A thin plano-convex lens of focal length f is split into two halves. One of the halves is shifted along the optical axis as shown in figure. The sepration between object and image planes is `1.8m.` The magnification of the image formed by one of the half lens is `2.` Find the focal length of the lens and separation between the two halves. Draw the ray diagram for image formation.

Explanation

Let magnification caused by the first lens be 2 and distance `OL_(1)=x.` Distance v of image from first lens`L_(1)` is given by
`m=(v)/(u)=2 rArr v=2u=2x.`
Clearly, `u+v=1.8rArr x+2x=1.8m `
or `3x = 1.8m rArr x=(1.8)/(3)=0.6m`
By sign convention,
`u=-x=-0.6m, upsilon=2x=1.2m`
Lens formula `(1)/(f)=(1)/(v)-(1)/(u)` gives
`(1)/(f)=(1)/(1.2)+(1)/(0.6)=(1+2)/(1.2)`
`:.` Focal length `f=(1.2)/(3) =0.4m`
For real image, lens formula takes the form
`(1)/(f)=(1)/(v)+(1)/(u)`
Clearly, u and v are interchangeable. Therefore, for lens `L_(2)`
`u^(')=v=1.2m` and `v^(')=0.6m`
`OL_(1)=L_(2)I_(2)=x`
If d is the separation between the lenses, then
`x+d+x=1.8m`
`:. d=1.8-2x=1.8-2xx0.6=0.6m`
Method-2 Since the magnification for `L_(1)` is 2
`rArr (v)/(u)=-2rArr ((D+d)/(2))/(-(D-d)/(2))=-2`
`(D+d)/(D-d)=2rArr D=1.8m,d=0.6m.`
`f=(D^(2)-d^(2))/(4D)=((1.8+0.6)(1.8-0.6))/(4xx1.8)=0.4m`