Q 12-09-131JEE MainJEE Main 2023 (1 Feb, Shift 1)Medium
A thin cylindrical rod of length $10$ cm is placed horizontally on the principal axis of a concave mirror of focal length $20$ cm. The rod is placed in such a way that mid point of the rod is at $40$ cm from the pole of mirror. The length of the image formed by the mirror will be $\dfrac x3$ cm. The value of $x$ is ______.
Numerical value type. Enter your answer.
Answer: 32
The ends are at $35$ cm and $45$ cm.
$u=35$: $\dfrac1v=\dfrac1{20}-\dfrac1{35}=\dfrac{3}{140}\Rightarrow v=\dfrac{140}{3}$ cm. $\ u=45$: $\dfrac1v=\dfrac1{20}-\dfrac1{45}=\dfrac1{36}\Rightarrow v=36$ cm.
Length $=\dfrac{140}{3}-36=\dfrac{32}{3}$ cm, so $x=32$.
Solution by Sreeraj P, M.Sc Physics