Cases of refraction through a concavo-convex lens

Question

First , let us consider a real object placed at some distance u from the concave side of a concavo-convex lens . Using the lens maker's equation , we get 4 cases for the sign of the focal length , them being as follows :

n2 is the refractive index of the lens , n1 is the refractive index of the surrounding medium . R1 is the radius of curvature of the side facing the object and R2 is the radius of curvature of the side away from the object.

If n₂>n₁ ; R₁>R₂ : f is positive.

                           ; R1<R2 : f is negative. 

If n₂<n₁ ; R₁>R₂ : f is negative.

                           ; R1<R2 : f is positive. 

But if we place the object at the same distance from the convex side of a similar concavo-convex lens , Shouldn't we get the exact opposite of what we got when we placed the object in the concave side? Considering R1 and R2 both would be negative as per the sign convention in case 1 and R1 and R2 both would be positive in case 2.

Answer 1

"But if we place the object at the same distance from the convex side of a similar concavo-convex lens , Shouldn't we get the exact opposite of what we got when we placed the object in the concave side?"

No we won't, it is due to the fact that reversibility is generally applicable when lenses are symmetric or have same focal length(with sign) for objects on both sides. In your case, its positive for object on one side and negative for object on the other side. You can verify all this using ray diagrams.