Question

An object moves with a uniform velocity $u_0=5m/s$ along the axis of a concave spherical mirror of focal length $f=-10cm$. If the object is at the centre of curvature $(C)$ at certain instant, then the magnitude of acceleration of image at this instant is $am/s^2$. Find $a$.

Answer 1

Answer: 5

$\frac{1}{v}+\frac{1}{u}=\frac{1}{f}$
Differentiating above equation w.r.t. time $t$,
$-\frac{1}{v^2}\frac{dv}{dt}-\frac{1}{u^2}\frac{du}{dt}=0$
or $\frac{dv}{dt}=-\frac{v^2}{u^2}\frac{du}{dt}$
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,
$a_i=\frac{dv}{dt}=\frac{-d}{dt}\left(m^2{u}_0\right)$
$=-2m\frac{dm}{dt}\cdot u_0$
Now, $m=-\frac{v}{u}$
$\Rightarrow \frac{dm}{dt}=\frac{d}{dt}\left(\frac{v}{u}\right)=\frac{1}{u}\frac{dv}{dt}-\frac{v}{u^2}\frac{du}{dt}$
$=\frac{1}{u}\left(-m^2{u}_0\right)-\frac{v{u}_0}{u^2}$
$=-\frac{v{u}_0}{u^2}\left(1+\frac{v}{u}\right)=-\frac{v{u}_0}{u^2}\left(\frac{v}{f}\right)=\frac{v^2{u}_0}{u^{2}f}$
Thus, instantaneous acceleration of the image
$a_i=-2m{u}_0\left(\frac{dm}{dt}\right)=-2m{u}_0\left(m^2\frac{u_0}{f}\right)$
$\Rightarrow \left|a_i\right|=\left|\frac{2m^3{u}_0^2}{f}\right|$
Now take $|m|=1$ for object at $C$.