In Young's double slit experiment, two slits are made 4 mm apart and the screen is placed 3 m away from slits. The fringe width when light of wavelength 600 nm is used will be
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 0.45 mm
In Young’s double slit experiment the fringe width is \( \beta = \lambda D/d \).
\( \lambda = 600\,\mathrm{nm} = 6\times10^{-7}\,\mathrm{m},\quad d = 4\,\mathrm{mm} = 4\times10^{-3}\,\mathrm{m},\quad D = 3\,\mathrm{m} \)
\( \displaystyle \beta = \frac{6\times10^{-7}\times3}{4\times10^{-3}}\,\mathrm{m} = 4.5\times10^{-4}\,\mathrm{m} = 0.45\,\mathrm{mm} \)