Question

Derive mirror formula.

Answer 1

The figure shows an object AB at a distance u from the pole of a concave mirror. The image $A_1{B}_1$ is formed at a distance v from the mirror. The position of the image is obtained by drawing a ray diagram.

Consider the $\Delta$$A_{1}C{B}_1$ and $\Delta$ACB
$\angle A_{1}C{B}_1=\angle ACB$ (vertically opposite angles)
$\angle A{B}_{1}C=\angle ABC$ (right angles)
$\angle B_1{A}_{1}C=\angle BAC$
[when two angles of $D{A}_{1}C{B}_1$ and D ACB are equal then the third angle $\angle B_1{A}_{1}C=\angle BAC$]
$\therefore \Delta A_{1}C{B}_1$ and $\Delta ACB$ are similar
$\therefore \frac{AB}{A_1{B}_1}=\frac{BC}{B_{1}C}\dots \dots \left(1\right)$
Similarly $\Delta F{B}_1{A}_1$ and $\Delta FED$ are similar
$\therefore \frac{ED}{A_1{B}_1}=\frac{EF}{F{B}_1}$
But ED = AB
$\frac{AB}{A_1{B}_1}=\frac{EF}{F{B}_1}\dots \dots \left(2\right)$
From equations (1) and (2)
$\frac{BC}{B_{1}C}=\frac{EF}{F{B}_1}$
If D is very close to P then EF = PF
$\frac{BC}{B_{1}C}=\frac{PF}{F{B}_1}$
BC = PC - PB
$B_{1}C=P{B}_1-PCF{B}_1=P{B}_1-PF\frac{PC-PB}{P{B}_1-PC}=\frac{PF}{P{B}_1-PF}\dots \dots \left(3\right)$
But $PC=R,PB=u,P{B}_1=v,PF=f$
By sign convention
$PC=-R,PB=-u,PF=-f$ and $P{B}_1=-v$
$\therefore$ Equation (3) can be written as
$\frac{-R-(-u)}{-v-(-R)}=\frac{-f}{-v-(-f)}\frac{-R+u}{-v+R}=\frac{-f}{-v+f}\frac{u-R}{R-v}=\frac{f}{v-f}uv-uf-Rv+Rf=Rf-vfuv-uf-Rv=-vfR=2fuv-uf-2fv=-vfuv-uf=2fv-fvuv-uf=fv\dots \dots \left(4\right)$
Dividing equation (4) throughout by uvf we get
$\frac{1}{f}-\frac{1}{v}=\frac{1}{u}\frac{1}{f}=\frac{1}{v}+\frac{1}{u}\dots \dots \left(5\right)$
Equation (5) gives the mirror formula.