A convex lens of focal length 40 cm is placed in contact with a concave lens of focal length 20 cm. If an object is placed at 40 cm in front of the combination, the position of the image formed by the combination is

20 cm, on the same side of object

The correct answer is Option (2) → 20 cm, on the same side of object

Focal length of convex lens: $f_{1} = +40 \, cm$

Focal length of concave lens: $f_{2} = -20 \, cm$

Equivalent focal length of combination:

$\frac{1}{f} = \frac{1}{f_{1}} + \frac{1}{f_{2}} = \frac{1}{40} - \frac{1}{20} = \frac{1}{40} - \frac{2}{40} = -\frac{1}{40}$

$f = -40 \, cm$

Object distance: $u = -40 \, cm$ (object on left of lens system)

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

$\frac{1}{-40} = \frac{1}{v} - \frac{1}{-40}$

$-\frac{1}{40} = \frac{1}{v} + \frac{1}{40}$

$\frac{1}{v} = -\frac{1}{40} - \frac{1}{40} = -\frac{2}{40} = -\frac{1}{20}$

$v = -20 \, cm$

Answer: Image is formed $20 \, cm$ on the same side as the object.