An equiconvex lens of refractive index `mu_(2)` is placed such that the refractive index of the surrounding media is as shown :

Explanation

Let a point object O be situated on the principal axis of the convex spherical surface. Let P be pole and C be centre of curvature and PC = R be the radius of curvature. Let `mu_(1)` and `mu_(2)` be refractive indices of rarer and denser medium respectively as shown in the figure.

The incident ray OA refracts at A and bends towards the normal along AQ and when produced in the backward direction, it meets at point I on the principal axis. Then I will be the virtual image of point O.
From Snell.s law for small i and r,
`(sin i)/(sin r)=(i)/(r)`
`=(mu_(2))/(mu_(1))`
or `mu_(1)i=mu_(2)r` ...(1)
In `Delta A O C`,
`i=alpha+beta`
In `Delta A I C`,
`r=beta+gamma`
Putting the values of i and r in Eq. (1), we get
`mu_(1)(alpha + gamma)=mu_(2)(beta+gamma)`
or `mu_(1)alpha+mu_(1)gamma=mu_(2)beta+mu_(2)gamma`
or `mu_(1)alpha-mu_(2)beta=(mu_(2)-mu_(1))gamma`
Here `alpha, beta` and `gamma` are small, so they can be replaced by their tangents.
`mu_(1) tan alpha-mu_(2)tan beta=(mu_(2)-mu_(1))tan gamma`
or `mu_(1)(AM)/(MO)-mu_(2)(AM)/(MI)=(mu_(2)-mu_(1))(AN)/(MC)` ...(2)
or `(mu_(1))/(MO)-(mu_(2))/(MI)=(mu_(2)-mu_(1))/(MC)`
Since M is very close to P, so
`MO ~= PO, MI ~= PI` and `MC ~= PC`
`:. (mu_(1))/(PO)-(mu_(2))/(PI)=(mu_(2)-mu_(1))/(PC)`
Using sign conventions,
PO = - u, PI = - v, PC = R, we get
`-(mu_(1))/(u)+(mu_(2))/(v)=(mu_(2)-mu_(1))/(R)`
or `(mu_(2))/(v)-(mu_(1))/(u)=(mu_(2)-mu_(1))/(R)`