JEE Main 2025 · 24 January · morning · Single correct · +4 / −1

What is the relative decrease in focal length of a lens for an increase in optical power by 0.1 D from 2.5D ? ['D' stands for dioptre]

  1. (a) 0.04
  2. (b) 0.1
  3. (c) 0.40
  4. (d) 0.01
Answer and solution

Answer (a) 0.04

Solution

Physics sir · 4 steps

Since f = 1/P, a small fractional rise in power gives the same fractional drop in focal length.

  1. Link f and P. Focal length and power are reciprocals: f = 1/P. Differentiate to see how a small change in P changes f.

    df = -1/P² dP ⇒ df/f = -dP/P

  2. Put in numbers. (Δ P)/P = 0.1/2.5 = 0.04. The minus sign means f decreases as P increases.
  3. Quick check. Directly, f goes from 1/2.5 = 0.4 m to 1/2.6 ≈ 0.385 m. The drop is about 0.0154 m, and 0.0154/0.4 ≈ 0.038, which is 0.04 to the given precision.
  4. So. The relative decrease in focal length is 0.04, option (a).

The idea to keep. For any reciprocal relation y = 1/x, the fractional change in y equals the fractional change in x with the opposite sign. No need to find f at all.

Answer from the NTA final answer key. Solution by PrepWiser.

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Asked about 200 times in JEE Main since 2020. This question is from 2025.