A linear object AB is placed along the axis of a concave mirror. The object is moving towards the mirror with speed u. The speed of the image of the point A is 4u and the speed of the image of B is also 4u but in opposite direction. If the centre of the line AB is at a distance L from the mirror then find out the length of the object.

For concave mirror, if x and y are object distance and image distance, respectively, we have

$$\begin{gathered} -\frac{1}{x}-\frac{1}{y}=-\frac{1}{\left|f\right|} \\ \Rightarrow \frac{1}{x}+\frac{1}{y}=-\frac{1}{\left|f\right|} \end{gathered}$$

$$\begin{gathered} \Rightarrow -\frac{1}{x^2}\frac{dx}{dt}-\frac{1}{y^2}\frac{dy}{dt}=0 \\ \Rightarrow \left|\frac{V_x}{V_y}\right|=\frac{x^2}{y^2} \end{gathered}$$

For, $\left|\frac{V_x}{V_y}\right|=\frac{1}{4},\frac{x}{y}=\pm \frac{1}{2}$

For $\raisebox{1ex}{$x$}\!\left/ \!\raisebox{-1ex}{$y$}\right.=\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.$, we get $x=\raisebox{1ex}{$3\left|f\right|$}\!\left/ \!\raisebox{-1ex}{$2$}\right.$ [for point A] and for $\raisebox{1ex}{$x$}\!\left/ \!\raisebox{-1ex}{$y$}\right.=-\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.$, we get $x=\raisebox{1ex}{$\left|f\right|$}\!\left/ \!\raisebox{-1ex}{$2$}\right.$ [for point B].

As the middle point happens to be focus of the mirror, we get $\left|f\right|=L$.

hence length of the object AB = L.