Q 12-09-153JEE MainJEE Main 2021 (31 Aug, Shift 1)Easy
An object is placed at the focus of a concave lens having focal length $f$. What is the magnification and distance of the image from the optical centre of the lens?
Answer: (B) $\dfrac12$, $\dfrac f2$
Concave lens: focal length $-f$, object at $u = -f$.
$$\frac1v - \frac1u = \frac{1}{(-f)} \Rightarrow \frac1v = -\frac1f - \frac1f = -\frac2f \Rightarrow v = -\frac f2$$
$m = \dfrac vu = \dfrac{-f/2}{-f} = \dfrac12$; the image is $\dfrac f2$ from the optical centre.
Solution by Sreeraj P, M.Sc Physics