The final image  of the object O shown in the figure is formed at point 20 cm below a thin equi-concave lens, which is at a depth of 65 cm from principal axis. From the given geometry, calculate the radius of curvature in cm of lens kept at "A". (Refractive index of equi-convex lens is 1.5 and placed in air.

Explanation

Image formation due to convex lens
`(1)/(v)-(1)/(-36)=(1)/(30)impliesv=(30xx36)/(6)=180cm`
This image will act like a virtual object for mirror and after reflection from mirror its image (shown by `I_(2)` ) will be formed at 80 cm below optical axis of convex lens. ltbr. For concave lens, this image will be object at a position of 15 cm below the lens for final image formed by concave lens
`(1)/(20)-(1)/(15)=(1)/(f)" "(1)/(f)=-(5)/(300)`
Also
`(1)/(f)=(mu-1)(-(1)/(R)-(1)/(R))`
or `-(5)/(300)=((3)/(2)-1)(-(2)/(R))impliesR=(300)/(5)`