In Young's double slit experiment,light of wavelength`5,000 overset @A` is used.The screen on which fringes are projected is 1.5 m from the centre of the narrow slits. The third bright band on the screen is formed at a distance of 1 cm from the central bright band calculate the separation between the slits.

Explanation

`lambda = 5000 A = 5000 xx 10^(-10) = 5 xx 10^(-7) m`
`D =1 cm = 10^(-2) m`
Distance of 3rd dark band is given by:
`y_(m) = (m + 1/2)(lambdaD)/d`
`therefore` m =2 (for 3rd dark band)
`therefore y_(3) = (2 + 1/2) (5 xx 10^(-7) xx 10^(-2))/d`
`y_(3) = (12.5 xx 10^(-9))/d`
Distance of first bright fringe is given by
`y_(m) = (mlambdaD)/d`
For first bright fringe, m=1
`y_(1) = ((1)5 xx 10^(-7) xx 10^(-2))/d`
`=(5 xx 10^(-9))/d`
According to question:
`y_(3) - y_(1) = 1 cm = 10^(-2) m`
`d = (7.5 xx 10^(-9))/10^(-2) = 7.5 xx 10^(-7) m`
`= 7.5 xx 10^(-4) m`
`= 0.00075 m`