Question

In the figure shown `L` is a converging lens of foacl length `10 cm` and `M` is a concave mirror of radius of curvature `20 cm`. A point object `O` is placed in front of the lens at a distance `15 cm.AB` and `CD` are optical axes of the lens and mirror repectively. Find the distance of the final image formed by this system from the optical centre of the lens. The distance between `CD & AB` is `1 cm`

Answer 1

Correct Answer - `6sqrt26 cm`
`I_(1)` is the iamge of object `O` formed by the lens.
`(1)/(v_(1))-(1)/(u_(1))=(1)/(f) u_(1)=-15 f_(1)=10`
Solving we get
`v_(1)=30cm`
`I_(1)` acts as source for mirror
`therefore u_(2)=-(45-v_(1))=-15 cm`
`I_(2)` is the iamge formed by the mirror
`therefore (1)/(v_(2))=(1)/(f_(m))-(1)/(u_(2))=-(1)/(10)+(1)/(15) thereforev_(2)=-30 cm`

The height of `I_(2)` above pricnipal axis of lins is `=(v_(2))/(u_(2))xx1+1=3cm`
`I_(2)` acts a source for lens `therefore u_(3)=-(45-v_(2))=-15 cm`
Hence the lens forms an image `I_(3)` at a distance `v_(3)=30 cm` to the left of lens and at a distance
The height of `I_(2)` above principal axis of lens is `|(v_(3))/(u_(3))|xx 3 cm =6 cm`
`therefore` required distance `=sqrt(30^(2)+6^(2))=6sqrt(26) cm`