An object 6 cm high is placed 30 cm in front of a concave lens of focal length 15 cm. The height of the image is
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 2 cm
First find where the image forms, using the thin lens formula. The magnification of a lens is \(m=v/u\), and the image height is \(h^{\prime}=mh\).
\( h=6\ \mathrm{cm} \)
\( u=-30\ \mathrm{cm} \)
\( f=-15\ \mathrm{cm} \)
\( \begin{aligned} \frac{1}{v} &= \frac{1}{f}+\frac{1}{u} \\ &= -\frac{1}{15}-\frac{1}{30} \\ &= -\frac{3}{30} \end{aligned} \)
\( v=-10\ \mathrm{cm} \)
\( \begin{aligned} m &= \frac{v}{u} \\ &= \frac{-10}{-30} \\ &= \frac{1}{3} \end{aligned} \)
\( \begin{aligned} h^{\prime} &= mh \\ &= \frac{1}{3}\times6 \\ &= 2\ \mathrm{cm} \end{aligned} \)
The magnification is positive and less than one, so the image is erect, virtual and smaller than the object. Hence, the image is 2 cm high.
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.”