Answer
$4.5$ cm/s; away
Work Step by Step
$\frac{1}{s} +\frac{1}{S} =\frac{1}{f} $................ eq (1)
$f=6$cm, $ \frac{ds}{dt}=-2$ cm/s, $ \frac{dS}{dt}=? $, $s=10$cm
Equation (1) may be written as
$s^ {-1} +S^ {-1} =f^ {-1} $
Taking derivative with respect to t
$\frac{d(s^{-1})}{dt} + \frac{d(S^{-1})}{dt} = \frac{d(f^{-1})}{dt} $
$\frac{d(s^{-1})}{ds}\frac{ds}{dt} + \frac{d(S^{-1})}{dS} \frac{dS}{dt} = \frac{d(f^{-1})}{df}\frac{df}{dt} $
As focal length f is constant
$-s^{-2}\frac{ds}{dt} - S^{-2} \frac{dS}{dt} = 0 $
$-s^{-2}\frac{ds}{dt} = S^{-2} \frac{dS}{dt} $ Or
$S^{-2} \frac{dS}{dt} =-s^{-2}\frac{ds}{dt} $ ................ eq (2)
Putting $s=10$cm, $f=6$cm in equation (1)
$\frac{1}{10}+\frac{1}{S}=\frac{1}{6}$
$\frac{1}{S}=\frac{1}{6}-\frac{1}{10}=\frac{1}{15}$
$\frac{1}{S}=\frac{1}{15}$
$S=15$ cm
Putting $ \frac{ds}{dt}=-2$ cm/s, $s=10$cm, $S=15$ cm in equation (2)
$15^{-2} \frac{dS}{dt} =- 10^{-2} \times (-2) $
$ \frac{1}{15^2} \frac{dS}{dt} =- \frac{1}{10^2} \times (-2) =\frac{1}{50}$
$ \frac{1}{225} \frac{dS}{dt} =\frac{1}{50}$
$ \frac{dS}{dt} =225 \times\frac{1}{50}$
$ \frac{dS}{dt} =4.5$ cm/s
Since image distance from the lens is positive, image is moving away from the lens.