In a Young's double slit experiment, the distance between slits is 2.5 mm and the screen is placed 2 m away from the slits. The fringe width for a light of wavelength 400 nm is
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \( 0.32\,\mathrm{mm} \)
The fringe width in Young's double-slit experiment is the distance between two consecutive bright (or dark) fringes, \( \beta = \lambda D/d \).
\( \lambda = 400\,\mathrm{nm} = 4\times10^{-7}\,\mathrm{m} \)
\( D = 2\,\mathrm{m} \)
\( d = 2.5\,\mathrm{mm} = 2.5\times10^{-3}\,\mathrm{m} \)
\( \displaystyle \beta = \frac{\lambda D}{d} = \frac{4\times10^{-7} \times 2}{2.5\times10^{-3}} = 3.2\times10^{-4}\,\mathrm{m} = 0.32\,\mathrm{mm} \)