JEE Main 2021 · 26 August · morning · Single correct · +4 / −1

Car B overtakes another car A at a relative speed of 40 ms⁻¹. How fast will the image of car B appear to move in the mirror of focal length 10 cm fitted in car A, when the car B is 1.9 m away from the car A?

  1. (a) 4 ms⁻¹
  2. (b) 0.2 ms⁻¹
  3. (c) 40 ms⁻¹
  4. (d) 0.1 ms⁻¹
Answer and solution

Answer (d) 0.1 ms⁻¹

Solution

Physics sir · 4 steps

Differentiate the mirror formula: the image speed is the object speed times the square of the magnification.

  1. Set up the mirror. A rear-view mirror is convex, so f = +10 cm = +0.1 m. Car B is behind it, so u = -1.9 m.
  2. Differentiate. Differentiate 1/v + 1/u = 1/f with respect to time. f is constant, so its term drops out.

    dv/dt = -(v/u)² du/dt

  3. Find the image. v = uf/(u - f) = (-1.9)(0.1)/-2.0 = 0.095 m. So v/u = 0.095/1.9 = 0.05, and its square is 0.0025.
  4. So. Image speed = 0.0025 × 40 = 0.1 m s⁻¹, which is option (d).

The idea to keep. For a moving object in front of a mirror, the image speed along the axis is m² times the object speed. A convex mirror makes m small, so the image barely moves.

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

Study Ray Optics

Asked about 200 times in JEE Main since 2020. This question is from 2021.