Question

A lens of focal length 12 cm forms an erect image three times the size of the object. The distance between the object and image is:

(a) 8 cm
(b) 16 cm
(c) 24 cm
(d) 36 cm

Answer 1

(b) 16 cm
Given:
Magnification, m = 3
Focal length f = 12 cm
Image distance v = ?
Object distance u = ?
We know that:
m = $\frac{v}{u}$
Therefore
3 = $\frac{\mathrm{v}}{\mathrm{u}}$
3u = v
Putting these values in lens formula, we get:
$$\begin{gathered} \frac{1}{v}-\frac{1}{u}=\frac{1}{f} \\ \Rightarrow \frac{1}{3u}-\frac{1}{u}=\frac{1}{12} \\ \Rightarrow \frac{1-3}{3u}=\frac{1}{12} \\ \Rightarrow \frac{-2}{3u}=\frac{1}{12} \\ \Rightarrow 3u=-24 \end{gathered}$$

$$\begin{gathered} \Rightarrow u=\frac{-24}{3} \\ \Rightarrow u=-6\mathrm{cm} \end{gathered}$$

v = 3u = $-$8 $\times$ 3 = $-$24 cm
Here, minus sign show that image is formed on the left side of the lens.
Distance between image and object = 24 $-$ 8 = 16 cm