Question

Deduce that an object placed between F and 2F of a concave mirror produces a real image beyond 2F.

Answer 1

u =-x
f =-y
y < x < 2y
let x =ky where 1 < k < 2
$\frac{1}{v}+\frac{1}{u}=\frac{1}{f}\Rightarrow \frac{1}{v}=\frac{1}{f}-\frac{1}{u}=\frac{u-f}{uf}$
$\therefore v=\frac{uf}{u-f}=\frac{-x(-y)}{-x-(-y)}=-\frac{xy}{x-y}${-ve sign shows image is real}
$=-\frac{k{y}^2}{(k-1)y}=-\frac{ky}{k-1}$
As 1 < k < 2.
$\frac{k}{k-1}=\frac{1}{1-\frac{1}{k}}>2$
$\frac{1}{2}<\frac{1}{k}<1$
$\therefore$ Image is formed beyond 2F.