Question

Starting with an expression for refraction at a single spherical surface, obtain Lens Maker's Formula.

Answer 1

Consider a convex lens (or concave lens) of absolute refractive index to be placed in a rarer medium of absolute refractive index. Considering the refraction of a point object on the surface $X{P}_{1}Y$, the image is formed at $I_1$ ;who is at a distance of $V_1$ .

$C{I}_1=P{I}_1=V_1\left(asthelensisthin\right)C{C}_1=P{C}_1=R_{1}CO=P_{1}O=u$

It follows from the diffraction due to convex spherical surface$X{P}_{1}Y$

$\left(\frac{{\mu }_1}{-u}\right)+\left(\frac{{\mu }_2}{v_1}\right)=\left(\frac{{{\mu }_2-\mu }_1}{R_1}\right)...\left(1\right)$

the refracted ray from A suffers a second refraction on the surface $X{P}_{2}Y$ and emerges along $BI$ . Therefore $I$ is the final real image of$O$ .

Here the object distance is

$u=C{I}_1\simeq {P_{2}I}_1=V$

Note that lens thickness is very small. It follows from the refraction due to convex spherical surface $X{P}_{1}Y$

$CI\simeq P_{2}I=V(-\frac{{\mu }_2}{v_1})+\left(\frac{{\mu }_2}{v}\right)=(-\frac{{{\mu }_2-\mu }_1}{R_2})...\left(2\right)$

$\frac{1}{v}-\frac{1}{u}=\frac{1}{f}...\left(3\right)$

adding equation $\left(1\right)$ and $\left(2\right)$

$\Rightarrow {\mu }_1[\frac{1}{v}-\frac{1}{u}]={\mu }_2-{\mu }_1[\frac{1}{R_1}-\frac{1}{R_2}]\Rightarrow \frac{1}{v}-\frac{1}{u}=\frac{{\mu }_2-{\mu }_1}{{\mu }_1}[\frac{1}{R_1}-\frac{1}{R_2}]$

$\Rightarrow \frac{1}{f}=\frac{{\mu }_2-{\mu }_1}{{\mu }_1}[\frac{1}{R_1}-\frac{1}{R_2}]$(replaced by equation $\left(3\right)$)

using $\frac{{\mu }_2}{{\mu }_1}=\mu$

$\Rightarrow \frac{1}{f}=(\mu -1)[\frac{1}{R_1}-\frac{1}{R_2}]$ this is called lens maker formula.