Question

Prove mirror formula in convex mirror​

Answer 1

$\color{darkblue}\underline{\underline{\sf Given-}}$

PB = -u

PB' = +v

PC = +R

PF = +f

$\underline{\rm In \:\triangle ABC\: and \: \triangle A'B'C}$

$\implies{\sf \dfrac{AB}{A'B'}=\dfrac{BC}{B'C}\:\:\:→(1) }$

$\underline{\rn In \: \triangle MNF \: and \: \triangle A'B'F-}$

$\implies{\sf \dfrac{MN}{A'B'}=\dfrac{NF}{B'F}}$

N = P

$\implies{\sf \dfrac{MP}{A'B'}=\dfrac{PF}{B'F}\:\:\:\:(2)}$

$\underline{\rm From \: equation \: 1\: and \:2}$

$\implies{\sf \dfrac{BC}{B'C}=\dfrac{PF}{B'F}}$

$\implies{\sf \dfrac{PB+PC}{PC-PB'}=\dfrac{PF}{PF-PB'}}$

$\implies{\sf \dfrac{-u+R}{+R-v}=\dfrac{+f}{+f-v}}$

We know that R = 2f

$\implies{\sf \dfrac{-u+2f}{2f-v}=\dfrac{+f}{+f-v}}$

$\implies{\sf (-u+2f)(f-v)=f(2f-v)}$

$\implies{\sf -uf+uv+2f^2-2fv=2f^2-vf}$

$\implies{\sf -uf+uv=vf }$

$\underline{\rm On\; dividing \: uvf \: both \: side }$

$\implies{\sf \dfrac{-uf}{uvf}+\dfrac{uv}{uvf}=\dfrac{vf}{uvf}}$

$\implies{\sf -\dfrac{1}{v}+\dfrac{1}{f}=\dfrac{1}{u}}$

$\large\color{red}\bullet\underline{\boxed{\sf \dfrac{1}{f}=\dfrac{1}{v}+\dfrac{1}{u}}}$

Answer 2

Answer:

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