JEE Main 2021 · 27 August · morning · Single correct · +4 / −1

An object is placed beyond the centre of curvature C of the given concave mirror. If the distance of the object is d₁ from C and the distance of the image formed is d₂ from C, the radius of curvature of this mirror is :

  1. (a) 2d₁d₂/(d₁ - d₂)
  2. (b) 2d₁d₂/(d₁ + d₂)
  3. (c) d₁d₂/(d₁ + d₂)
  4. (d) d₁d₂/(d₁ - d₂)
Answer and solution

Answer (a) 2d₁d₂/(d₁ - d₂)

Solution

Physics sir · 4 steps

Write both distances from the pole in terms of R, then let the mirror formula solve for R.

  1. Locate the image. For an object beyond C, a concave mirror forms a real image between F and C. So the object is d₁ farther than C from the pole, and the image is d₂ nearer than C.
  2. Distances from pole. Taking distances in front of the mirror as negative, u = -(R + d₁) and v = -(R - d₂). The focal length is f = -R/2.

    1/(R - d₂) + 1/(R + d₁) = 2/R

  3. Clear the fractions. Cross-multiplying gives R(2R + d₁ - d₂) = 2(R - d₂)(R + d₁). The 2R² terms cancel, which leaves R d₁ - R d₂ = 2 d₁ d₂.
  4. So. R = (2 d₁ d₂)/(d₁ - d₂), which is option (a).

The idea to keep. When distances are measured from C or F, rewrite them as distances from the pole first. Only the pole is the origin of the mirror formula.

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.