Question

DERIVATION OF MIRROR FORMULA.

Answer 1

Hey mate

your answer is..

Mirror formula is 1/v + 1/u = 1/f.

hope it helps you

Answer 2

Answer:

Considering similar triangles ABF and FPO'

    $\frac{AB}{BF} =\frac{PO'}{PF}$

⇒ $\frac{AB}{PO'} =\frac{BF}{PF}$

PO' = A'B' therefore

⇒ $\frac{AB}{A'B'} =\frac{u-f}{f}$ ------ 1

Considering similar triangles A'B'F and FPO

    $\frac{A'B'}{B'F} =\frac{PO}{PF}$

⇒ $\frac{PO}{A'B'} =\frac{PF}{B'F}$

PO' = A'B' therefore

⇒ $\frac{AB}{A'B'} =\frac{f}{v-f}$ -------- 2

From 1 and 2,

$\frac{AB}{A'B'} =\frac{u-f}{f} =\frac{f}{v-f}$

      ⇒ $(u-f)(v-f)=f^2$

      ⇒ $uv-uf-vf+f^2=f^2$

      ⇒ $uv-uf-vf=0$

      ⇒ $uv=uf+vf$

Now let us divide it by uv. It becomes

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