Consider the situation shown in figure. ADBE is a thin lens. An object 0 is placed on its principal axis. The two spherical surfaces of the lens have their centres at C1 and C2. The optical centre is at P and the principal axis cuts the two spherical surfaces at D and E.
As shown in the figure, after the first refraction image is formed at O1 and after second refraction the final image is formed at I.
As
the lens is thin, the points D,PandE are all close to each other and we may take the origin at P for both these refractions.
The general equation for refraction at a spherical surface is
μ2v−μ1u=μ2−μ1R..........(i)
For the first refraction, the object is at O, the
image is at O1 and the centre of curvature is at C1. If
u, v, and R, denote their x-coordinates,
μ2v1−μ1u=μ2−μ1R1.......(ii)
For the second refraction at AEB, the incident rays GHand DE diverge from O1. Thus, O1 is the object for
this refraction and its x-coordinate is v1. The image is
formed at I and the centre of curvature is at C2. Their
x-coordinates are v and R respectively. The light goes
from the medium μ2 to medium μ1.
Hence
μ1v−μ2v2=μ1−μ2R2.......(iii)
Adding (ii) and (iii),
1v−1u=(μ2μ1−1)×(1R1−1R2)
If the object O is taken far away from the lens, the image is formed close to the focus. Thus, for u= ∞, v=f