A Fraunhofer diffraction pattern due to a single slit of width 0.2 mm is being obtained on a screen placed at a distance of 2 m from the slit. The first minimum lies at 5 mm on either side of the central maximum on the screen. The wavelength of light is:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \( 5000\,\mathring{\mathrm{A}} \)
In single-slit diffraction the first minimum is at \( a\sin\theta = \lambda \). The angle is small, so \( \sin\theta \approx \tan\theta = y/D \), where \( y \) is the distance of the first minimum from the centre of the screen.
\( \displaystyle \frac{a\,y}{D} = \lambda \;\Rightarrow\; \lambda = \frac{a\,y}{D} \)
\( a = 0.2\,\mathrm{mm} = 2\times10^{-4}\,\mathrm{m} \)
\( y = 5\,\mathrm{mm} = 5\times10^{-3}\,\mathrm{m} \)
\( D = 2\,\mathrm{m} \)
\( \displaystyle \lambda = \frac{2\times10^{-4} \times 5\times10^{-3}}{2} = 5\times10^{-7}\,\mathrm{m} = 5000\,\mathring{\mathrm{A}} \)