In a Young's double slit experiment, interference fringes were produced on a screen placed 1.8 m from the two slits 0.3 mm apart and illuminated by light of wavelength 600 nm. The fringe width is
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \( 3.6\,\mathrm{mm} \)
In Young's double slit experiment the fringe width is \( \beta = \lambda D/d \), where \( D \) is the distance of the screen and \( d \) the slit separation.
\( \lambda = 600\,\mathrm{nm} = 6\times10^{-7}\,\mathrm{m} \)
\( D = 1.8\,\mathrm{m} \)
\( d = 0.3\,\mathrm{mm} = 3\times10^{-4}\,\mathrm{m} \)
\( \displaystyle \beta = \frac{6\times10^{-7} \times 1.8}{3\times10^{-4}} = 3.6\times10^{-3}\,\mathrm{m} = 3.6\,\mathrm{mm} \)