The lens combination shown consists of two lenses L₁ (focal length +10 cm) and L₂ (focal length -10 cm). An object O is placed 30 cm to the left of L₁, and L₂ is 3 cm to the right of L₁. The position of the image formed is:

Solution & Explanation

Treat the two lenses in succession; the image from L₁ becomes the object for L₂. Lens L₁ (convex, f = +10 cm), object 30 cm to its left so u = −30 cm. 1/v − 1/u = 1/f → 1/v = 1/10 + 1/(−30) = 3/30 − 1/30 = 2/30, so v = +15 cm. This image lies 15 cm to the right of L₁; since L₂ is 3 cm right of L₁, the image is 15 − 3 = 12 cm to the right of L₂. That point is on the far side of L₂, so it acts as a virtual object for L₂: u′ = +12 cm, with L₂ concave, f′ = −10 cm. 1/v′ = 1/f′ + 1/u′ = −1/10 + 1/12 = −6/60 + 5/60 = −1/60, so v′ = −60 cm. The negative sign means the final image is 60 cm to the left of the concave lens (a virtual image), matching answer (B).