Question

An object and a concave mirror of focal length $f=10 \,cm$ both move along the principal axis of the mirror with constant speeds. The object moves with speed $V_0=15\, cm s ^{-1}$ 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 $u=30\, cm$, 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

$u =-30 \,cm $
$ f =-10 \,cm$
$ v =\frac{ f _0}{ u - f }=-15 \, cm $
$ \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=15 \, cm / 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{15 cm / s \hat{ i }}{5}=3 m / s \hat{ i } $
$\left|\overrightarrow{ v }_{ m }\right|_{ m } cm / s =3$