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

Step-by-step solution

According to the first condition f = 25 cm , V = 75 cm 1 f = 1 v - 1 u 1 25 = 1 75 - 1 u 1 u = 1 75 - 1 25 1 u = 1 - 3 75 u = - 75 2 = - 37 .5 cm According to the second condition v 1 = 50 cm , f = 25 cm 1 f = 1 v 1 - 1 u 1 1 25 = 1 50 - 1 u 1 1 u 1 = 1 50 - 1 25 ⇒ 1 u 1 = 1 - 2 50 ⇒ u 1 = - 50 cm So, the image is sharp again for, Δu = u 1 - u = 50 - 37 .5 = 12 .5 cm