In Figure ., L is a converging lens of focal length 10cm and M Iis a concave mirror of radius of curvature 20cm. A point object O is placed in front of the lens at a distance of 15cm. Ab and CD are optical axes of the lens and mirror, respectively. Find the distance of the final image formed by this system from the optical center of the lens. The distance between CD and AB is 1cm.

Answer

`6sqrt26 cm`

Explanation

`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)/(1\5) 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`