An upright object is placed at a distance of 40 cm in front of a convergent lens of focal length 20 cm. A convergent mirror of focal length 10 cm is placed at a distance of 60 cm on the other side of the lens. The position and size of the final image will be:

We take the usual Cartesian sign-convention in which, at every stage, the positive direction is chosen along the direction in which the light is travelling at that instant. All distances are measured from the respective optical element (lens or mirror) towards the incident light; a distance measured opposite to the incident light is therefore negative.

First passage through the convergent lens
At the very beginning the light travels from left to right, so the positive axis is towards the right. The object is 40 cm to the left of the lens, hence

$$u_1=-40\ \text{cm}, \qquad f_L=+20\ \text{cm}$$

The thin-lens formula is stated first:

$$\frac1v-\frac1u=\frac1f$$

Substituting the numbers for the first passage, we have

$$\frac1{v_1}-\frac1{(-40)}=\frac1{20}\; ,$$ $$\frac1{v_1}+\frac1{40}=\frac1{20}\; ,$$ $$\frac1{v_1}=\frac1{20}-\frac1{40}=\frac1{40}\; ,$$ $$v_1=+40\ \text{cm}.$$

The image $$I_1$$ is therefore formed 40 cm to the right of the lens. The linear magnification in this first step is

$$m_1=\frac{v_1}{u_1}=\frac{+40}{-40}=-1.$$

Thus $$I_1$$ is real, inverted and of the same size as the object.

Reflection from the concave mirror
The mirror is 60 cm to the right of the lens, hence the distance of $$I_1$$ from the mirror is

$$u_2=-(60-40)=-20\ \text{cm}$$

(negative because the light, still travelling to the right, meets the mirror from its left side). For a mirror the formula is

$$\frac1v+\frac1u=\frac1f.$$

With the concave mirror focal length $$f_M=+10\ \text{cm}$$ we get

$$\frac1{v_2}+\frac1{(-20)}=\frac1{10}\; ,$$ $$\frac1{v_2}-\frac1{20}=\frac1{10}\; ,$$ $$\frac1{v_2}=\frac1{10}-\frac1{20}=\frac1{20}\; ,$$ $$v_2=+20\ \text{cm}.$$

The mirror therefore produces the image $$I_2$$ 20 cm in front of it, i.e. 20 cm to the left of the mirror and consequently 40 cm to the right of the lens once more. Its magnification is

$$m_2=-\frac{v_2}{u_2}=-\frac{+20}{-20}=-1.$$

So $$I_2$$ is again real, inverted and of the same size as $$I_1$$; overall it is upright with respect to the original object because it has undergone two inversions.

Second passage through the convergent lens
After reflection, the light now travels from right to left; the positive axis is consequently towards the left. The object for this second passage is $$I_2$$, situated 40 cm to the right of the lens, so with the new positive direction we must put

$$u_3=-40\ \text{cm}.$$

Applying the lens formula once more,

$$\frac1{v_3}-\frac1{(-40)}=\frac1{20}\; ,$$ $$\frac1{v_3}+\frac1{40}=\frac1{20}\; ,$$ $$\frac1{v_3}=\frac1{20}-\frac1{40}=\frac1{40}\; ,$$ $$v_3=+40\ \text{cm}.$$

The positive sign (with the light now moving to the left) tells us that the final image $$I_3$$ lies 40 cm to the left of the lens, that is, on the same side as the original object.

Overall magnification
The three individual magnifications are

$$m_1=-1,\qquad m_2=-1,\qquad m_3=-1.$$

Therefore the combined magnification is

$$M=m_1\,m_2\,m_3=(-1)\times(-1)\times(-1)=-1.$$

The modulus $$|M|=1$$ shows that the final image is exactly the same size as the original object (the negative sign merely indicates that it is inverted).

Thus the optical combination produces a final image 40 cm from the convergent lens and of the same size as the object.

Hence, the correct answer is Option C.