Question

Define lens formula derive $\frac{1}{v}-\frac{1}{u}=\frac{1}{f}$ for convex lens?

Answer 1

Let $AB$ represent an object places at right angles to the principal axis at a distance greater that the focal length $f$ of the convex lens. The image $A^1{B}^1$ is formed beyond $2F_2$ and is real and inverted.

$OA=$ Object distance $=u$
$O{A}^1=$ Image distance $=v$
$O{F}_2=$Focal length $=f$
$\Delta OAB$ and $\Delta O{A}^1{B}^1$ are simila
$[\because \angle BAO=\angle B^1{A}^{1}O={90}^o,\angle AOB=\angle A^{1}O{B}^1$ as they are vertically opposite $\therefore \angle ABO=\angle A^1{B}^{1}O]$
$\therefore \frac{A^1{B}^1}{AB}=\frac{O{A}^1}{OA}$ ....(1)
similarly $\Delta OC{F}_2$ and $\Delta F_2{A}^1{B}^1$ are similar
$\therefore \frac{A^1{B}^1}{OC}=\frac{F_2{A}^1}{O{F}_2}$
But we know that $OC=AB$
$\therefore$ teh above equation can be written as
$\frac{A^1{B}^1}{OC}=\frac{A^1{B}^1}{AB}=\frac{F_2{A}^1}{O{F}_2}$
$\frac{A^1{B}^1}{AB}=\frac{F_2{A}^1}{O{F}_2}$ ....(2)
From equation (1) and (2), we get
$\frac{O{A}^1}{OA}=\frac{F_2{A}^1}{O{F}_2}=\frac{O{A}^1-O{F}_2}{O{F}_2}$
$\frac{v}{-u}=\frac{v-f}{f}$
$vf=-u(v-f)$
$vf=-uv+uf$ ....(3)
Deviding equation (3) throughout by $uvf$
$\frac{1}{u}=-\frac{1}{f}+\frac{1}{v}$
$\frac{1}{f}=\frac{1}{v}-\frac{1}{u}$