For a given lens, the magnification was found twice as large when as when the object was 0.15 m distance from it as when the distance was 0.2 m. Find power of the lens.

Explanation

Let as shown 1 and 2 are positions of objects and image in two different situations

It is given,`abs(v_(1)/u_(1))=2abs(v_(2)/u_(2))`
Here,`u_(1)=-15 cmand u_(2)=-20 cm`
`v_(1)=2v_(2)xxu_(1)/u_(2)=2v_(2)xx15/20=3/2v_(2)`
Now, `1/f=1/v-1/u rArr1/f=1/v_(1)-1/u_(1)and1/f=1/v_(2)-1/u_(2)`
So, `1/v_(1)-1/u_(1)=1/v_(2)-1/u_(2)`
`rArr2/(3v_(2))+1/15=1/v_(2)+1/20rArrv_(2)=20 cm`
`v_(1)/v_(1)=2v_(2)/v_(2)=2xx20/20=2`
`rArrv_(1)=2u_(1)=2xx15=30 cm`
`therefore1/f=1/v_(1)-1/u_(1)=1/15+1/30=3/30`
`rArrf=10 cm-0.10 m`
`therefore` Power of the lens,`P=1/f=1/0.1=+10 D`