A thin converging lens with focal length `f = 25 cm` projects the image of an object on a screen removed from the lens by a distance `l == 5.0 m`. Then the screen was draws closer to the lens by a distance `Delta l = 18 cm`. By what distance should the object be shifted for its image to become sharp again ?

Explanation

Lens formula can be written as follows :
`(1)/(v)-(1)/(u)=(1)/(f)`
Hence for the first case we can substitute has values.
`rArr" "(1)/(+25)=(1)/(+500)-(1)/(u)`
`rArr" "(1)/(u)=(1)/(500)-(1)/(25)`
`rArr" "(1)/(u)=(1-20)/(500)`
`rArr" "u=-(500)/(19)cm=-26.31cm`
We can see that object is very close to focus because screen is at 500 cm which is large compared to focal length of lens.
Now for the second case screen is moved closer to lens by 18 cm, hence for lens formula `v=+(500-18)=482cm`
Let u. represent new location of object.
`(1)/(v)-(1)/(u.)=(1)/(f)`
`rArr" "(1)/(482)-(1)/(u.)=(1)/(25)`
`rArr" "(1)/(u.)=(1)/(482)-(1)/(25)=(25-482)/(482xx25)`
`rArr" "u.=-(482xx25)/(457)=-26.368cm`
Hence object needs to be moved little away from focus by a distance as follow :
Distance moved by object `=26.368-26.316=0.052cm`.
Alternate method :
Lens formula can be written as follows :
`(1)/(v)-(1)/(u)=(1)/(f)`
Hence for the first case we can substitute the values.
`rArr" "(1)/(+25)=(1)/(+500)-(1)/(u)`
`rArr" "(1)/(u)=(1)/(500)-(1)/(25)`
`rArr" "(1)/(u)=(1-20)/(500)`
`rArr" "u=-(500)/(19)cm`
We have lens formula as follows :
`(1)/(v)-(1)/(u)=(1)/(f)`
On differentiating the above formula we get the following :
`-(1)/(v^(2))(dv)/(dt)+(1)/(u^(2))(du)/(dt)=0`
Focal length is constant hence its differential is zero.
From the above reation we can write the following :
`(dv)/(v^(2))=(du)/(u^(2))`
`rArr" "du=((u)/(v))^(2)dv`
`rArr" "Deltau=((u)/(v))^(2)Deltav`
Here `Deltav=-18cm` because screen is moved closer to lens.
`rArr" "Deltau=((500)/(19xx500))^(2)(-18)=-0.049cm`
Hence object is to be moved appromiately by 5 mm away from the lens.