A finite size object is placed normal to the principal axis at a distance of 30 cm from a convex mirror of…
- 45 cm
- 7.5 cm
- 22.5 cm
- 15 cm
Solution

For Convex mirror
$\begin{aligned}
& \frac{1}{\mathrm{v}}+\frac{1}{\mathrm{u}}=\frac{1}{\mathrm{f}} \\ & \frac{1}{\mathrm{v}}-\frac{1}{30}=\frac{1}{30} \\ & \frac{1}{\mathrm{v}}=\frac{2}{30}=\frac{1}{15} \Rightarrow \mathrm{v}=15 \mathrm{~cm}
\end{aligned}$
Image formed by convex mirror is at 45 cm from object so plane mirror should be placed midway at 22.5 cm from object so that both of their images may coinside,
Therefore distance between both mirrors
$=30-22.5=7.5 \mathrm{~cm}$
Correct Answer : Option 2
Asked in: JEE Main 2025 (04 Apr Shift 2)