An air bubble inside glass slab n = 1.5 appear from one side at 6 cm and from other side at 4 cm. Then the thickness of the glass slab is __cm

Given that
An air bubble inside the glass slab
Let the thickness of glass slab = T
Position of air bubble from  A =X
 

It appears to be at 4 cm from A
we know $\mathrm{\mu }=\frac{\text{real\hspace{0.17em}depth}}{\text{apparant\hspace{0.17em}dept}}$
$\frac{3}{2}=\frac{\mathrm{x}}{4}\to \left(1\right)$
It appears to be at 6 cm from B
$\mathrm{\mu }=\frac{\text{real\hspace{0.17em}depth}}{\text{apparant\hspace{0.17em}dept}}$
$$\begin{gathered} \frac{3}{2}=\frac{\mathrm{T}-\mathrm{x}}{6}\to \left(2\right) \\ \mathrm{From} \left(1\right) \& \left(2\right) \\ \frac{\mathrm{x}}{4}=\frac{\mathrm{T}-\mathrm{x}}{6} \\ 6\mathrm{x}=4\mathrm{T}-4\mathrm{x} \\ 10\mathrm{x}=4\mathrm{T} \\ 5\mathrm{x}=2\mathrm{T} \end{gathered}$$
Substituting $\mathrm{x}=\frac{2\mathrm{T}}{5}$ in eq (1)
$$\begin{gathered} \frac{3}{2}=\frac{2\mathrm{T}}{20} \\ 30=2\mathrm{T} \\ \therefore \mathrm{T}=15\text{\hspace{0.17em}cm} \\ \mathrm{The} \mathrm{thickness} \mathrm{of} \mathrm{glass} \mathrm{slab} = 15 \mathrm{cm}. \end{gathered}$$