NEETPhysicsRay Optics
An object is placed at x = -30 cm on the principal axis of a coordinate system. A convex lens L₁ of focal length 20 cm is fixed at the origin ( x = 0 ). A second lens L₂ is placed at x = +30 cm. Match the specifications of lens L₂ in List-I with the x-coordinate of the final image in List-II. List-I (Lens L₂ ) List-II (Final image position) (A) Convex, f = +15 cm (I) x = 0 cm (B) Concave, f = -15 cm (II) x = +40 cm (
Options
- A(A)-(I), (B)-(II), (C)-(III), (D)-(IV)
- B(A)-(III), (B)-(IV), (C)-(I), (D)-(II)
- C(A)-(II), (B)-(IV), (C)-(I), (D)-(III)
- D(A)-(II), (B)-(I), (C)-(IV), (D)-(III)
Correct answer
D. (A)-(II), (B)-(I), (C)-(IV), (D)-(III)
Step-by-step solution
For the first lens L₁ (convex lens at x = 0 ): Object distance, u₁ = -30 cm Focal length, f₁ = +20 cm Using the lens formula: 1 v₁ - 1 -30 = 1 20 1 v₁ = 1 20 - 1 30 = 3 - 2 60 = 1 60 v₁ = +60 cm The image from L₁ is formed at x = +60 cm. For the second lens L₂ (placed at x = +30 cm): The image formed by L₁ acts as a virtual object for L₂ . Object distance for L₂ , u₂ = +60 - 30 = +30 cm. Evaluating each case in List-I: (A) Convex, f = +15 cm: 1 v₂ - 1 30 = 1 15 1 v₂ = 1 15 + 1 30 = 3 30 = 1 10 v₂ = +10 cm. Final co