In a Young's double slit experiment, a light of wavelength 550 nm is used to get fringes on screen 1.2 m away from the slits. The distance between the two slits is 1.5 mm. The fringe width is
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 0.44 mm
In Young's double slit experiment the fringe width is
\( \displaystyle \beta = \frac{\lambda D}{d} \)
\( \lambda = 550\,\mathrm{nm} = 5.5\times10^{-7}\,\mathrm{m} \)
\( D = 1.2\,\mathrm{m} \)
\( d = 1.5\,\mathrm{mm} = 1.5\times10^{-3}\,\mathrm{m} \)
\( \displaystyle \beta = \frac{5.5\times10^{-7} \times 1.2}{1.5\times10^{-3}} = 4.4\times10^{-4}\,\mathrm{m} = 0.44\,\mathrm{mm} \)