Question

An object moves with a uniform velocity $u_{0}=5 \,m / s$ along the axis of a concave spherical mirror of focal length $f=-10\, cm$. If the object is at the centre of curvature $(C)$ at certain instant, then the magnitude of acceleration of image at this instant is $a m / s ^{2}$. Find $a$.

Answer 1

$\frac{1}{v}+\frac{1}{u}=\frac{1}{f}$
Differentiating above equation w.r.t. time $t$,
$-\frac{1}{v^{2}} \frac{d v}{d t}-\frac{1}{u^{2}} \frac{d u}{d t}=0$
or $\frac{d v}{d t}=-\frac{v^{2}}{u^{2}} \frac{d u}{d t}$
Instantaneous velocity of the image,
$v_{i}=-\frac{v^{2}}{u^{2}} u_{0}=-m^{2} u_{0}$
Now, the instantaneous acceleration of the image,
$a_{i}=\frac{d v}{d t}=\frac{-d}{d t}\left(m^{2} u_{0}\right)$
$=-2 m \frac{d m}{d t} \cdot u_{0}$
Now, $m=-\frac{v}{u}$
$\Rightarrow \frac{d m}{d t} =\frac{d}{d t}\left(\frac{v}{u}\right)=\frac{1}{u} \frac{d v}{d t}-\frac{v}{u^{2}} \frac{d u}{d t} $
$=\frac{1}{u}\left(-m^{2} u_{0}\right)-\frac{v u_{0}}{u^{2}}$
$=-\frac{v u_{0}}{u^{2}}\left(1+\frac{v}{u}\right)=-\frac{v u_{0}}{u^{2}}\left(\frac{v}{f}\right)=\frac{v^{2} u_{0}}{u^{2} f}$
Thus, instantaneous acceleration of the image
$a_{i}=-2 m u_{0}\left(\frac{d m}{d t}\right)=-2 m u_{0}\left(m^{2} \frac{u_{0}}{f}\right) $
$\Rightarrow\left|a_{i}\right|=\left|\frac{2 m^{3} u_{0}^{2}}{f}\right|$
Now take $|m|=1$ for object at $C$.