Obtain the relative between object distance and image distance in terms of refractive index of medium and radius of curvature for spherically curved surface.

Explanation

Figure shows the geometry of formation of image I of an object O on the principal axis of a spherical surface with centre of curvature C and radius of curvature R.

The rays are incident from a medium of refractive index n, to another of refractive index `n_2`.
NM will be taken to be nearly equal to the length of the perpendicular from the point N on the principal axis.
We have for small angles,
`tan angleNOM=(MN)/(OM) ~~ angle NOM=(MN)/(OM)`
`tan angle NCM=(MN)/(MC) ~~ angle NCM =(MN)/(MC)`
`tan angle NIM = (MN)/(MI) ~~ angle NIM =(MN)/(MI)`
Now, for `triangle`NOC i is the exterior angle. Therefore,
`i= angle NOM+ angle NCM`
`therefore i =(MN)/(OM)+(MN)/(MC)` ... (1)
`implies` Similarly, for `triangleNCI,angle NCM=angle NIM+angle INC`
`therefore angle NCM=angle NIM+angle INC`
`(MN)/(MC)=(MN)/(MI)+r`
`therefore r =(MN)/(MC)-(MN)/(MI)` ... (2)
Now,by Snell.s law,
`n_1sin i=n_2 sin r`
Substituting i and r from equation (1) and (2), For small angles sin i `~~` i and sin r `~~` r
`therefore n_1i=n_2r`
`therefore n_1((MN)/(OM)+(MN)/(MC)) = n_2((MN)/(MC)-(MN)/(MI))`
`therefore(n_1)/(OM)+(n_1)/(MC)=(n_2)/(MC)-(n_2)/(MI)` [ `therefore` Dividing by MN]
Now taking OM =-u, MC=+R and MI =+v
`therefore (n_1)/(-u)+(n_1)/(+R)=(n_2)/(+R)-(n_2)/(+v)`
`therefore (n_1)/(-u)+(n_2)/(+v)=(n_2-n_1)/(+R)`
or `(n_2)/(v)-(n_1)/(u)=(n_2-n_1)/(R)`
If is true for any curved spherical surface.