A concave mirror of focal length $2 \mathrm{cm}$ is placed on a glass slab as shown in the figure. Then the image of object $\mathrm{O}$ formed due to reflection at mirror and then refraction by the slab :

Given Data:

concave mirror focal length:$\mathrm{f}=-2 \mathrm{cm}$

object distance:$\mathrm{u}=-1 \mathrm{cm}$

using mirror formula for image formation for image(1):

$\frac{1}{\mathrm{f}}=\frac{1}{\mathrm{v}}+\frac{1}{\mathrm{u}}$

Here, $\mathrm{f}=$focal length

          $\mathrm{v}=$Image distance from pole $\mathrm{P}$

          $\mathrm{u}=$object distance from pole $\mathrm{P}$

$\frac{1}{-2}=\frac{1}{\mathrm{v}}+\frac{1}{-1} \Rightarrow \frac{1}{\mathrm{v}}=\frac{1}{-2}+\frac{1}{1}$

$\mathrm{v}=+2 \mathrm{cm}$

so, image(1) will be form at $+2 \mathrm{cm}$.

Now, find shift due to the glass slab:

$∆=\left(1-\frac{1}{\mathrm{\mu }}\right) \mathrm{t}$

Here,

$∆=$Shift

$\mathrm{\mu }=$refractive Index of glass slab

$\mathrm{t}=$height of glass slab

$∆=\left(1-\frac{1}{{\displaystyle \raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}\right)·9$

$∆=3 \mathrm{cm}$

this shift due to glass slab gives the position of image(2).so Image (2) will forms at $3 \mathrm{cm}$ from Image(1). that means Image(2) will forms at point $\mathrm{O}$. Point $\mathrm{O}$ is behind the slab, so Image(2) will be virtual.

Option 4 is correct.