Question

"when an object is at distance u1 and u2 from lens ,a real image and a virtual image is formed respectively haveing same magnification. the focal lenght of the lens is

Answer 1

Let us say it is a convex lens of focal length f. It can form both real and virtual images.

1/v1 - 1/u = 1/f,      here v & f are positive and u is negative.
1/v1 + 1/u1 = 1/f   --- (1)
=> u1/v1 + 1 = u1/f   -- (2)

The virtual image is formed in front of the lens. So v and u1 are negative.
- 1/v2 + 1/u2 = 1/f    --- (3)
- u2/v2 + 1 = u2/f      --(4)

Magnification = v1/u1 = v2/u2   ---(5)

Add (2) and (4) and use (5) to get:    2 = u1/f + u2/f
  => f = (u1 + u2)/2

Answer 2

1/v1 - 1/u = 1/f,      here v & f are positive and u is negative.

1/v1 + 1/u1 = 1/f   --- (1)

=> u1/v1 + 1 = u1/f   -- (2)The virtual image is formed in front of the lens. So v and u1 are negative.

- 1/v2 + 1/u2 = 1/f    --- (3)

- u2/v2 + 1 = u2/f      --(4)

Magnification = v1/u1 = v2/u2   ---(5)

Add (2) and (4) and use (5) to get:    2 = u1/f + u2/f=(u1+u2)/f.