An object placed 30 cm in front of a concave mirror forms a real imag…

The correct answer is Option (3) → 20 cm behind the mirror

The first position gives the focal length of the mirror. A real image lies in front of the mirror, so both distances are negative. The same mirror equation then gives the new image.

    \( \begin{aligned} \frac{1}{f} &= \frac{1}{v}+\frac{1}{u} \\ &= -\frac{1}{60}-\frac{1}{30} \\ &= -\frac{3}{60} \end{aligned} \)

    \( f=-20\ \mathrm{cm} \)

For the new position, \( u=-10\ \mathrm{cm} \).

    \( \begin{aligned} \frac{1}{v} &= \frac{1}{f}-\frac{1}{u} \\ &= -\frac{1}{20}+\frac{1}{10} \\ &= \frac{1}{20} \end{aligned} \)

    \( v=+20\ \mathrm{cm} \)

The object is now inside the focal length, so the image is virtual. Hence, the new image is formed 20 cm behind the mirror.

NCERT (2026-27):
The mirror equation – “Dividing it by v, we get”

    \( \displaystyle \frac{1}{v}+\frac{1}{u}=\frac{1}{f} \)\( \qquad (9.7) \)

“This relation is known as the mirror equation.”