Q.

The diagram shows a cuboidal container with a non–viscous liquid of nonuniform density but a uniform refractive index  μ0. The height of container is 0.4 SI units and the density of liquid varies linearly according to the relation:
ρ=ρ 0(14y)
Where ρ 0 is a constant and y is the vertical distance measured from the midplane (ABPQ) of the container. (y is positive when measured upward, and negative when measured downward).
There are two horizontal slits, covered with perfectly transparent films, in one of the vertical faces as shown in the figure. A thin transparent strip of material having superficial dimensions equal to the slits and thickness equal to d (volume V0) is released from rest from an initial depths below the mid–plane (ABPQ) (i.e., from 
y = – s) close to the slits. The strip has a constant density  ρ 0, uniform refractive index  μ(>μ0) and it  move due to buoyancy keeping its plane parallel to the plane of the slits; always remaining very close to the plane of the slits).
Monochromatic plane wavefronts (of light) parallel to the plane of the slits enter into the liquid through the face opposite the double–slit. The intensity of light is observed on the screen placed outside the container parallel to the plane of the slits.
Question Image

COLUMN-ICOLUMN-II
A)s = 2d and 0 < t < 1 secP)Central bright fringe is always at C
B)s = 3d and 0 < t < 2 secQ)Central bright fringe is below C four times
C)s = d/4 and 0 < t < 6 secR)Central bright fringe is above C twice
D)s = d and 0 < t < 3/4 secS)Central bright fringe is above C four times
  T)Central bright fringe is below C twice

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a

A–RS, B–QR, C–P, D–R

b

A–RT, B–QS, C–P, D–R

c

A–RP, B–QT, C–P, D–R

d

A–RT, B–QS, C–PQ, D–RS

answer is D.

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Detailed Solution

a=FBmgm           a=ρ0(14g)γgρ0γgρ0γ      a = - 4g y           ω=4g            T=2π4g=1 sec

 In one oscillation the slab would have gone infront of the slit twice so the central maxima shifts 

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