Question

mirror formula derivation​

Answer 1

Answer:

the derivation is in the picture..

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Answer 2

$\Large\color{darkblue}\underline{\underline{\sf Mirror \: Formula-}}$

Refer to the attachment for the Figure.

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Given

PB = - u

PC = - R

PB' = - v

PF = -f

$\color{darkblue}\underline{\sf In\: \triangle ABC \: and \: \triangle A'B'C}$

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

$\color{darkblue}\underline{\sf In \: \triangle MNF \: and \: \triangle A'B'F-}$

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

N = P

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

$\color{darkblue}\underline{\underline{\sf From\: equation \: 1 \; and \;2-}}$

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

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

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

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

$\color{green}\underline{\sf We \; know \; that \: R = 2f}$

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

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

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

$\implies{\sf uv-fu=fv}$

$\underline{\sf On \: dividing\:uvf\:both\:side-}}$

$\implies{\sf \dfrac{uv}{uvf}-\dfrac{fu}{uvf}=\dfrac{fv}{uvf}}$

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

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