Q 12-09-234JEE MainJEE Main 2025 (4 Apr, Shift 1)Medium
The distance between an object and its image (magnified by $-\dfrac13$) is $30\ \text{cm}$. The focal length of the mirror used is $\left(\dfrac x4\right)\ \text{cm}$, where the magnitude of the value of $x$ is ______.
Numerical value type. Enter your answer.
Answer: 45
A real, inverted, diminished image is formed by a concave mirror with the object beyond C. $m = -\dfrac vu = -\dfrac13$, so both are on the same side and $|v| = \dfrac{|u|}{3}$.
The object–image distance: $|u| - |v| = \dfrac23|u| = 30 \Rightarrow |u| = 45\ \text{cm}$, $|v| = 15\ \text{cm}$.
$$\frac1f = \frac1v + \frac1u = -\frac1{15} - \frac1{45} = -\frac{4}{45} \Rightarrow |f| = \frac{45}{4}\ \text{cm}$$
So $x = 45$.
Solution by Sreeraj P, M.Sc Physics