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Q.
If a lens is moved towards the object from a distance of $20 \,cm$ to $15\, cm$, the magnification of the image remains the same. The focal length of the lens is
AMUAMU 2003
Solution:
From lens formula
$\frac{1}{f}=\frac{1}{v}-\frac{1}{u}$
where $f$ is focal length, $v$ the image distance and $u$ the object distance.
For convex lens
$\frac{1}{f}=\frac{1}{v}-\frac{1}{(-u)}=\frac{1}{v}+\frac{1}{u}$
Multiplying both sides by, $u$ we get
$\frac{u}{v}+1=\frac{u}{f}$
$\Rightarrow \quad \frac{u}{v}=\frac{u-f}{f}$
or $\frac{v}{u}=\frac{f}{u-f}$
or $m=\frac{f}{u-f}$
According to question
$ m_{1} =-m_{2} $
$ \therefore \frac{f}{u_{1}-f} =\frac{f}{f-u_{2}} $
$\Rightarrow \frac{f}{20-f} =\frac{f}{f-15} $
$\Rightarrow 2 f =20+15 $
$ \Rightarrow f =\frac{35}{2} = 17.5 \,cm$