Ferive the relation between u,v,f,n and …

When, the object lies beyond the C a real and inverted image II’ is formed between C and F after reflection from the concave mirror.
From similar triangles BOA and BFP
{tex}\frac{{{\text{AB}}}}{{{\text{PB}}}}{\text{ = }}\frac{{{\text{AO}}}}{{\text{P}}}{/tex}
{tex}\frac{{{\text{AB}}}}{{{\text{AO}}}}{\text{ = }}\frac{{{\text{PB}}}}{{{\text{PF}}}}{/tex}.......(i)

From similar triangles AIB and AFP
{tex}\frac{{{\text{AB}}}}{{{\text{PS}}}}{\text{ = }}\frac{{{\text{BI}}}}{{{\text{PF}}}}{/tex}
(or) {tex}\frac{{{\text{AB}}}}{{{\text{BI}}}}{\text{ = }}\frac{{{\text{PA}}}}{{{\text{PF}}}}{/tex}...........(ii)

Adding (i) and (ii)
{tex}\frac{{{\text{AB}}}}{{{\text{AO}}}}{\text{ + }}\frac{{{\text{AB}}}}{{{\text{BI}}}}{\text{ = }}\frac{{{\text{PB}}}}{{{\text{PF}}}}{\text{ + }}\frac{{{\text{PA}}}}{{{\text{PF}}}}{/tex}
={tex}\frac{{{\text{PB + PA}}}}{{{\text{PF}}}}{/tex}={tex}\frac{{{\text{AB}}}}{{{\text{PF}}}}{/tex}
{tex}\frac{{\text{1}}}{{{\text{AO}}}}{\text{ + }}\frac{{\text{1}}}{{{\text{BI}}}}{\text{ = }}\frac{{\text{1}}}{{{\text{PF}}}}{/tex}
Using cartesian sign convention
AO = Object distance  -u
BI= Image distance  -v
PF = Focal length = -f
{tex}\frac{1}{{ - u}} + \frac{1}{{ - v}} = - \frac{1}{f}{/tex}
{tex}\frac{1}{u} + \frac{1}{v} = \frac{1}{f}{/tex}
This is the mirror formula.