First refraction through lens

Here, u = $$-$$30 cm, f = +15 cm

Using $${1 \over v} - {1 \over u} = {1 \over f}$$

$${1 \over v} - {1 \over { - 30}} = {1 \over { + 15}}$$ or $${1 \over v} = {1 \over {15}} - {1 \over {30}}$$

v = +30 cm

$$\therefore$$ Image I₁ is real and formed at a distance of 30 cm on the right side of the lens.

Distance of image I₁ from the mirror is

30 cm $$-$$ 10 cm = 20 cm

The image I₁ acts as an virtual object for the mirror. The mirror forms an image I₂ at a distance of 20 cm in front of it.

The image I₂ acts as an object for the lens.

Second refraction through lens

Here, u = +10 cm, f = +15 cm

$${1 \over v} - {1 \over { + 10}} = {1 \over { + 15}}$$ or $${1 \over v} = {1 \over {15}} + {1 \over {10}}$$ $$\Rightarrow$$ v = +6 cm

The final image I₃ is real and at a distance of 16 cm from the mirror.