Mirror Formula and Magnification for Convex Mirror :
mirror formula for a convex mirror.
Let AB be an object lying on the principal axis of the convex mirror of small aperture. A'B' is the virtual image of the object lying behind the convex mirror.
As $\triangle ABC$ and $\triangle A'B'C$ are similar
$\frac{A'B'}{AB} = \frac{CA'}{CA} \quad \cdots (i)$
As $\triangle ABP$ and $\triangle A'B'P$ are similar
$\frac{A'B'}{AB} = \frac{PA'}{PA} \quad \cdots (ii)$
From eqns. (i) and (ii), we get,
$\frac{CA'}{CA} = \frac{PA'}{PA}$
According to the new Cartesian sign convention all distances are to be measured from pole so
$\frac{PC + PA'}{PC - PA} = \frac{PA'}{PA} \quad \cdots (iii)$
Applying sign conventions,
$PC = R$, $PA' = v$; $PA = -u$
Hence, eqn. (iii) becomes,
$\frac{R - v}{R - u} = \frac{v}{-u}$
$-uR + uv = vR - vu$ or $vR + uR = 2vu$
Dividing both sides by $uvR$, we get
$\frac{1}{v} + \frac{1}{u} = \frac{2}{R}$
But
R = 2f
$\frac{1}{f} = \frac{1}{u} + \frac{1}{v} \quad \cdots (iv)$
Magnification Produced by Convex Mirror :
Formation of the Virtual image A'B' of the object AB by the concave mirror.
$\triangle ABP$ and $\triangle A'B'P$ are similar.
$\frac{A'B'}{AB} = \frac{PA'}{PA} \quad \cdots (i)$
Applying New Cartesian sign conventions:
$AB = h$; $A'B' = h'$; $PA = -u$; and $PA' = +v$
$\therefore$ Equ(i) becomes,
$\frac{h'}{h} = \frac{v}{-u}$
$m = \frac{h'}{h} = -\frac{v}{u} \quad \cdots (iii)$
Note: Magnification is positive where because height of object is positive and height of image is positive as per New Cartesian sign conventions.