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=+m0 ∵vu=+m0orv=−10m0Or   1v−1u=1for−110m0+110=1FOr   −1+m010m0=1F.................................iIf  u'=−30cm,f'=F ∴m'=−m0=v'u'∴v'=−m0u' =−m0−30=30m0 ∴130m0+130=1F...............................ii  From Eqs. (i) and (ii) we get,  −1+m010m0=130m0+130Or   -1+m010m0=1+m030m0Or −30m0+30m02=10m0+10m02Or 20m02=40m0 ∴m0=2 ∴1F=160+130=1+260=120          P=1/F = 100/20=5D