In a diffraction experiment, the screen is placed 2 m away from a narrow slit. The first minimum observed is at 5 mm on either side of the central maximum for wavelength of \( 5 \times 10^{-5}\,\mathrm{cm} \). The width of the slit is:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 0.02 cm
In single-slit diffraction the first minimum is at \( a\sin\theta = \lambda \). For small angles \( \sin\theta \approx \tan\theta = y/D \), so
\( \displaystyle y = \frac{\lambda D}{a} \;\Rightarrow\; a = \frac{\lambda D}{y} \)
\( \lambda = 5\times10^{-5}\,\mathrm{cm} = 5\times10^{-7}\,\mathrm{m} \)
\( D = 2\,\mathrm{m} \)
\( y = 5\,\mathrm{mm} = 5\times10^{-3}\,\mathrm{m} \)
\( \displaystyle a = \frac{5\times10^{-7} \times 2}{5\times10^{-3}} = 2\times10^{-4}\,\mathrm{m} = 0.02\,\mathrm{cm} \)