In a Young's double slit experiment, two slits are separated by 3 mm and the screen is placed 1.5 m away. The distance between two consecutive minima for a light of wavelength 420 nm is
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → 0.21 mm
In Young's experiment the bright and dark fringes are equally spaced, and the distance between two consecutive minima (or maxima) is the fringe width.
\( \displaystyle \beta = \frac{\lambda D}{d} \)
\( \lambda = 420\,\mathrm{nm} = 420\times10^{-9}\,\mathrm{m} \)
\( D = 1.5\,\mathrm{m} \)
\( d = 3\,\mathrm{mm} = 3\times10^{-3}\,\mathrm{m} \)
\( \displaystyle \beta = \frac{420\times10^{-9} \times 1.5}{3\times10^{-3}} = 2.1\times10^{-4}\,\mathrm{m} = 0.21\,\mathrm{mm} \)