**6.99: A bi-convex lens is formed with two thin plano-convex lenses as shown in the figure. Refractive index $n$ of the first lens is $1.5$ and that of the second lens is $1.2$. Both the curved surface are of the same radius of curvature $R \:=\: 14\:cm$. Find the effective focal length of the combination.**

**Solution:**

To find the position of image, effective focal length of the lens must be calculated.

We will assume that the lens is placed in air.

It is the combination of two plano convex lenses.

Focal length of a lens made out of a material of refractive index $n_1$ in a medium of refractive index $n_2$ is given by the equation $\frac{1}{f}\:=\:(\frac{n_2}{n_1}\:-\:1)(\frac{1}{R_1}\:-\:\frac{1}{R_2})$ where $R_1$ and $R_2$ are the radius of curvature of the two surfaces.

Problem in ray optics needs sign convention. In this problem Cartesian sign convention is followed.

For first lens,

The first surface is spherical with radius of curvature of 14 cm and the second surface is plane with infinite $(\:=\:\alpha)$ radius.

$n_1\:=\:1$, $n_2\:=\:1.5$,$R_1\:=\:+14$,$R_2\:=\:\alpha$.

Therefore

$\frac{1}{f_1}\:=\:(\frac{1.5}{1}\:-\:1)(\frac{1}{14}\:-\:\frac{1}{\alpha})$

Or, $f_1\:=\:28\:cm$.

For second lens,

The first surface is plane with infinite $(\:=\:\alpha)$ radius and the second surface is spherical with radius of curvature of 14 cm.

$n_1\:=\:1$, $n_2\:=\:1.2$,$R_1\:=\:\alpha$, $R_2\:=\:-14$.

Therefore

$\frac{1}{f_2}\:=\:(\frac{1.2}{1}\:-\:1)(\frac{1}{\alpha}\:-\:\frac{1}{-14})$

Or, $f_1\:=\:70\:cm$.

The two lenses are in contact.

Therefore effective focal length is given by,

$\frac{1}{F}\:=\:\frac{1}{f_1}\:+\:\frac{1}{f_2}$

Or, $\frac{1}{F}\:=\:\frac{1}{28}\:+\:\frac{1}{70}$

Or effective focal length is $F\:=\:20\:cm$.