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
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 20 cm, on the same side of object
For two thin lenses in contact the powers add, \( 1/f = 1/f_1 + 1/f_2 \). With the Cartesian sign convention a convex lens has \( f_1 = +40\,\mathrm{cm} \) and a concave lens has \( f_2 = -20\,\mathrm{cm} \).
\( \displaystyle \frac{1}{f} = \frac{1}{40} - \frac{1}{20} = -\frac{1}{40} \;\Rightarrow\; f = -40\,\mathrm{cm} \)
The combination is diverging. The object is at \( u = -40\,\mathrm{cm} \) (in front of the lens). By the lens formula \( 1/v - 1/u = 1/f \):
\( \displaystyle \frac{1}{v} = \frac{1}{f} + \frac{1}{u} = -\frac{1}{40} - \frac{1}{40} = -\frac{1}{20} \)
\( \displaystyle v = -20\,\mathrm{cm} \)
The negative sign means the image is virtual, \( 20\,\mathrm{cm} \) from the lens, on the same side as the object.