A parallel beam of light of wavelength \( 700\,\mathrm{nm} \) falls on a narrow slit and the resulting diffraction pattern is observed on a screen \( 1.5\,\mathrm{m} \) away. It is observed that the first minimum is at a distance of \( 3\,\mathrm{mm} \) from the centre of the screen. The width of the slit is :
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \( 0.35\,\mathrm{mm} \)
In single-slit diffraction the first minimum occurs where \( a\sin\theta = \lambda \). The angle is small, so \( \sin\theta \approx \tan\theta = y/D \).
\( \displaystyle \frac{y}{D} = \frac{\lambda}{a} \Rightarrow a = \frac{\lambda D}{y} \)
\( \lambda = 700\,\mathrm{nm} = 7\times10^{-7}\,\mathrm{m} \)
\( y = 3\,\mathrm{mm} = 3\times10^{-3}\,\mathrm{m} \)
\( \displaystyle a = \frac{7\times10^{-7}\,\mathrm{m}\times1.5\,\mathrm{m}}{3\times10^{-3}\,\mathrm{m}} = 3.5\times10^{-4}\,\mathrm{m} = 0.35\,\mathrm{mm} \)