Question

A short linear object of length $L$ lies on the axis of a spherical mirror of focal length $f$ at a distance $b$ from the mirror. Find the size of the image.

Answer 1

$\frac{1}{v}+\frac{1}{u}=\frac{1}{f}$
By differentiating both sides
$\Rightarrow \frac{dv}{v^2}=-\frac{du}{u^2}$
$du$: size of object $=L$; $dv$:size of image;
$u$: object distance
$v$: image distance
$dv=\frac{v^2}{u^2}du$ $(\frac{1}{v}=\frac{1}{f}-\frac{1}{u}\Rightarrow v=\frac{fu}{u-f})$
$=-\frac{f^2{u}^2}{{(u-f)}^2{u}^2}du=-\frac{f^2}{{(b-f)}^2}L$
Negative sing implies that object is lying between $u$ and $u+$ and the image will lie between $v$ and $v-dv$