Q125 / 180 · KCET 2022 · Physics
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
- A$\frac{m x}{(m-1)^2}$
- B$\frac{(m+1)^2 x}{m}$
- C$\frac{(m-1)^2 x}{m}$
- Correct answerD$\frac{m x}{(m+1)^2}$
Solution
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)]
$$
\begin{aligned}
& =\frac{m+1}{m x}+\frac{m+1}{x}=\frac{m+1}{x}\left[\frac{1}{m}+1\right] \\
& =\frac{m+1}{x}\left(\frac{1+m}{m}\right)=\frac{(m+1)^2}{m x} \Rightarrow f=\frac{m x}{(m+1)^2}
\end{aligned}
$$
Hence, the correct answer is option: 'D'.