Suppose an object has thickness du so that it extends from object distance u to `u+du.` Prove that the thickness dv of its image is given by `(-v^2/u^2)du,` so the longitudinal magnification `(dv)/(du)=-m^2,` where m is the lateral magnification.

Answer

A

Explanation

`:' 1/v-1/u=1/f`
Differentiating this equation, we get
`-v^-2.dv+u^-2.du=0 (as `f="constant"`)`
`:. dv=(-v^2/u^2).du`