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Question:

A 4 cm tall object is placed on the principal axis of a convex lens. The distance of the object from the optical centre of the lens is 12 cm and its sharp image is formed at a distance of 24 cm from it on a screen on the other side of the lens. If the object is now moved a little away from the lens, in which way (towards the lens or away from the lens) will he have to move the screen to get a sharp image of the object on it again?

Solution:

From the lens formula, 1/v - 1/u = 1/f ⇒ 1/f = 1/24 - 1/(-12) = 1/8 ⇒ f = 8

When u = -12, the object is placed between F and 2F and the image is formed beyond 2F on the other side. This is an enlarged inverted image. If the object is moved away from the lens, the image moves towards the lens. So the screen should be moved towards the lens.

Magnification = v/u = 24/(-12) = -2; When the denominator increases, v decreases and hence magnification decreases.