Question

An object is placed $15\text{cm}$ from a thin convex lens on its principle axis and a virtual image of certain size is formed. Now the object is moved $10\text{cm}$ away from the lens and real image of the same size as that of the previous) virtual image is formed. The focal length (in cm) of the lens is .

Answer 1

Let the focal length of the lens is $f$ and magnification is $m$.

Case-I: when object is at $15\text{cm}$, image is virtual.
$u=-15\text{cm}$
Using, $m=\frac{v}{u}$, $v=-15m$

Lens formula, $\frac{1}{f}=\frac{1}{v}-\frac{1}{u}$
$\frac{1}{f}=\frac{1}{-15m}+\frac{1}{15}$

$\frac{15}{f}=1-\frac{1}{m}$ .....(1)

Case-II: when object further moved $10\text{cm}$ away from lens, image is real (magnification is negative).
$u=-(15+10)=-25\text{cm}$
Using, $m=\frac{v}{u}$, $v=25m$

Again by lens formula, $\frac{1}{f}=\frac{1}{v}-\frac{1}{u}$
$\frac{1}{f}=\frac{1}{25m}+\frac{1}{25}$

or, $\frac{25}{f}=1+\frac{1}{m}$ .....(2)

Now, equation (1) + (2):
$\frac{40}{f}=2\Rightarrow f=20\text{cm}$