An extended real object of size `2cm` is palced perpendicular to the principle axis of a converging lens of focal length `20cm`. The distance between the object and the lens is `30cm`
(i) Find the laterial magnification produced by the lens.
(ii) Find the height of the image.
(iii) Find the change in laterial magnification, if the object is brought closer to the lens by `1mm` along the principle axis.

Explanation

Using `(1)/(v) - (1)/(u) = (1)/(f)`
and `m = (v)/(u)` we get `m = (f)/(f+u).......(A)`
`:. M = (+20)/(20+(-30)) = (+20)/(-10) = -2`
`-ve` sign implies that the image is inverted.
(ii) `(h_(2))/(h_(1)) = m`
`:. h_(2) = mh_(1) = (-2)(2) =- -4cm`
(iii) Differentiating `(A)` we get
`dm = (-f)/((f+u)^(2)) du = (-(20))/((-10)^(2)) (0.1) = (-2)/(100) = -0.2`
Note that the method of different is valid only when changes are small.
Alternate method:
`u` (after displacing the object)
`= -(30 + 0.1) = -29.9cm`
Applying the formula
`m = (f)/(f+u)`
`m = (20)/(20 + 29.9) = -2.02`
`:.` change in `'m' = -0.2`
Since in this mehod difference is not used, this method can be used for any changes, small or large.