An object is placed infront of a glass slab $(\mu =1.5)$ of thickness 6cm at a distance 28cm from it. Other face of glass slab is silvered. Find the position of final image of object from the silvered face of slab.

Let d₂ is the distance from first face.

${\mathrm{d}}_2=\frac{{\mathrm{d}}_1}{{\mathrm{n}}_{\text{relative }}}=\frac{{\mathrm{d}}_1}{(1/\mathrm{\mu })}=28\times 1.5=42 \mathrm{cm}$

For silvered face, object distance is 42+6=48 cm.

So, image distance behind the silvered face is also 48cm.

For second refraction, consider d₃ as 48+6=54 cm.

${\mathrm{d}}_4=\frac{{\mathrm{d}}_3}{\left(\mathrm{\mu }\right)}=\frac{54}{1.5}=36 \mathrm{cm}$

SO, distance of final image from silvered face is 36-6=30 cm.