ΔABC is similar to ΔA’B’C
Therefore, \(frac{AB}{A’B’}=frac{CB}{CB’}\) (equ.1)
Draw DC = AB
In ΔDCF~ ΔA’B’F
\(frac{DC}{A’B’}=frac{CF}{B’F}\) \(frac{AB}{A’B’}=frac{CF}{B’F}\) (equ.2)From equ.1 and equ.2,
\(frac{CB}{CB’}=frac{CF}{CB’+CF}\) \(frac{+u}{+v}=frac{f}{-v+f}\)u(-f-v) = fv
Uf – uv = fv
Dividing by u,v,f we get,
\(frac{1}{v}-frac{1}{f}=frac{1}{u}\) \(frac{1}{f}=frac{1}{v}-frac{1}{u}\)Therefore, above is the derivation of the lens formula for a convex lens.
