The focal length of a concave mirror is ‘f’ and the distance of the object to principal focus is ‘p’. Then the ratio of the size of the image to the size of the object is                                  

$\text{v}=?$

$\frac{\text{1}}{\text{f}}=\frac{\text{1}}{\text{u}}+\frac{\text{1}}{\text{v}}$

$-\frac{1}{\text{f}}=\frac{-1}{\text{p}+\text{f}}+\frac{\text{1}}{\text{v}}$

$\frac{\text{1}}{\text{v}}=\frac{-1}{\text{f}}+\frac{1}{\text{p}+\text{f}}$

$\frac{\text{1}}{\text{v}}=\frac{-\text{p}-\text{f}+\text{f}}{\text{f}\left(\text{p}+\text{f}\right)}$

 $\frac{\text{1}}{\text{v}}=\frac{-\text{p}}{\text{f}\left(\text{p}+\text{f}\right)}$

$\text{v}=\frac{-\text{f}\left(\text{p}+\text{f}\right)}{\text{p}}$

But magnification is

$\text{m}=\frac{\text{size}\text{of}\text{the}\text{image}}{\text{size}\text{of}\text{the}\text{object}}$

$=\frac{\text{image}\text{distance}}{\text{object}\text{distance}}$

$=-\left[\frac{-\text{f}\left(\text{p}+\text{f}\right)}{\text{p}\left[-\left(\text{p}+\text{f}\right)\right]}\right]=\frac{-\text{f}}{\text{p}}$

$\therefore$  Ratio is $\frac{\text{f}}{\text{p}}$