Q 12-09-078JEE MainJEE Main 2025 (29 Jan, Shift 1)Medium
Let $u$ and $v$ be the distances of the object and the image from a lens of focal length $f$. The correct graphical representation of $u$ and $v$ for a convex lens when $|u| > f$ is
Answer: (C) see figure
With the sign convention, $u$ is negative and $\dfrac{1}{v} = \dfrac{1}{f} + \dfrac{1}{u}$, i.e.
$$v = \frac{f|u|}{|u| - f}$$
For $|u| > f$ the image is real ($v > 0$):
- as $|u| \to f$ (the dashed line at $u = -f$), $v \to \infty$;
- as $|u|$ grows, $v$ decreases steadily towards $f$.
So the curve rises steeply next to the line $u = -f$ and flattens towards $v = f$ further away: graph (3).
Solution by Sreeraj P, M.Sc Physics