Question

Establish the formula $\frac{1}{f}=\frac{1}{u}+\frac{1}{v}$ for a concave mirror, where symbols have their usual meanings.

Answer 1

in the above ray diagram (ray diagram for concave mirror forming real image) we can see two triangles $\Delta A^{\prime }{B}^{\prime }F$ and $\Delta MPF$ are similar triangles (for paraxial rays, MN can be considered as straight line and perpendicular to the line FP). Because $\angle A^{\prime }{B}^{\prime }F=\angle MPF$ (right angles )

and $\angle B^{\prime }F{A}^{\prime }=\angle PFM$ (opposite angles)

so if two angles are equal then third angle will also be equal.

hence both triangles are similar.

therefore, $\frac{B^{\prime }{A}^{\prime }}{PM}=\frac{B^{\prime }F}{FP}$

$\frac{B^{\prime }{A}^{\prime }}{BA}=\frac{B^{\prime }F}{FP}$ because as we all know $MP=AB$

similarly since $\angle APB=\angle A^{\prime }P{B}^{\prime }$ then right angle triangles $\Delta A^{\prime }{B}^{\prime }P$ and $\Delta ABP$ are also similar.

therefore, $\frac{B^{\prime }{A}^{\prime }}{BA}=\frac{B^{\prime }P}{BP}$

by comparing both the above equations we can say that $\frac{B^{\prime }F}{FP}=\frac{B^{\prime }P}{BP}=\frac{B^{\prime }P-FP}{FP}$

now applying sign convention (light travels from left to right so that direction taken as positive ) on all the distances involved we get $B^{\prime }P=-v$, $FP=-f$, $BP=-u$

Hence we get, $\frac{-v+f}{-f}=\frac{-v}{-u}$ or $\frac{v-f}{f}=\frac{v}{u}$

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

hence proved.