Q 12-09-187JEE MainJEE Main 2020 (3 Sep, Shift 2)Medium
When an object is kept at a distance of $30\ \text{cm}$ from a concave mirror, the image is formed at a distance of $10\ \text{cm}$ from the mirror. If the object is moved with a speed of $9\ \text{cm s}^{-1}$, the speed (in $\text{cm s}^{-1}$) with which image moves at that instant is ______.
Numerical value type. Enter your answer.
Answer: 1
Differentiating the mirror equation $\dfrac1v + \dfrac1u = \dfrac1f$: $\dfrac{dv}{dt} = -\dfrac{v^{2}}{u^{2}}\dfrac{du}{dt}$.
$$|v_{image}| = \left(\frac{10}{30}\right)^{2}\times9 = 1\ \text{cm s}^{-1}$$
Solution by Sreeraj P, M.Sc Physics