Light is incident on the thin lens as shown in figure. The refractive index of the material of the lens is ${\mathrm{\mu }}_2$. The radii of curvature of both of the surfaces are R. The refractive index of the medium of left side is ${\mathrm{\mu }}_1$ , while that of right side is ${\mathrm{\mu }}_3$. What is the focal length of the lens?

As the object is at infinity on principal axis, the distance of final image will be the focal length.

at ${\mathrm{P}}_1:$

${\mathrm{\mu }}_1={\mathrm{\mu }}_1,{\mathrm{\mu }}_2={\mathrm{\mu }}_2,{\mathrm{R}}_1=+\mathrm{R},{\mathrm{u}}_1=\mathrm{\infty }$

$\frac{{\mathrm{\mu }}_2}{\mathrm{V}}-\frac{{\mathrm{\mu }}_1}{\mathrm{u}}=\frac{{\mathrm{\mu }}_2-{\mathrm{\mu }}_1}{\mathrm{R}}\Rightarrow \frac{{\mathrm{\mu }}_2}{{\mathrm{V}}_1}-\frac{{\mathrm{\mu }}_1}{\mathrm{\infty }}=\frac{{\mathrm{\mu }}_2-{\mathrm{\mu }}_1}{+\mathrm{R}}$

$\Rightarrow \mathrm{V}=\frac{{\mathrm{\mu }}_2\mathrm{R}}{{\mathrm{\mu }}_2-{\mathrm{\mu }}_1}$

This will be the object distance to the second surface at ${\mathrm{P}}_2:$

${\mathrm{\mu }}_1={\mathrm{\mu }}_2, {\mathrm{\mu }}_2={\mathrm{\mu }}_3,{\mathrm{u}}_2=+\frac{{\mathrm{\mu }}_2\mathrm{R}}{{\mathrm{\mu }}_2-{\mathrm{\mu }}_1}$

${\mathrm{R}}_2=+\mathrm{R} \left[{\mathrm{P}}_2\to {\mathrm{C}}_2\right]$

$\frac{{\mathrm{\mu }}_2}{\mathrm{V}}-\frac{{\mathrm{\mu }}_1}{\mathrm{u}}=\frac{{\mathrm{\mu }}_2-{\mathrm{\mu }}_1}{\mathrm{R}}$

$\Rightarrow \frac{{\mathrm{\mu }}_3}{{\mathrm{V}}_2}-\frac{{\mathrm{\mu }}_2}{\left(\frac{{\mathrm{\mu }}_2\mathrm{R}}{{\mathrm{\mu }}_2-{\mathrm{\mu }}_1}\right)}=\frac{{\mathrm{\mu }}_3-{\mathrm{\mu }}_2}{\mathrm{R}}$

$\Rightarrow {\mathrm{V}}_2=\frac{{\mathrm{\mu }}_3\mathrm{R}}{{\mathrm{\mu }}_3-{\mathrm{\mu }}_1}=\text{ focal length. }$