A small object is placed 30 cm in front of a thin lens of power −5 D. Where is the image formed?
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → 12 cm from the lens, on the same side as the object
A negative power belongs to a diverging (concave) lens. Change the power into a focal length in metres first. Then the thin lens formula gives the image distance.
\( u=-30\ \mathrm{cm} \)
\( \begin{aligned} f &= \frac{1}{P} \\ &= \frac{1}{-5}\ \mathrm{m} \\ &= -0.2\ \mathrm{m}=-20\ \mathrm{cm} \end{aligned} \)
\( \begin{aligned} \frac{1}{v} &= \frac{1}{f}+\frac{1}{u} \\ &= -\frac{1}{20}-\frac{1}{30} \\ &= -\frac{5}{60} \end{aligned} \)
\( v=-12\ \mathrm{cm} \)
The minus sign of \(v\) places the image on the same side as the object, so the image is virtual. Hence, the image is formed 12 cm from the lens, on the side of the object.
NCERT (2026-27):
Power of a lens – “… Thus,”
\( \displaystyle P=\frac{1}{f} \)\( \qquad (9.25) \)
Refraction by a lens – “… Applying the sign convention, BO = – u, DI = +v, we get”
\( \displaystyle \frac{1}{v}-\frac{1}{u}=\frac{1}{f} \)\( \qquad (9.23) \)