Consider a convex lens (or concave lens) of absolute refractive index $μ_{2}$ to be placed in a rarer medium of absolute refractive index $μ_{1}$. Considering the refraction of a point object on the surface $XP_{1}Y$, the image is formed at $I_{1}$ who is at a distance of $V_{1}$.

$CI_{1}=P_{1}I_{1}=V_{1}$ (as the lens is thin)

$CC_{1}=P_{1}C_{1}=R_{1}$

$CO=P_{1}O=u$

It follows from the refraction due to convex spherical surface $XP_{1}Y$

$−uμ_{1} +v_{1}μ_{2} =R_{1}μ_{2}−μ_{1} ..........(i)$

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

Here the object distance is

$u=CI_{1}≈P_{2}I_{1}=V$

$P_{1}P_{2}$ is very small

Let $CI≈P_{2}I=V$

(Final image distance)

Let $R_{2}$ be radius of curvature of second surface of the lens. It follows from refraction due to concave spherical surface from denser to rarer medium that

$v_{1}−μ_{2} +vμ_{1} =R_{2}μ_{1}−μ_{2} =−R_{2}μ_{2}−μ_{1} ............(ii)$

Adding $(i)and(ii)$

$u−μ_{1} +vμ_{1} =(μ_{2}−μ_{1})(R_{1}1 −R_{2}1 )$

or

$μ_{1}(v1 −u1 )=(μ_{2}−μ_{1})(R_{1}1 −R_{2}1 )$

But $v1 −u1 =f1 $ and $μ_{1}μ_{2} =μ$

Thus we get=

$f1 =(μ−1)(R_{1}1 −R_{2}1 )$