Three identical thin convex lenses are placed in contact. The combination forms the image of a distant object 10 cm from it. One of the three lenses is now removed. The image of the same object is now formed at a distance from the combination of
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 15 cm
The image of a distant object is formed at the focus. So in both cases the image distance equals the focal length of the combination. For thin lenses in contact, the reciprocals of the focal lengths add. Let \(f\) be the focal length of one lens.
\( F_3=10\ \mathrm{cm} \)
\( \begin{aligned} \frac{1}{F_3} &= \frac{1}{f}+\frac{1}{f}+\frac{1}{f} \\ \frac{1}{10} &= \frac{3}{f} \\ f &= 30\ \mathrm{cm} \end{aligned} \)
\( \begin{aligned} \frac{1}{F_2} &= \frac{1}{f}+\frac{1}{f} \\ &= \frac{2}{30} \\ F_2 &= 15\ \mathrm{cm} \end{aligned} \)
With one lens fewer, the combination is weaker, so its focal length is longer. Hence, the image is now formed 15 cm from the combination.
NCERT (2026-27):
Refraction by a lens – “The point where image of an object placed at infinity is formed is called the focus F, of the lens and the distance f gives its focal length.”
Combination of thin lenses in contact – “The derivation is valid for any number of thin lenses in contact. If several thin lenses of focal length f1, f2, f3,... are in contact, the effective focal length of their combination is given by”
\( \displaystyle \frac{1}{f}=\frac{1}{f_1}+\frac{1}{f_2}+\frac{1}{f_3}+\ldots \)\( \qquad (9.31) \)