A parallel beam of light of 600 nm falls on a narrow slit and the resulting diffraction pattern is observed on a screen 1.2 m away. It is observed that the first minimum is at a distance of 3 mm from the centre of the screen. The width of the slit is
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → 0.24 mm
In single slit diffraction the first minimum occurs where \( a\sin\theta = \lambda \). The screen is far away, so the small-angle approximation \( \sin\theta \approx \tan\theta = y/D \) applies.
\( \displaystyle a\,\frac{y}{D} = \lambda \)
\( \displaystyle a = \frac{\lambda D}{y} \)
\( \displaystyle a = \frac{600\times10^{-9}\,\mathrm{m} \times 1.2\,\mathrm{m}}{3\times10^{-3}\,\mathrm{m}} \)
\( \displaystyle a = 2.4\times10^{-4}\,\mathrm{m} = 0.24\,\mathrm{mm} \)