimum is at
\( \displaystyle y_n = \frac{n\lambda D}{d} \;\Rightarrow\; d = \frac{n\lambda D}{y_n} \)
\( n = 2 \)
\( \lambda = 500\,\mathrm{nm} = 5\times10^{-7}\,\mathrm{m} \)
\( D = 0.6\,\mathrm{m} \)
\( y_2 = 6.0\,\mathrm{mm} = 6\times10^{-3}\,\mathrm{m} \)
\( \displaystyle d = \frac{2 \times 5\times10^{-7} \times 0.6}{6\times10^{-3}} = 10^{-4}\,\mathrm{m} = 0.1\,\mathrm{mm} \)