In a modified YDSE the region between screen and slits is immersed in a liquid whose refractive index with time as ${\mu }_1=(5/2)-(T/4)$ until it reaches a steady state value $5/4$. A glass plate of thickness $36\mu m$ of refractive index $3/2$ is introduced of one of the slits.

Find the position of central maxima as a function of time and the time when is is at point $O$, symmetrically on the x- axis.

We consider a point $P$ on the screen.

Optical path length, 

${\left[S_{1}P\right]}_{liquid}={\left[{\mu }_t{S}_{1}P\right]}_{air}$

and  ${\left[S_{2}P-t\right]}_{liquid}+t_{glass}={\mu }_t{\left[S_{2}P-t\right]}_{air}+{\left[{\mu }_g\right]}_{air}={\left[{\mu }_t{S}_{2}P+({\mu }_g-{\mu }_t)t\right]}_{air}$

Hence optical path difference at $P,$

$∆x=\left[{\mu }_g{S}_{2}P+({\mu }_g-{\mu }_t)t\right]-{\mu }_g{S}_{1}P={\mu }_t\left[S_{2}P+S_{1}P\right]+({\mu }_g-{\mu }_t)t$

For a point $P$ at the screen in the absence of liquid,

$S_{2}P-S_{1}P=\frac{yd}{D}$

If a liquid is filled,

${\left[S_{2}P-S_{1}P\right]}_{liquid}=\frac{{\mu }_{1}yd}{D}$

Thus $∆x=\frac{{\mu }_{t}yd}{D}+\left[{\mu }_g-{\mu }_t\right]t$

For a central maxima,

$∆x=0$

$y=-\left[\frac{{\mu }_g-{\mu }_t}{{\mu }_t}\right]\frac{tD}{d}$

$=\frac{(4-T)tD}{(10-T)d}$

At point O, $y=0$

Thus  $T=4 sec$