Question: Derive mirror formula and Magnification for a concave mirror when a real and virtual image is formed by it.
1. mirror formula and Magnification for a concave mirror when a real image is formed by it.
As $\triangle ABC$ and $\triangle A'B'C$ are similar,
$\frac{A'B'}{AB} = \frac{CA'}{CA} \quad \cdots (i)$
Also $\triangle ABP$ and $\triangle A'B'P$ are similar,
$\frac{A'B'}{AB}= \frac{PA'}{PA} \quad \cdots \text(ii)$
From eqns. (i) and (ii), we get
$\frac{CA'}{CA}= \frac{PA'}{PA} \quad \cdots (iii)$
According to the new cartesian sign conventions, all the distances are measured from the pole of the mirror.
$CA' = (PC - PA')$ and $CA = (PA - PC)\quad \cdots (iv)$
Substituting the values of eqn. (iv) in eqn. (iii), we get,
$\frac{(PC - PA')}{(PA - PC)}= \frac{PA'}{PA}$
Applying new cartesian sign convention, $PA' = -v$; $PC = -R$ and $PA = -u$
Hence, eqn. (v) becomes,
$\frac{-R + v}{-u + R} = \frac{-v}{-u} \quad \cdots (vi)$
$-uR + uv = -vu + vR$
or $uR + vR = 2uv \quad \cdots (vii)$
Dividing both sides by $uvR$, we get,
$\frac{1}{v} + \frac{1}{u} = \frac{2}{R}$
$R=2f$
$\frac{1}{f} = \frac{1}{v} + \frac{1}{u}\quad \cdots (viii)$
Equation (viii) is the mirror formula i.e. mirror equation.
Magnification Produced by a Concave Mirror For real image :
Magnification produced by a concave mirror may be positive or negative depending upon the nature of the image of an object formed by it.
Formation of the real 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,
$\frac{-h'}{h} = \frac{-v}{-u}$
$\frac{h'}{h} = -\frac{v}{u} \quad \cdots (ii)$
$m = \frac{h'}{h} = -\frac{v}{u} \quad \cdots (iii)$
Note: Magnification is negative here because height of object is positive and height of image is negative as per New Cartesian sign conventions.
2.Mirror Formula and Magnification for Concave Mirror when Virtual Image
When an object is placed between the pole and the focus of a concave mirror, virtual, erect and enlarged image is formed behind the mirror.
$\triangle ABC$ and $\triangle A'B'C$ are similar
$\because \frac{A'B'}{AB} = \frac{CA'}{CA} \quad \cdots (i)$
Also $\triangle ABP$ and $\triangle A'B'P$ are similar, therefore,
$\frac{A'B'}{AB} = \frac{PA'}{PA} \quad \cdots (ii)$
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 new Cartesian sign conventions,
$PA' = +v$; $PC = -R$ and $PA = -u$
Hence, eqn. (iii) becomes
$\frac{-R + v}{-R + u} = \frac{v}{-u}$
Ru - uv = -Rv + uv
i.e. Ru + Rv = 2uv
Dividing both sides by $uvR$, we get,
$\frac{1}{v} + \frac{1}{u} = \frac{2}{R}$
Since $R = 2f$
$\frac{1}{f} = \frac{1}{u} + \frac{1}{v} \quad \cdots (iv)$
Magnification Produced by a Concave Mirror for Virtual Image
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.