A convex lens of focal length 20 cm is used as a magnifying glass. The lens forms an erect image 60 cm from it, on the same side as the object. The distance of the object from the lens and the magnification are
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 15 cm and \(+4\)
An erect image formed on the same side as the object is a virtual image. Its distance is taken as negative, \(v=-60\) cm. The lens formula gives \(u\), and the ratio \(v/u\) gives the magnification.
\( f=+20\ \mathrm{cm} \)
\( v=-60\ \mathrm{cm} \)
\( \begin{aligned} \frac{1}{u} &= \frac{1}{v}-\frac{1}{f} \\ &= -\frac{1}{60}-\frac{1}{20} \\ &= -\frac{4}{60} \end{aligned} \)
\( u=-15\ \mathrm{cm} \)
\( \begin{aligned} m &= \frac{v}{u} \\ &= \frac{-60}{-15} \\ &= +4 \end{aligned} \)
A positive magnification fits the erect image. Hence, the object is 15 cm from the lens and the magnification is +4.
NCERT (2026-27):
Refraction by a lens – “… Applying the sign convention, BO = – u, DI = +v, we get”
\( \displaystyle \frac{1}{v}-\frac{1}{u}=\frac{1}{f} \)\( \qquad (9.23) \)
Refraction by a lens – “Magnification (m) produced by a lens is defined, like that for a mirror, as the ratio of the size of the image to that of the object. … it is easily seen that for a lens”
\( \displaystyle m=\frac{h^{\prime}}{h}=\frac{v}{u} \)\( \qquad (9.24) \)
“When we apply the sign convention, we see that, for erect (and virtual) image formed by a convex or concave lens, m is positive, while for an inverted (and real) image, m is negative.”