Question

Starting with an expression for refraction at a single spherical surface, obtain an expression for lens maker's formula.

Answer 1

Let a convex lens is taken which is made up of glass of refractive index $n_2$. Let ${}^{\prime }{O}^{\prime }$ is the point object placed on the principal axis and ${}^{\prime }{u}^{\prime }$ is the distance of the object from $P_1$.
For refraction at the first surface
$\frac{n_2}{v^{\prime }}-\frac{n_1}{u}=\frac{n_2-n_1}{R_1}....\left(1\right)$
For refraction at the second surface
$\frac{n_1}{v}-\frac{n_2}{v^{\prime }-t}=\frac{n_1-n_2}{R_2}$
If the lens is thin $t<<v^{\prime }$, so
$\frac{n_1}{v}-\frac{n_2}{v^{\prime }}=\frac{n_2-n_1}{R_2}....\left(2\right)$
Adding $\left(1\right)$ and $\left(2\right)$, we get $\frac{n_1}{v}-\frac{n_1}{u}=(n_2-n_1)(\frac{1}{R_1}-\frac{1}{R_2})$
Dividing both sides by $n_1$, we get
$\frac{1}{v}-\frac{1}{u}=(\frac{n_2}{n_1}-1)(\frac{1}{R_1}-\frac{1}{R_2})$
Putting $\frac{n_2}{n_1}=n$ we get $\frac{1}{v}-\frac{1}{u}=(n-1)(\frac{1}{R_1}-\frac{1}{R_2})$
When object is at infinity image will be at focus
So, $\frac{1}{f}-\frac{1}{\infty }=(n-1)(\frac{1}{R_1}-\frac{1}{R_2})$
$\frac{1}{f}=(n-1)(\frac{1}{R_1}-\frac{1}{R_2})$
This is lens makers formula.