A thin converging lens of focal length (25 ~cm ) forms a sharp image of an object on a screen placed at a distance of (75 ~cm ) from the lens. Later the screen is moved closer to the lens by a distance (25 ~cm ). The distance through which the object is to be shifted so that its image on the screen is sharp again is

Step-by-step solution

According to the question, Given, focal length, (f=25 ~cm ) and distance between image of an object and screen, (v=75 ~cm ) Now, By lens formula, ( 1 f = 1 v - 1 u ) [ ( ) Because screen is moved closer to the lens.] ( 1 u = 1 f - 1 v ) ( aligned 1 u & = 1 25 - 1 75 u & = 75 25 50 = 75 2 ~cm aligned ) When the screen shift upto by (25 ~cm ), then the screen will be at (2 f ). ( ) For to get sharp image, object has to be at (2 f ). So, the distances is (v-u=f ). (50- 75 2 = 25 2 =12.5 ~cm ) ( [ array l at 2 f v=50 ~