When Young's double slit experiment is conducted in vacuum, fringe width is found to be β. When separation between the slits is halved, distance between the slits and the screen is doubled and the whole experiment is conducted in water (μ=43), the new fringe width will be
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a
β3
b
3β
c
2β3
d
4β3
answer is B.
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Detailed Solution
β=λDd and β1=λ1D1d1=(λμ)(2D)(d2)=4μ(λDd)∴ β1=44/3β=3β
When Young's double slit experiment is conducted in vacuum, fringe width is found to be β. When separation between the slits is halved, distance between the slits and the screen is doubled and the whole experiment is conducted in water (μ=43), the new fringe width will be