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Question:

A diverging lens with magnitude of focal length 25 cm is placed at a distance of 15 cm from a converging lens of magnitude of focal length 20 cm. A beam of parallel light falls on the diverging lens. The final image formed is :

Real and at a distance of 40cm from convergent lens.

Virtual and at a distance of 40cm from convergent lens.

Real and at a distance of 6cm from the convergent lens.

Real and at a distance of 40cm from the divergent lens.

Solution:

For concave lens, 1/f₁ = 1/V₁ - 1/u₁ ⇒ 1/25 = 1/V₁ - 1/∞ ⇒ V₁ = -25cm
For Convex lens 1/f₂ = 1/V₂ - 1/u₂ ⇒ 1/20 = 1/V₂ - 1/(-25 - 15) ⇒ 1/20 = 1/V₂ + 1/40 ⇒ 1/V₂ = 1/20 - 1/40 ⇒ V₂ = 40cm
So, answer is (b)