Question

When the object is at distances $u_1$ and $u_2$ from the optical center of a convex lens, a real and a virtual image of the same magnification are obtained. The focal length of the lens is

• A. $\frac{u_1-u_2}{2}$
• B. $u_1+u_2$
• C. $\sqrt{u_1+u_2}$
• D. $\frac{u_1+u_2}{2}$

Answer 1

The correct option is D $\frac{u_1+u_2}{2}$

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

1/v₁ - 1/u₁ = 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 = v_2/u₂ ---(5)

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

=> f = (u₁ + u₂)/2