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′?
- (a)1
- (b)2
- (c)1/2
- (d)2/3
Correct — C, 1/2. AB lies along the principal axis with A at 2f and B at 3f, so the two vertical sides of the rectangle sit at object distances 2f (side AD) and 3f (side BC). The image height of each side scales with the mirror's transverse magnification m = v/u there. At u = 2f the image forms at v = 2f, so |m| = 1 and A′D′ equals AD. At u = 3f the mirror formula 1/v + 1/u = 1/f gives v = 3f/2, so |m| = (3f/2)/(3f) = 1/2 and B′C′ = ½·BC. Since AD = BC in a rectangle, B′C′ / A′D′ = (1/2) / 1 = 1/2.
- (a)1 — A ratio of 1 would need equal magnification at both ends, but AD and BC lie at different object distances (2f and 3f), so their images are magnified by different factors.
- (b)2 — This reverses the ratio. The farther side (at 3f) is magnified less than the nearer side (at 2f), so B′C′ is smaller than A′D′, not larger.
- (d)2/3 — 2/3 comes from mishandling the mirror formula (for example, taking a ratio of object distances 2f/3f) instead of computing the two transverse magnifications correctly.
For a concave mirror the object distance u and image distance v satisfy 1/v + 1/u = 1/f, and the transverse (linear) magnification is m = -v/u. A transverse object placed at the centre of curvature (u = 2f) images to the same size (|m| = 1); placed farther away it images smaller. Here the rectangle's two vertical edges act as two transverse objects at u = 2f and u = 3f.
Although a figure accompanies the item, the geometry is fully fixed by the text: A at 2f, B at 3f, AB on the axis. So the two upright sides AD and BC are transverse segments at 2f and 3f, and the whole problem reduces to comparing the mirror's magnification at those two distances.
- Mirror formula: 1/v + 1/u = 1/f; transverse magnification m = -v/u.
- Object at u = 2f (the centre of curvature) → image at v = 2f, |m| = 1.
- Object at u = 3f → v = 3f/2, |m| = 1/2.
- Image-size ratio equals the ratio of magnifications: (1/2) / 1 = 1/2.
The farther edge (at 3f) is magnified half as much as the edge at 2f, giving the ratio 1/2 — option (c).
- Taking a simple ratio of object distances (2f : 3f) instead of computing each magnification.
- Forgetting that an object at the centre of curvature (2f) images to the same size (|m| = 1).
Asked as a numerical mirror-magnification problem where two parts of one object sit at different distances.
No directly related past PYQ was found.
- practice — not a real PYQ
An object placed at the centre of curvature (u = 2f) of a concave mirror forms an image whose magnitude of magnification is:
- (a)2
- (b)1
- (c)1/2
- (d)0
Answer(b) 1 — the image is the same size as the object.
- practice — not a real PYQ
For a concave mirror of focal length f, an object at u = 3f forms its image at:
- (a)3f
- (b)2f
- (c)3f/2
- (d)f
Answer(c) 3f/2 — from 1/v + 1/u = 1/f.