Physics - Geometrical Optics Question with Solution | TestHub

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 $0=-\frac{1}{v^2}dv-\frac{1}{u^2}du$ So, $dv=-\frac{v^2}{u^2}\times b$ $\dots (2)$ (Here : $du=b$ ) From equation (1) $\begin{array}{rl}\frac{1}{v}& =\frac{1}{f}-\frac{1}{u}=\frac{u-f}{fu}\\ \frac{u}{v}& =\frac{u-f}{f}\end{array}$ $\frac{v}{u}=\frac{f}{u-f}$ $\dots (3)$ Now, from equations (2) and (3), we get $dv=-{\left(\frac{f}{u-f}\right)}^{2}b$ Therefore, size or image is $={\left(\frac{f}{u-f}\right)}^{2}b$