A convex lens forms an $8.0\ \mathrm{cm}$ long image of a $2.0\ \mathrm{cm}$ long object which is kept at a distance of $6.0\ \mathrm{cm}$ from the optical centre of the lens. If the object and the image are on the same side of the lens, find (i) the nature of the image, (ii) the position of the image, and (iii) the focal length of the lens.

Since the object and image are on the same side of the lens, the image is virtual and erect (this matches the chapter's Table 9.4: when the object lies between focus $F_1$ and optical centre O, the image forms on the same side of the lens as the object, and is virtual, erect and enlarged).

Given object height $h=2.0\ \mathrm{cm}$, image height $h'=8.0\ \mathrm{cm}$, object distance $u=-6.0\ \mathrm{cm}$ (New Cartesian Sign Convention).

Magnification: $m=\dfrac{h'}{h}=\dfrac{8.0}{2.0}=4$ (positive, since image is virtual and erect, on the same side as the object).

Using $m=\dfrac{v}{u}$:

$$4=\frac{v}{-6.0}\ \Rightarrow\ v=-24.0\ \mathrm{cm}$$

(i) Nature of image: virtual, erect and enlarged (4 times the object size), on the same side of the lens as the object.

(ii) Position of image: $24.0\ \mathrm{cm}$ from the optical centre, on the same side as the object.

(iii) Focal length: using the lens formula $\dfrac{1}{v}-\dfrac{1}{u}=\dfrac{1}{f}$:

$$\frac{1}{f}=\frac{1}{-24.0}-\frac{1}{-6.0}=\frac{-1}{24.0}+\frac{1}{6.0}=\frac{-1+4}{24.0}=\frac{3}{24.0}=\frac{1}{8.0}$$

$$f=+8.0\ \mathrm{cm}$$

The positive focal length confirms it is a convex lens, and since $u\,(6\ \mathrm{cm}) < f\,(8\ \mathrm{cm})$, the object lies between the optical centre and the principal focus — consistent with the convex lens acting as a simple magnifier, exactly as described in Table 9.4 of the chapter for that object position.