A point object is moving at a constant speed of \(1 \mathrm{~ms}^{-1}\) along the principal axis of a convex lens of focal length \(10 \mathrm{~cm}\). The speed of the image is also \(1 \mathrm{~ms}^{-1}\), when the object is at \(\mathrm{cm}\) from the optical centre of the lens.

Given, \(f=10 \mathrm{~cm}\)
\(\frac{d u}{d t}=1 \mathrm{~ms}^{-1}, \frac{d v}{d t}=1 \mathrm{~ms}^{-1}\)
From lens formula
\(\begin{gathered}\frac{1}{f}=\frac{1}{u}+\frac{1}{v} \\ \frac{1}{v}=\frac{1}{f}-\frac{1}{u}=\frac{u-f}{f u} \Rightarrow v=\frac{f u}{u-f}\end{gathered}\)
Differentiating Eq. (i) w.r.t. \(t\),
\(\begin{aligned} \frac{d\left(\frac{1}{f}\right)}{d t} & =\frac{d\left(\frac{1}{u}\right)}{d t}+\frac{d\left(\frac{1}{v}\right)}{d t} \\ 0 & =-\frac{1}{u^2} \frac{d u}{d t}-\frac{1}{v^2} \frac{d v}{d t} \Rightarrow \frac{1}{u^2} \frac{d u}{d t}=-\frac{1}{v^2} \frac{d v}{d t} \\ \Rightarrow-\frac{v^2}{u^2} & =\frac{d v}{d t} \times \frac{d t}{d u}\end{aligned}\)
Putting Eq. (ii) values in the above equation,
\(-\frac{(f u)^2}{(u-f)^2} \frac{1}{u^2}=\frac{d v}{d t} \times \frac{d t}{d u} \Rightarrow(u-f)^2=f^2 \times \frac{d u}{d t} \times \frac{d t}{d v}\)
Substituting given values in above equation,
\((u-10)^2=(10)^2 \times 1 \times \frac{1}{1}=10^2\)
or \(u-10=10 \Rightarrow u=20 \mathrm{~cm}\)