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
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → - 40 cm
Magnification formula: $m = \frac{v}{u} = \frac{1}{2} $ i.e. $ u = 2v$
The physical distance between them is 20 cm, so taking magnitudes:
u - v = 20
2v - v = 20
v = 20 cm
Thus, $u = 2 \times 20 = 40\text{ cm}$
Since a concave lens forms a virtual image on the same side as the object, both values are negative: $u = -40\text{ cm}$ and $v = -20\text{ cm}$.
Now, use the standard lens formula: $\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$
$\frac{1}{f} = \frac{1}{-20} - \frac{1}{-40} = -\frac{1}{20} + \frac{1}{40} = -\frac{1}{40}$
$f = -40\text{ cm}$