For a given lens, the magnification was found to be twice as large as when the object was 0.15 m distant from it as when the distance was 0.2 m . The focal length of the lens is

Step-by-step solution

Let as shown, 1 and 2 are positions of objects and images in two different situations. It is given | v₁ u₁ |=2 | v₂ u₂ | Here, u₁=-15 ~cm , u₂=-20 ~cm v₁=2 v₂ u₁ u₂ =2 v₂ 15 20 = 3 2 v₂ Now, 1 f = 1 v - 1 u 1 f = 1 v₁ - 1 u₁ and 1 f = 1 v₂ - 1 u₂ So, 1 v₁ - 1 u₁ = 1 v₂ - 1 u₂ 2 3 v₂ + 1 15 = 1 v₂ + 1 20 v=20 ~cm