The image formed by a concave lens of focal length 20 cm is 15 cm from the lens. The magnification of the image is
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 0.25
A concave lens forms a virtual image of a real object, on the same side as the object. So the image distance is negative. The focal length of a concave lens is negative too. The lens formula gives \(u\), and \(m=v/u\) then gives the magnification.
\( f=-20\ \mathrm{cm} \)
\( v=-15\ \mathrm{cm} \)
\( \begin{aligned} \frac{1}{u} &= \frac{1}{v}-\frac{1}{f} \\ &= -\frac{1}{15}+\frac{1}{20} \\ &= -\frac{1}{60} \end{aligned} \)
\( u=-60\ \mathrm{cm} \)
\( \begin{aligned} m &= \frac{v}{u} \\ &= \frac{-15}{-60} \\ &= 0.25 \end{aligned} \)
The value is positive and less than one. So the image is erect and smaller than the object. Hence, the magnification of the image is 0.25.
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”
\( \displaystyle m=\frac{h^{\prime}}{h}=\frac{v}{u} \)\( \qquad (9.24) \)