When an object is at distance 10 cm and 30 cm from a thin converging lens, virtual and real image of same size is formed. The power of lens is

Here,  $u=-10cm,f=+F,m=+m_0$
 $\because \frac{v}{u}=+m_{0}orv=-10m_0$
Or   $\frac{1}{v}-\frac{1}{u}=\frac{1}{f}or-\frac{1}{10m_0}+\frac{1}{10}=\frac{1}{F}$
Or   $\frac{-1+m_0}{10m_0}=\frac{1}{F}\mathrm{.................................}\left(i\right)$
If  $u\text{'}=-30cm,f\text{'}=F$
 $\therefore m\text{'}=-m_0=\frac{v\text{'}}{u\text{'}}$

$\therefore v\text{'}=-m_{0}u\text{'}$
 $$\begin{gathered} =-m_0\left(-30\right)=30m_0 \\ \therefore \frac{1}{30m_0}+\frac{1}{30}=\frac{1}{F}\mathrm{...............................}\left(ii\right) \end{gathered}$$
 From Eqs. (i) and (ii) we get,  $\frac{-1+m_0}{10m_0}=\frac{1}{30m_0}+\frac{1}{30}$
Or   $\frac{-1+m_0}{10m_0}=\frac{1+m_0}{30m_0}$
Or $-30m_0+30m_0^2=10m_0+10m_0^2$
Or $20m_0^2=40m_0$
 $\therefore m_0=2$
 $\therefore \frac{1}{F}=\frac{1}{60}+\frac{1}{30}=\frac{1+2}{60}=\frac{1}{20}$
          P=1/F = 100/20=5D