Q 12-09-083JEE MainJEE Main 2024 (1 Feb, Shift 1)Easy
The distance between an object and its 3 times magnified virtual image as produced by a convex lens is $20\ \text{cm}$. The focal length of the lens used is ______ cm.
Numerical value type. Enter your answer.
Answer: 15
For a virtual image, $v = 3u$ on the same side as the object. Let $u = -x$, $v = -3x$.
Distance between them: $3x - x = 2x = 20 \Rightarrow x = 10\ \text{cm}$. So $u = -10\ \text{cm}$, $v = -30\ \text{cm}$.
$$\frac1f = \frac1v - \frac1u = -\frac1{30} + \frac1{10} = \frac{2}{30} \Rightarrow f = 15\ \text{cm}$$
Solution by Sreeraj P, M.Sc Physics