Q 12-09-233JEE MainJEE Main 2025 (4 Apr, Shift 1)Medium
When an object is placed $40\ \text{cm}$ away from a spherical mirror, an image of magnification $\dfrac12$ is produced. To obtain an image with magnification of $\dfrac13$, the object is to be moved:
Answer: (A) $40\ \text{cm}$ away from the mirror.
A real object giving an erect, diminished image ($m = +\tfrac12$) means a convex mirror. For a mirror $m = \dfrac{f}{f - u}$:
$$\frac12 = \frac{f}{f + 40} \Rightarrow f = +40\ \text{cm}$$
For $m = \tfrac13$:
$$\frac13 = \frac{40}{40 - u} \Rightarrow 40 - u = 120 \Rightarrow u = -80\ \text{cm}$$
The object moves from $40\ \text{cm}$ to $80\ \text{cm}$, i.e. $40\ \text{cm}$ away from the mirror.
Solution by Sreeraj P, M.Sc Physics