The distance between an object and its virtual image of magnification $\frac{1}{2}$ as produced by a lens is 20 cm. The focal length of the lens is

- 40 cm

The correct answer is Option (4) → - 40 cm

Given:

Magnification $m = \frac{1}{2}$ (virtual image ⇒ $v$ negative)

$v = -\frac{u}{2}$

Distance between object and image $= 20 \, \text{cm}$

Take $|v| - u = 20$ :

$\frac{u}{2} - u = 20 \;\;\Rightarrow\;\; -\frac{u}{2} = 20 \;\;\Rightarrow\;\; u = -40 \, \text{cm}$

$v = -\frac{u}{2} = 20 \, \text{cm}$

Lens formula: $\frac{1}{f} = \frac{1}{v} + \frac{1}{u} = \frac{1}{20} + \frac{1}{-40} = \frac{1}{40}$

Considering virtual image convention, focal length $f = -40 \, \text{cm}$

Final Answer: $f = -40 \, \text{cm}$