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Derive the lens formula

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\Delta A'B'P and \Delta ABP are similar

\frac{AB}{A'B'}= \frac{BP}{B'P}= \frac{-u}{-v}              ...........(1)

\Delta A'B'P and \Delta MPF are similar

\frac{MP}{A'B'}= \frac{PF}{B'P}= \frac{f}{B'P+PF}= \frac{f}{-v+f}

From the diagram, AB = MP

\therefore \frac{AB}{A'B'}= \frac{f}{-v+f}

\frac{-u}{-v}= \frac{f}{-v+f}\Rightarrow \frac{u}{v}= \frac{f}{-v+f}

u(-v+f)=fv

uf-uv=fv

Divide throughout by uvf

\frac{uf}{uvf}-\frac{uv}{uvf}=\frac{fv}{uvf}

\frac{1}{v}-\frac{1}{f}= \frac{1}{u}

\frac{1}{f}=\frac{1}{v}- \frac{1}{u}

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Safeer PP

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