An object is placed at a distance 24 cm in front of a convex lens of focal length 8 cm.

(a) As, the object distance (= 24 cm) is greater than the focal length (= 8 cm) of the convex lens which means the object placed beyond it's focus so the lens will form a real and inverted image.

(b)

Given,

Object distance (u) = -24 cm

Focal length (f) = +8 cm

From lens formula,

$$\begin{aligned}\frac{1}{\text f} = \frac{1}{\text v} - \frac{1}{\text u} \\[1 em] ⇒ \frac{1}{\text v} = \frac{1}{\text f} + \frac{1}{\text u} \\[1 em] ⇒ \frac{1}{\text v} = \frac{1}{8}+\frac{1}{(-24)} \\[1 em] ⇒\frac{1}{8}-\frac{1}{24}= \frac{3-1}{24}\\[1 em] ⇒\frac{1}{\text v} = \frac{2}{24} \\[1 em] ⇒ \text v = \frac{24}{2} = 12 \text { cm}\end{aligned}$$

∴ The image is formed at a distance of 12 cm behind the lens.