Example: $n$ transparent slabs of refractive index $1.5$ each, having thickness $1cm,2cm,..........$ to $ncm$ are arranged one over another. A point object is seen through this combination with the perpendicular light. If the shift of object by the combination is $1cm$ then find the value of $n$.

Solution:
As the refractive index of all the slabs are $1.5$ so those may be a considered as a single slab with the thickness summed up.
Summed up thickness of $n$ slabs $\Delta t=1+2+3+...+n$$=\frac{n(n+1)}{2}$
Given, for small angle of incidence, shift of object: $s=\left(1-\frac{1}{\mu }\right)\Delta t$
so, $\Rightarrow 1cm=\left(1-\frac{1}{1.5}\right)\frac{n(n+1)}{2}$
$\Rightarrow 6=n(n+1)$
$\Rightarrow n^2+n-6=0$
$\Rightarrow n^2+3n-2n-6=0$
$\Rightarrow n(n+3)-2(n+3)=0$
$\Rightarrow (n+3)(n-2)=0$
So, $n=-3$ or $n=2$
As, $n$ can not be negative. so $n=2$