Question

Derive lens formula $\frac{1}{v}-\frac{1}{u}=\frac{1}{f}$ .

Answer 1

The figure above shows that the formation of a real, inverted and diminished image AB of the object AB placed beyond the centre of curvature at a distance u from the convex lens. Let v be the image distance,
According to Cartesian sign convention
Object distance $\left(OB\right)=-u$
Image distance $\left(OB\right)=+v$
Focal length $(OF1=OF2)=+f$
From the geometry of the figure above, right angled $△$ ABO and $△$ ABO are similar.
$\therefore \frac{AB}{A^{\prime }{B}^{\prime }}=\frac{OB}{O^{\prime }{B}^{\prime }}=-v$ ........ (i)
Similarly, from the geometry of the figure above, right angled $△OD{F}_2$ and $△BA{F}_2$ are similar.
$\therefore \frac{OD}{A^{\prime }{B}^{\prime }}=\frac{O{F}_2}{F_2{B}^2}$
$\therefore \frac{AB}{A^{\prime }{B}^{\prime }}=\frac{O{F}_2}{F_2{B}^2}$ ( $\because$ OD = AB, are the opposite sides of $\square$ ABOD )
$\therefore \frac{AB}{A^{\prime }{B}^{\prime }}=\frac{O{F}_2}{O{B}^{\prime }-O{F}_2}$
$\therefore \frac{AB}{A^{\prime }{B}^{\prime }}=\frac{f}{v-f}$ .......(ii)
From (i) and (ii),
$-\frac{u}{v}=\frac{v}{v-f}$
$\Rightarrow -u(v-f)=vf$
$\Rightarrow -uv+uf=vf$
Dividing each term by $uvf,$

$-\frac{1}{f}+\frac{1}{v}=\frac{1}{u}$
$\frac{1}{v}-\frac{1}{u}=\frac{1}{f}$
This equation is called the 'Lens formula'.