Question

Derive the thin lens formula i.e.
1/v - 1/u = 1/f for a convex lens when it forms a real image . ​

Answer 1

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Answer 2

Derivation :

We know that ∆ A'B'O and ∆ABO are similar

$\sf \therefore \dfrac{A'B'}{AB}=\dfrac{OB'}{BO} \longrightarrow (1)$

Also ∆A'B'F and ∆MOF are similar

$\sf \therefore \dfrac{A'B'}{MO}=\dfrac{FB'}{OF} \\\\ \sf \because \: MO = AB \: \: thus \: we \: have \\\\ \sf \implies \dfrac{A'B'}{AB} = \dfrac{FB'}{OF} \longrightarrow (2)$

Comparing (1) and (2) we get

$\sf \implies \dfrac{OB'}{BO}=\dfrac{FB'}{OF} \\\\ \sf \implies \dfrac{OB'}{BO}=\dfrac{OB'-OF}{OF}$

Using new cartesian sign convention ,

Object distance , BO = -u

Image distance , OB'= +v

Focal length , OF = +f

Therefore we have ,

$\sf \implies \dfrac{v}{-u}=\dfrac{v-f}{f}\\\\ \sf \implies vf = -uv + uf \\\\ \sf \implies -vf = uv - uf \\\\ \sf \implies uf - vf = uv \\\\ \sf \implies \dfrac{uf}{uvf}-\dfrac{vf}{uvf}=\dfrac{uv}{uvf}\\\\ \sf \implies \dfrac{1}{v}-\dfrac{1}{u}=\dfrac{1}{f}$

Hence Derived