In Young's double-slit experiment, the fringe pattern shifts by a distance `x_(0)` when a mica sheet, `1.964 mu m` thich and of refractive index 1.6, covers one of the slits. If the mica sheet is removed and the slite-to-screen distance doubled, the new fringe width is equal to `x_(0)`. Find the wavelength of light used.

Explanation

Data: `n_(m) = 1.6, b=1.964, mu m= 1.964 xx 10^(-6)m, D_(2) = 2D_(1), X_(2) = x_(0)`
The fringe shift with the mica sheet,
`x_(0) = D_(1)/d (n_(m)-1)b`
Subsequent to the removal of the mica sheet and doubling the slits-to-screen distance, the new fringe width is
`X_(2) = (lambdaD_(2))/d = (lambda(2D_(1)))/(d)`
Since, `X_(2) = x_(0)`,
`(2lambdaD_(1))/(d) = D_(1)/d(n_(m)-1)b`
`therefore` The wavelength of the light used,
`lambda= (n_(m)-1)/(2).b-(1.6-1)/2.(1.964 xx 10^(-6))`
`=0.3 xx 1.964 xx 10^(-6) = 5.892 xx 10^(-7)m = 5892 Å`