In Young's double slit experiment, the two slits are covered by slabs of same thickness but refractive index `1.4` and `1.7` . The distance between slits and screen is 1 m and distance between slits is 1mm and wavelength of coherent source used is `4000 overset(@)Delta` and the central fringe shifts to the 3rd bright fringe positions, then

Options

A. Shift will be toward slab of `R.I. -1.7 by 1.2mm`
B. Shift will be towards slab of `R.I.-1.4 by 1.2mm`
C. Slabs are of thickness `4mu m`
D. slabs are of thickness `Å`

Answer

A, C

Explanation

Shift due to the introduction of slab is `(D)/(d)(mu-1)t` . On the side of the slit where the slab is placed. On palcing the two slabs of same thicknes, the shift should be
`(D)/(d) (mu_(2) - mu_(1))t` . Since there bright fringes shift is seen,
`(D)/(d)(mu_(2)-mu_(1))t = (3lambda D)/(d)`
`t = (3lambda D)/((mu_(2)-mu_(1))) = (3xx4000xx10^(-10))/((1-7-1.4))`
`= (3xx4xx10^(-7))/(0.3) = 4xx10^(-6) = 4 mu m`
`t = 4mu m` Shift
`=(D)/(d)(mu_(2)=mu_(1))t = (1)/(10^(-3))xx0.3xx4xx10^(-6) = 1.2xx10^(-3)`
Shift `= 1.2mm`