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 BC' to AD'?

The problem requires calculating the magnification of the vertical sides of the rectangle using the mirror formula (1/v + 1/u = 1/f) and the magnification formula (m = -v/u = h'/h).

• At Point A: The distance uA = 2f (Center of Curvature). For an object at 2f, the image is formed at vA = 2f. The magnification |mA| = |v/u| = 2f/2f = 1. Thus, the height of the image A'D' = 1 × AD.
• At Point B: The distance uB = 3f. Using the mirror formula: 1/v + 1/(-3f) = 1/(-f) ⇒ 1/v = -1/f + 1/3f = -2/3f. Thus, vB = 1.5f. The magnification |mB| = |v/u| = 1.5f/3f = 0.5. Thus, the height of the image B'C' = 0.5 × BC.

Since ABCD is a rectangle, the object heights are equal (AD = BC). Therefore, the ratio of the image heights B'C' to A'D' is (0.5 × BC) / (1 × AD) = 1/2. This corresponds to Option C.