An object AB of height 2 cm is placed at distance 63 cm from a concave mirror of radius of curvature 40 cm. A slab of refractive index $\frac{4}{3}$ and thickness 12 cm is placed between the object and the mirror.

The object shift due to slab is $x=t\left(1-\frac{1}{u}\right)=12\left(1-\frac{3}{4}\right)=3cm$

 
The image $A_1{B}_1$ formed due to refraction through slab behaves as object for concave mirror.
For concave mirror, $u=-60cm,f=-20cm$   
According to mirror formula,
$\because \frac{1}{v}+\frac{1}{u}=\frac{1}{f}$ or $\frac{1}{v}-\frac{1}{60}=-\frac{1}{20}$
Or  $\frac{1}{v}=\frac{1}{60}-\frac{1}{20}=\frac{1-3}{60}$
 $\therefore v=-30cm$
 $m=\frac{A_1{B}_2}{A_1{B}_1}=-\frac{v}{u}=-\frac{\left(-30\right)}{-60}$
 $\therefore A_2{B}_2=-1cm$
 $A_2{B}_2$behaves as virtual object for second times refraction through slab. The final image  is shifted 3cm leftward from $A_2{B}_2$ due to second times refraction through slab. Thus, final image is real, 1 cm high inverted and 33 cm from the concave mirror. The ray diagram is shown in the figure.