A thin converging lens is cut into two halves. They are placed at positions separated by 60 cm along the optical axis as shown, and give real images at same position. The ratio of image heights is 9:1. If original lens focal length f = (22 + 1/x) cm, find x. [See Image.]
Physics · Mixed Chapter
JEE Main Practice

Question

A thin converging lens is cut into two halves. They are placed at positions separated by 60 cm along the optical axis as shown, and give real images at same position. The ratio of image heights is 9:1. If original lens focal length f = (22 + 1/x) cm, find x. [See Image.]

Physics · Mixed Chapter

G1 Physics · Accepted Answer

Step 1 — Apply thin-lens formula 1/f = 1/v − 1/u for each half, keeping the two lens halves at 60 cm spacing. Step 2 — The final real image is common, and magnifications are |m₁|/|m₂| = 9, where |m| = |v/u|. Step 3 — Solve the two lens-position equations with common image coordinate. The source geometry gives f = 22.5 cm. Step 4 — From f = 22 + 1/x, 1/x = 0.5 ⇒ x = 2. [Verify the left/right lens placement against Image.]