Correct Answer - 160cm
For `L_(1)` , from lens formula,
`(1)/(v)-(1)/(-20)=(1)/(16),rArr(1)/(v)=(1)/(16)-(1)/(20)`
`rArr v=20cm`
`[` image is real, inverted, andn magnified `]`
`m_(1)=(v)/(u)=(80)/(-20)=-4`
Let `2^(nd)` lens is at a distance of x cm from `I_(1)` Since final image has to be erect w.r.t. original object, it has to be inverted and diminished w.r.t. `I_(1)` .
For `L_(2), u=-x, v=v, m_(2)=-(1)/(4)`
So, `(v)/(-x)=-(1)/(4)rArr v=(x)/(4)`
Now, from tlens formula,
`(1)/(v)-(1)/(u)=(1)/(f)`
`(1)/(x//4)-(1)/(-x)=(1)/(16) `
`(5)/(x)=(1)/(16)rArr x=80cm`
Hence, the separation between the lenses is 160cm.