A short linear object of length b lies along the axis of a concave...

Using the relation for the focal length of concave mirror $\frac{1}{f}=\frac{1}{v}+\frac{1}{u}$ ...(1) Differentiating equation (1), we obtain $0=-\frac{1}{v^2} d v-\frac{1}{u^2} d u$ So, $\quad d v=-\frac{v^2}{u^2} \times b \ldots(2)$ (Here : $d u=b$ ) From equation (1) $$\begin{aligned} \frac{1}{v} & =\frac{1}{f}-\frac{1}{u}=\frac{u-f}{f u} \\ \frac{u}{v} & =\frac{u-f}{f} \\ \end{aligned}$$\frac{v}{u}=\frac{f}{u-f} \ldots(3)$Now, from equations (2) and (3), we get$d v=-\left(\frac{f}{u-f}\right)^2 b$Therefore, size or image is$=\left(\frac{f}{u-f}\right)^2 b$