In a Young’s double slit experiment. Let A and B be the two slits. The films are of thickness t_A and t_B having refractive indices ${\mathrm{\mu }}_{\mathrm{A}}\text{ and }{\mathrm{\mu }}_{\mathrm{B}}$ are paced in front of A and B respectively. If ${\mathrm{\mu }}_{\mathrm{A}}{\mathrm{t}}_{\mathrm{A}}={\mathrm{\mu }}_{\mathrm{B}}{\mathrm{t}}_{\mathrm{B}}$, the central maxima will

Net path difference
$\begin{array}{l}=\left(\mathrm{BP}-{\mathrm{t}}_{\mathrm{B}}+{\mathrm{\mu }}_{\mathrm{B}}{\mathrm{t}}_{\mathrm{B}}\right)-\left(\mathrm{AP}-{\mathrm{t}}_{\mathrm{A}}+{\mathrm{\mu }}_{\mathrm{A}}{\mathrm{t}}_{\mathrm{A}}\right)\\ =(\mathrm{BP}-\mathrm{AP})-\left({\mathrm{t}}_{\mathrm{B}}-{\mathrm{t}}_{\mathrm{A}}\right)+{\mathrm{\mu }}_{\mathrm{B}}{\mathrm{t}}_{\mathrm{B}}-{\mathrm{\mu }}_{\mathrm{A}}{\mathrm{t}}_{\mathrm{A}}\\ =\frac{\mathrm{xd}}{\mathrm{D}}-\left({\mathrm{t}}_{\mathrm{B}}-{\mathrm{t}}_{\mathrm{A}}\right)+0\\ {\left({\mathrm{x}}_{\mathrm{n}}\right)}_{\mathrm{t}=0}\frac{\mathrm{d}}{\mathrm{D}}=\mathrm{n\lambda }\dots \dots \left(1\right)\\ {\left({\mathrm{x}}_{\mathrm{n}}\right)}_{\mathrm{t}}\frac{\mathrm{d}}{\mathrm{D}}-\left({\mathrm{t}}_{\mathrm{B}}-{\mathrm{t}}_{\mathrm{A}}\right)=\mathrm{n\lambda }\dots \dots \dots \left(2\right)\end{array}$
So, from (1) and (2)
${\left({\mathrm{x}}_{\mathrm{n}}\right)}_{\mathrm{t}}\frac{\mathrm{d}}{\mathrm{D}}-{\left({\mathrm{x}}_{\mathrm{n}}\right)}_{\mathrm{t}=0}\frac{\mathrm{d}}{\mathrm{D}}={\mathrm{t}}_{\mathrm{B}}-{\mathrm{t}}_{\mathrm{A}}$
So, shift, $\mathrm{\Delta X}={\left({\mathrm{x}}_{\mathrm{n}}\right)}_{\mathrm{t}}-{\left({\mathrm{x}}_{\mathrm{n}}\right)}_0$
So, $\mathrm{\Delta X}\frac{\mathrm{d}}{\mathrm{D}}={\mathrm{t}}_{\mathrm{B}}-{\mathrm{t}}_{\mathrm{A}}\Rightarrow \mathrm{\Delta X}=\left({\mathrm{t}}_{\mathrm{B}}-{\mathrm{t}}_{\mathrm{A}}\right)\frac{\mathrm{D}}{\mathrm{d}}$
If ${\mathrm{t}}_{\mathrm{B}}>{\mathrm{t}}_{\mathrm{A}}, \mathrm{\Delta X}$ is positive. So shift is towards A.
So, ${\mathrm{t}}_{\mathrm{B}}<{\mathrm{t}}_{\mathrm{A}}, \mathrm{\Delta X}$ is negative so shift is towards B.