In given system of lens and mirror, position of object O in front of lens is at a distance 2f. i.e. u = 2f = 40 cm

So, image (I₁) formed is real, inverted and at a distance, v = 2f = 2 $$\times$$ 20 = 40 cm, (behind lens) magnification, $${m_1} = {v \over u} = {{40} \over {40}} = 1$$

Thus, size of image is same as that of that of object. This image (I₁) acts like a real object for mirror.

As object distance for mirror is

u = C = 2f = $$-$$ 20 cm

where, C = centre of curvature.

So, image (I₂) formed by mirror is at 2f.

$$\therefore$$ For mirror v = 2f = 2($$-$$ 10) = $$-$$ 20 cm

Magnification, $${m_2} = - {v \over u} = - {{( - 20)} \over {( - 20)}} = - 1$$

Thus, image size is same as that of object.

The image I₂ formed by the mirror will act like an object for lens.

As the object is at 2f distance from lens, so image (I₃) will be formed at a distance 2f or 40 cm. Thus, magnification,

$${m_3} = {v \over u} = {{40} \over {40}} = 1$$

So, final magnification, $$m = {m_1} \times {m_2} \times {m_3} = - 1$$

Hence, final image (I₃) is real, inverted of same size as that of object and coinciding with object.