A thin convex lens of focal length 20 cm and a thin concave lens of focal length 60 cm are kept in contact. A sharp image of a small object is formed on a screen placed 90 cm from the pair. The distance of the object from the pair is
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 45 cm
Two thin lenses in contact act as one lens of focal length \(F\), with \(1/F=1/f_1+1/f_2\). The image on the screen is real, so it lies on the other side and \(v\) is positive. The thin lens formula for the pair then gives \(u\).
\( f_1=+20\ \mathrm{cm} \)
\( f_2=-60\ \mathrm{cm} \)
\( v=+90\ \mathrm{cm} \)
\( \begin{aligned} \frac{1}{F} &= \frac{1}{f_1}+\frac{1}{f_2} \\ &= \frac{1}{20}-\frac{1}{60} \\ &= \frac{1}{30} \end{aligned} \)
\( F=+30\ \mathrm{cm} \)
\( \begin{aligned} \frac{1}{u} &= \frac{1}{v}-\frac{1}{F} \\ &= \frac{1}{90}-\frac{1}{30} \\ &= -\frac{2}{90} \end{aligned} \)
\( u=-45\ \mathrm{cm} \)
The minus sign puts the object in front of the pair. Hence, the object is 45 cm from the pair.
NCERT (2026-27):
Combination of thin lenses in contact – “If the two lens-system is regarded as equivalent to a single lens of focal length f, we have”
\( \displaystyle \frac{1}{v}-\frac{1}{u}=\frac{1}{f} \)
“so that we get”
\( \displaystyle \frac{1}{f}=\frac{1}{f_1}+\frac{1}{f_2} \)\( \qquad (9.30) \)