A 20cm thick glass slab of refractive index 1.5 is kept in front of a plane mirror. Find the position of the image (relative to mirror) as seen by an observer through the glass slab when a point object is kept in air at a distance of 40cm from the mirror.

As thickness of glass slab is 20 cm and its $\mathrm{\mu }=(3/2),$  so shift produced by it will be $\mathrm{x}=\mathrm{d}[1-\frac{1}{\mathrm{\mu }}]=20[1-\frac{2}{3}]=\frac{20}{3}\mathrm{cm}$  (towards the mirror)
 

The image of object in the mirror is  $\frac{100}{3}$ cm from the mirror. The glass slab shift this image towards the observer by  $\frac{20}{3}$cm.
Therefore the position of the final image from the mirror w.r.t observer is $\frac{100}{3}-\frac{20}{3}=\frac{80}{3}$ cm