A rectangle ABCD is kept in front of a concave mirror of focal length f with its corners A and B being, respectively, at distances 2f and 3f from the mirror with AB along the principal axis as shown in the figure. It forms an image A’B’C’D’ in front of the mirror. What is the ratio of B’C’ to A’D’ ?
[amp_mcq option1=”1″ option2=”2″ option3=”1/2″ option4=”2/3″ correct=”option3″]
This question was previously asked in
UPSC NDA-1 – 2023
For point A at u_A = -2f (assuming standard sign convention where f is negative for concave mirror, but distances are given as 2f and 3f. Let’s assume f > 0 is magnitude, and u is negative): u_A = -2f. 1/f = 1/(-2f) + 1/v_A => 1/v_A = 1/f + 1/(2f) = 3/(2f) => v_A = 2f/3. No, if A is at 2f from the mirror, and f is the focal length, then 2f is the radius of curvature C. Object at C forms image at C. So u_A = -2f, then v_A = -2f. Magnification m_A = -v_A/u_A = -(-2f)/(-2f) = -1. A’D’ = |m_A| * AD = AD.
For point B at u_B = -3f: 1/f = 1/(-3f) + 1/v_B => 1/v_B = 1/f + 1/(3f) = 4/(3f) => v_B = 3f/4. Wait, the standard formula for concave mirror with positive f is 1/f = 1/u + 1/v. Let’s use this and assume distances are positive. u_A = 2f, u_B = 3f, f=f. 1/v_A = 1/f – 1/u_A = 1/f – 1/(2f) = (2-1)/(2f) = 1/(2f) => v_A = 2f. Magnification m_A = -v_A/u_A = -(2f)/(2f) = -1. A’D’ = |-1| * AD = AD.
1/v_B = 1/f – 1/u_B = 1/f – 1/(3f) = (3-1)/(3f) = 2/(3f) => v_B = 3f/2. Magnification m_B = -v_B/u_B = -(3f/2)/(3f) = -1/2. B’C’ = |-1/2| * BC = 1/2 * BC.
Since AD = BC, the ratio B’C’ / A’D’ = (1/2 * BC) / AD = 1/2 * (BC/AD) = 1/2 * 1 = 1/2.