The minimum distance between an object and its real image formed by a convex lens of focal length \(f\) is

We know:

Let the seperation between object and image by \(x\),
\(
x=u-v
\)
Using equation (1), on rewriting,
\(
\begin{aligned}
& x=u-\frac{x f}{u} \\
& \Rightarrow x=\frac{u^2}{(u+f)}---(1)
\end{aligned}
\)
we first take log of equation (1)
\(
\Rightarrow \ln (x)=2 \ln (u)-\ln (u+f)
\)
Subsequently, to minimize \(x\) w.r.t. \(u\), take the derivative of \(x\) w.r.t. \(u\) :
\(
\begin{aligned}
& \Rightarrow \frac{1}{x} \frac{d x}{d u}=\frac{2}{u}-\frac{1}{(u+f)} \\
& \Rightarrow \frac{d x}{d u}=\frac{x(u+2 f)}{u(u+f)}---(2)
\end{aligned}
\)
Condition for minimum \(x\) requires, \(\frac{d x}{d u}=0\) and \(\frac{d^2 x}{\partial u^2}<0\)
\(
\Rightarrow \frac{d x}{d u}=\frac{x(u+2 f)}{u(u+f)}=0
\)
So, \(x=0\) \{is the trivial solution \(\}\) and the real solution occurs for \(u=-2 f=-R\). Lets check the minimum value using equation (1),
\(
\Rightarrow x_{\min }=\frac{u^2}{(u+f)}=\frac{(-2 f)^2}{(-2 f+f)}=4 f
\)