since the lens is equiconvex, the radius of curvature of each half is same, say R. We know from Lens maker's formula
1
f
=(µ−1)(
1
R1
−
1
R2
) (considering the lens to be placed in air). Here R1=R R2=−R by convention ∴
1
f
=(µ−1)
2
R
⇒(µ−1)
1
R
=
1
2f
......(i) If we cut the lens along XOXprime then the two halves of the lens will be having the same radii of curvature and so, focal length f′=f But when we cut it along YOY′ then, we will have R1=RbutR2=∞ ∴