In YDSE arrangement as shown in the figure, fringes are seen on screen using monochromatic source S having wavelength `3000Å` (in the air). `S_(1) and S_(2)` are two slits separated by d = 1 mm and D = 1 m. Left of slits `s_(1) and s_(2)` medium of refractive index `n_(1)=2` is present and to the right of `S_(1) and S_(2)` medium of `n_(2)=(3)/(2)`, is present. A thin slab of thickness 't' is placed in front of `S_(1)`. The refractive index of `n_(3)` of the slab varies with distance from it's starting face as shown in figure.

In order of get central maxima at the centre of the screen, the thickness of the slab `("in "mum)` required is :

Options

A. 1/3 mm
B. `-1/3 mm`
C. 1/6 mm
D. `-1/6 mm`

Explanation

From previous equation :
`0=1 mu m +(3yd)/(2D)-0.5 t`
`3/2 (yd)/(D)=-0.5 mu m`
`y=-((10^(-6))/3)((1m)/(1xx10^(-3)))=(10^(-3))/3=(-1)/3 mm`, below centre.