A convex lens of focal length \(f\) is placed somewhere in between an object and a screen. The distance between the object and the screen is \(x\). If the numerical value of the magnification produced by the lens is \(m\), then the focal length of the lens is

Given, distance between object and screen for convex lens \(=x\)
i.c \(\quad u+v=x\)
but magnification, \(m=\frac{v}{u}\)
From Eqs. (i) and (ii), we get
\[
\begin{array}{rlrl}
& & u+m u & =x \Rightarrow u(m+1)=x \\
\Rightarrow & u & =\frac{x}{m+1}
\end{array}
\]
Again from Eqs. (i) and (ii), we get
\[
\begin{aligned}
\frac{v}{m}+v & =x \Rightarrow v\left(\frac{1}{m}+1\right)=x \\
\Rightarrow \quad & =\frac{m x}{m+1}
\end{aligned}
\]
Using lens formula ,
\[
\frac{1}{f}=\frac{1}{v}-\frac{1}{u}=\frac{1}{\left(\frac{m x}{m+1}\right)}-\frac{1}{\left(\frac{-x}{m+1}\right)}
\]
[from Eqs. (iii) and (iv)]
\[
\begin{aligned}
& =\frac{m+1}{m x}+\frac{m+1}{x}=\frac{m+1}{x}\left[\frac{1}{m}+1\right] \\
& =\frac{m+1}{x}\left(\frac{1+m}{m}\right)=\frac{(m+1)^2}{m x} \Rightarrow f=\frac{m x}{(m+1)^2}
\end{aligned}
\]