In a Young’s double slit experiment, two slits are placed \( 0.9 \) mm apart and the fringes are observed on a screen \( 1 \) m away from the slits. If the second dark fringe is obtained at distance of \( 1 \) mm from the central bright fringe, then the wavelength of light would be
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \( 6\times10^{-7} \) m
In Young’s experiment the n-th dark fringe is at \( y = (2n-1)\lambda D/2d \), so the second dark fringe (n = 2) is at \( 3\lambda D/2d \), one and a half fringe widths from the centre.
\( d = 0.9\,\mathrm{mm} = 0.9\times10^{-3}\,\mathrm{m} \)
\( y = 1\,\mathrm{mm} = 1\times10^{-3}\,\mathrm{m}, \quad D = 1\,\mathrm{m} \)
\( \displaystyle y = \frac{3\lambda D}{2d} \)
\( \displaystyle \lambda = \frac{2yd}{3D} \)
\( \displaystyle \lambda = \frac{2\times(1\times10^{-3}\,\mathrm{m})\times(0.9\times10^{-3}\,\mathrm{m})}{3\times1\,\mathrm{m}} \)
\( \displaystyle \lambda = 6\times10^{-7}\,\mathrm{m} \)
This is 600 nm, in the visible range, which is a quick check that the powers of ten are right.