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

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Correct answer: B. 0.20 m

Let as shown, 1 and 2 are positions of objects and images in two different situations. It is given |(v_1)/(u_1)|=2|(v_2)/(u_2)| Here, u_1=-15 ~cm, u_2=-20 ~cm v_1=2 v_2 × (u_1)/(u_2) =2 v_2 × (15)/(20)=(3)/(2) v_2 Now, (1)/(f)=(1)/(v)-(1)/(u) (1)/(f)=(1)/(v_1)-(1)/(u_1) and (1)/(f)=(1)/(v_2)-(1)/(u_2) So, (1)/(v_1)-(1)/(u_1)=(1)/(v_2)-(1)/(u_2) (2)/(3 v_2)+(1)/(15)=(1)/(v_2)+(1)/(20) v=20 ~cm

For a given lens, the magnification was found to be twice as large...

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