For a given lens, the magnification was found to be twice as large when the object was 0.15 m distance from it as when the distance was 0.2 m. The focal length of the lens is

Options

A. 0.15 m
B. 0.20 m
C. 0.10 m
D. 0.05 m

Answer

A

Explanation

1 and 2 are positions of objects and images in two positions. `|(v_(1))/(u_(1))|=2|(v_(2))/(u_(2))|`
`u_(1)=15cm, u_(2)=20 cm, therefore v_(1)=2. v_(2).(u_(1))/(u_(2))`
`v_(1)=v_(2)xx2xx(15)/(20)=(3)/(2)v_(2)`
`(1)/(v_(1))+(1)/(u_(1))=(1)/(f)` (convention real and virtual)
`(1)/(f)=(1)/(v_(2))+(1)/(u_(2))rArr (1)/(v_(1))+(1)/(u_(1))=(1)/(v_(2))+(1)/(u_(2))`
`rArr (2)/(3v_(2))+(1)/(15)=(1)/(v_(2))+(1)/(20)rArr v_(2)=20 m`
`therefore |(v_(2))/(u_(2))|=1, (v_(1))/(u_(1))=2 rArr v_(1)=2xx15=30 cm`
`therefore (1)/(f)=(1)/(u_(1))+(1)/(v_(1))=(1)/(15)+(1)/(30)=(3)/(30)`
`rArr f=10 cm 0.10 m`.