An object is placed 15 cm in front of a convex lens of focal length 10 cm. The distance between the object and its image is
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → 45 cm
The object is beyond the focus of the convex lens, so the image is real and forms on the other side. The object and the image then lie on opposite sides, and their distances from the lens add up.
\( u=-15\ \mathrm{cm} \)
\( f=+10\ \mathrm{cm} \)
\( \begin{aligned} \frac{1}{v} &= \frac{1}{f}+\frac{1}{u} \\ &= \frac{1}{10}-\frac{1}{15} \\ &= \frac{1}{30} \end{aligned} \)
\( v=+30\ \mathrm{cm} \)
\( \begin{aligned} \text{Separation} &= |u|+v \\ &= 15+30 \\ &= 45\ \mathrm{cm} \end{aligned} \)
Hence, the object and its image are 45 cm apart.
NCERT (2026-27):
Refraction by a lens – “… Applying the sign convention, BO = – u, DI = +v, we get”
\( \displaystyle \frac{1}{v}-\frac{1}{u}=\frac{1}{f} \)\( \qquad (9.23) \)