A point object $O$ is placed in front of a concave mirror of focal length $10\text{cm}$. A glass slab of refractive index $\mu =\frac{3}{2}$ and thickness $3\text{cm}$ is inserted between object and mirror. Find the position of final image when the object is placed at a distance of $31\text{cm}$ from the mirror as shown.

The correct option is A $16\text{cm}$ from the mirror
The normal shift (s) produced by the glass slab in the direction of incident ray is
$s=t(1-\frac{1}{\mu })$
$=3(1-\frac{1}{\frac{3}{2}})$
$=3(1-\frac{2}{3})=1\text{cm}$
Thus for the mirror, object placed at a distance $31\text{cm}$ will appear at,
$u^{{}^{\prime }}=31-s=31-1=30\text{cm}$
Applying mirror formula,
$\frac{1}{v}+\frac{1}{u}=\frac{1}{f}$
Substituting the given values we get,
$\frac{1}{v}+\frac{1}{(-30)}=+\frac{1}{(-10)}$
or $v=-15\text{cm}$
When the light falls on slab after getting reflected from mirror, the slab will again cause a shift of $s=1\text{cm}$ in direction of light ray. Thus final image will form at
$v^{{}^{\prime }}=-(15+1)=-16\text{cm}$