Question

A convex lens of focal length $f$ is placed somewhere between an object and a screen. The distance between the object and the screen is $x$. If the magnification is $m$, focal length of the lens is given as :

• A. $\frac{mx}{(m+1{)}^2}$
• B. $\frac{(m-1{)}^{2}x}{m}$
• C. $\frac{mx}{(m-1{)}^2}$
• D. $\frac{(m+1{)}^2}{m}x$

Answer 1

The correct option is A $\frac{mx}{(m+1{)}^2}$

Given,

Focal length $=f$

$v+u=x....\left(1\right)$

magnification $=m$

magnification, $m=\frac{v}{u}$

$v=um.....\left(2\right)$

Put (2) in (1)

$um+u=x$

$\Rightarrow$ $u=\frac{x}{m+1}.....\left(3\right)$

from lens formula,

$\frac{1}{f}=\frac{1}{v}-\frac{1}{u}$

but $u=-u$

$\frac{1}{f}=\frac{1}{v}+\frac{1}{u}$

$=\frac{1}{um}+\frac{1}{u}$

$\frac{1}{f}=\frac{1+m}{um}....\left(4\right)$

put (3) in (4)

$\Rightarrow$ $\frac{1}{f}=\frac{{(1+m)}^2}{xm}$

$\Rightarrow$ $f=\frac{xm}{{(1+m)}^2}$