Light of wavelength \( \lambda \) is incident on a single slit. The first minimum of the diffraction pattern is found at a distance of 6 mm from the central maximum on a screen placed 2 m from the slit. If the slit width is 0.2 mm, the wavelength of the light used is
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 6000 Å
In single-slit diffraction the first minimum lies at the angle \( \theta\approx\lambda/a \), where \( a \) is the slit width. For a screen at distance \( D \), this minimum is at a distance \( y=D\theta \) from the centre of the central maximum, so \( \lambda=ya/D \).
\( y=6\,\mathrm{mm}=6\times10^{-3}\,\mathrm{m} \)
\( a=0.2\,\mathrm{mm}=0.2\times10^{-3}\,\mathrm{m} \)
\( D=2\,\mathrm{m} \)
\( \begin{aligned} \lambda &= \frac{ya}{D} \\ &= \frac{6\times10^{-3}\,\mathrm{m}\times0.2\times10^{-3}\,\mathrm{m}}{2\,\mathrm{m}} \\ &= 6\times10^{-7}\,\mathrm{m} \end{aligned} \)
Since 1 Å = \( 10^{-10}\,\mathrm{m} \), \( \lambda=6\times10^{-7}\,\mathrm{m} \) = 6000 Å.