Question

Derive the Lens Maker's formula.

Answer 1

Consider the situation shown in figure. $ADBE$ is a thin lens. An object 0 is placed on its principal axis. The two spherical surfaces of the lens have their centres at $C_1$ and $C_2$. The optical centre is at $P$ and the principal axis cuts the two spherical surfaces at $D$ and $E$.

As shown in the figure, after the first refraction image is formed at $O_1$ and after second refraction the final image is formed at $I$.

As

the lens is thin, the points $D,PandE$ are all close to each other and we may take the origin at $P$ for both these refractions.

The general equation for refraction at a spherical surface is

$\frac{{\mu }_2}{v}-\frac{{\mu }_1}{u}=\frac{{\mu }_2-{\mu }_1}{R}$..........(i)

For the first refraction, the object is at $O$, the

image is at $O_1$ and the centre of curvature is at $C_1$. If

$u$, $v$, and $R$, denote their x-coordinates,

$\frac{{\mu }_2}{v_1}-\frac{{\mu }_1}{u}=\frac{{\mu }_2-{\mu }_1}{R_1}$.......(ii)

For the second refraction at $AEB$, the incident rays $GH$ and $DE$ diverge from $O_1$. Thus, $O_1$ is the object for

this refraction and its x-coordinate is $v_1$. The image is

formed at $I$ and the centre of curvature is at $C_2$. Their

x-coordinates are $v$ and $R$ respectively. The light goes

from the medium ${\mu }_2$ to medium ${\mu }_1$.

Hence

$\frac{{\mu }_1}{v}-\frac{{\mu }_2}{v_2}=\frac{{\mu }_1-{\mu }_2}{R_2}$.......(iii)

Adding (ii) and (iii),

$\frac{1}{v}-\frac{1}{u}=(\frac{{\mu }_2}{{\mu }_1}-1)\times (\frac{1}{R_1}-\frac{1}{R_2})$

If the object $O$ is taken far away from the lens, the image is formed close to the focus. Thus, for $u$= $\infty$ , $v=f$

Hence,

$\frac{1}{f}=(\frac{{\mu }_2}{{\mu }_1}-1)\times (\frac{1}{R_1}-\frac{1}{R_2})$

This is lens makers formula.