Question

Derive Lens Maker's formula for a convex lens.

Answer 1

The formula for refraction of light from the spherical surface is

$\frac{n_2}{v}-\frac{n_1}{u}=\frac{n_2-n_1}{R}$ where $OB=-u,OI=+v,BC=+R$

$n_2$ is refractive index of medium 2 i.e. spherical glass

and $n_1$ is refractive index of medium 1.

Figure (a) shows the geometry of image formation by a double convex lens. the image formation can be seen in two steps, the first refracting surface forms an image $I_1$ which acts as a virtual object for the second surface. applying the equation above written for surface 1

$\frac{n_1}{OB}+\frac{n_2}{B{I}_1}=\frac{n_2-n_1}{B{C}_2}$

similarly, form the second interface

$-\frac{n_2}{D{I}_1}+\frac{n_1}{DI}=\frac{n_2-n_1}{D{C}_2}$ (refractive index of right side medium is $n_1$ and for left side medium is $n_2$, and also $D{I}_1$is negative.)

now for thin lens $B{I}_1=D{I}_1$. adding both the equations we get

$\frac{n_1}{OB}+\frac{n_1}{DI}=(n_2-n_1)(\frac{1}{B{C}_1}+\frac{1}{D{C}_2})$

and assuming the object at infinity means OB will be very large

we get $\frac{n_1}{DI}=(n_2-n_1)(\frac{1}{B{C}_1}+\frac{1}{D{C}_2})$

and dividing both sides with $n_1$

$\frac{1}{DI}=(n_{21}-1)(\frac{1}{B{C}_1}+\frac{1}{D{C}_2})$

replacing $DI=f,B{C}_1=+R_1,D{C}_2=-R_2$

Finally, we get $\frac{1}{f}=(n_{21}-1)(\frac{1}{R_1}+\frac{1}{R_2})$.

This equation is known as lens makers formula.