Question

An object and a concave mirror of focal length $f=10cm$ both move along the principal axis of the mirror with constant speeds. The object moves with speed $V_0=15cm{s}^{-1}$ towards the mirror with respect to a laboratory frame. The distance between the object and the mirror at a given moment is denoted by $u$. When $u=30cm$, the speed of the mirror $V_m$ is such that the image is instantaneously at rest with respect to the laboratory frame, and the object forms a real image. The magnitude of $V_m$ is _______ $cm{s}^{-1}$.

Answer 1

Answer: 3

$u=-30cm$
$f=-10cm$
$v=\frac{f_0}{u-f}=-15cm$
$\frac{1}{v}+\frac{1}{u}=\frac{1}{f}$
$\frac{du}{dt}=-\frac{v^2}{u^2}\frac{du}{dt}$
${\overrightarrow{v}}_{lm}=-{\left(\frac{v}{u}\right)}^2{\overrightarrow{v}}_{om}$
Given ${\vec{v}}_1=\overrightarrow{0}$
${\overrightarrow{v}}_1-{\overrightarrow{v}}_m=-{\left(\frac{-15}{-30}\right)}^2\left({\overrightarrow{v}}_{O/m}\right)$
${\overrightarrow{v}}_1-{\overrightarrow{v}}_m=-\frac{1}{4}{\overrightarrow{v}}_0+\frac{1}{4}{\overrightarrow{v}}_m$
${\overrightarrow{v}}_0=15cm/s\hat{i}$
${\overrightarrow{v}}_I=\overrightarrow{0}cm/s$
$\frac{5}{4}{\overrightarrow{v}}_m=\frac{{\overrightarrow{v}}_0}{4}$
${\overrightarrow{v}}_m=\frac{{\overrightarrow{v}}_0}{4}=\frac{15cm/s\hat{i}}{5}=3m/s\hat{i}$
${\left|{\overrightarrow{v}}_m\right|}_{m}cm/s=3$