a. The first image is formed at the focus of the first lens. This is at 20cm from the first lens and hence at `u=-40cm ` from the second. Using the lens formula for the second lens,
`(1)/(v)=(1)/(u)+(1)/(f)=-(1)/(40)+(1)/(20) or v=40cm`
The final image is formed 40cm to the right of the second lens.
b. The equivalent focal length is
`(1)/(F)=(1)/(f_(1))+(1)/(f_(2))-(d)/(f_(1)f_(2))=(1)/(20)+(1)/(20)-(60)/((20)^2)`
or `F=-20cm`
It is a divergent lens. It should be kept at a distance
`D=(dF)/(f_(1)) ` behind the second lens.
Here, `D=(60xx(-20))/(20)=60cm`
Thus, the equivalent divergent lens should be placed at a distance of 60cm to the right of the second lens. The final image is formed at the focus of this divergent lens, i.e., 20cm of the left of it. It is, therefore, 40 cm to the right of the second lens.