A lens is 5cm thick and the radii of curvature of its object is placed at a distance of 12cm from the surface whose radius of curvature is 10cm. How far beyond the other surface is the image formed?

Explanation

We know that `(mu_(2))/(v)-(mu_(1))/(u)=(mu_(2)-mu_(1))/(R)`
`u=-12cm, R=10cm, mu_(1)=1, mu_(2)=1.5`
`rArr (1.5)/(v)-(1)/(-12)=(1.5-1)/(10)rArr v=-45 cm`
This iamge will serve as an object for the second surface.
For the second surface, object distance, `u=5+45=50cm`
For the second surface again, `u=-50 cm, R=-25, mu=1.5, mu_(2)=1`
`(mu_(2))/(v)-(mu_(1))/(u)=(mu_(2)-mu_(1))/(R)=(1)/(v)-(1.5)/(-50)=(1-1.5)/(-25)`
or `v=-100 cm`
Find image will be at a distance of `-95 cm` from the first surface on the same side as the object.