An object is placed 40 cm in front of a thin lens. The lens forms a virtual image 8 cm from it, on the same side as the object. The power of the lens is
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → −10 D
The image lies on the object side, so \(v\) is negative, like \(u\). The lens formula gives the focal length. The power is then \(P=1/f\), with \(f\) in metres.
\( u=-40\ \mathrm{cm} \)
\( v=-8\ \mathrm{cm} \)
\( \begin{aligned} \frac{1}{f} &= \frac{1}{v}-\frac{1}{u} \\ &= -\frac{1}{8}+\frac{1}{40} \\ &= -\frac{4}{40} \end{aligned} \)
\( f=-10\ \mathrm{cm} \)
\( \begin{aligned} P &= \frac{1}{f} \\ &= \frac{1}{-0.1\ \mathrm{m}} \\ &= -10\ \mathrm{D} \end{aligned} \)
The negative power marks a diverging (concave) lens. Therefore, the power of the lens is −10 D.
NCERT (2026-27):
Refraction by a lens – “… Applying the sign convention, BO = – u, DI = +v, we get”
\( \displaystyle \frac{1}{v}-\frac{1}{u}=\frac{1}{f} \)\( \qquad (9.23) \)
Power of a lens – “… Thus,”
\( \displaystyle P=\frac{1}{f} \)\( \qquad (9.25) \)
“The SI unit for power of a lens is dioptre (D): 1D = 1m–1. The power of a lens of focal length of 1 metre is one dioptre. Power of a lens is positive for a converging lens and negative for a diverging lens.”