Light of wavelength 6000 `overset(@)A` is used to obtain interference fringe of width 6 mm in a young's double slit experiment. Calculate the wavelength of light required to obtain fringe of width 4 mm if the distance between the screen and slits is reduced to half of its initial value.

Explanation

Given `lambda=6xx10^(-7) m, beta=6xx10^(-3) m `
`beta'=4xx10^(-3) m D'=(D)/(2)`
We know that, `beta alpha lambda D` ,
Hence `(beta')/(beta)=(lambda')/(lambda)(D')/(D)`
i.e., `(beta')/(beta)=(lambda')/(6xx10%^(-7))xx(1)/(2)`
i.e., `lambda' =(12xx10^(-7)xx4xx10^(-3))/(6xx10^(-3))=8xx10^(-7) m`