An object is kept at a distance of 16 cm from a thin lens and the image formed is real. If the object is kept at a distance of 6 cm from the same lens, the image formed is virtual. If the sizes of the images formed are equal the focal length of the lens will be

Step-by-step solution

As a convex lens alone can form a real images as well as a virtual images, therefore, the lens in the present question is a convex lens. Let f be the focal length of the lens and m be the magnification produced In the first case, when image is real, u = - 16 cm, v = ( m × 16 ) cm As 1 v - 1 u = 1 f ∴ 1 16 m + 1 16 = 1 f or 1 + 1 m = 16 f ...(i) In the second case, when image is virtual u = - 6 c m , v = ( - 6 m)cm From 1 v - 1 u = 1 f 1 - 6 m + 1 6 = 1 f or 1 - 1 m = 6 f , ...(ii) Add Eq.(i) and Eq.(ii) we have 2 =