Question

Rate at which magnification is changing is given as $x\times {10}^3/s$. Find $x$. Assume that police jeep is on axis of the mirror.

Answer 1

f = 10 m
We have $m=\frac{f}{f-u}=\frac{1}{10}$
Let $u=-d$ where $d$ is a positive quantity
$⟹\frac{10}{10-(-d)}=\frac{1}{10}$
$⟹d=90$ and hence $u=-90$
$v=-mv=\frac{-1}{10}\times (-90)=+9$
Now $\frac{dm}{dt}=\left[\frac{(-90)(-1\times {10}^{-2})-9\left(1\right)}{{90}^2}\right]$ per sec
$⟹\frac{dm}{dt}=\left[\frac{-81}{10\times 8100}\right]=+1\times {10}^{-3}$per sec.
Hence magnification is changing at the rate of $1\times {10}^{-3}$ per sec