AI-generated practice question — model-generated for extra practice, not a previous-year CBSE board question.
For a convex mirror, the centre of curvature (C) lies behind the mirror (virtual). A ray directed towards C, upon striking the mirror, is reflected back along the same path (i.e., the reflected ray retraces the incident ray and appears to diverge away from C when extended behind the mirror).
How to draw it: Draw the incident ray from the object tip aimed at C (behind the mirror). At the point where it hits the mirror surface, draw the reflected ray going straight back along the same direction as the incident ray, retracing the path.
Principle: The incident ray falls on the mirror along the normal to the reflecting surface (since every radius passes through the centre of curvature). The normal to a spherical mirror at any point passes through C. Therefore, the angle of incidence = 0°, so the angle of reflection = 0°, and the ray reflects back along the same path. This is consistent with the laws of reflection.
Source: Chapter 9, Section 9.2.2 — Rule (iii) for ray directed towards centre of curvature of a convex mirror
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Examiners expect two things: (1) a clear description of how to draw the reflected ray (back along the same path / retracing) and (2) the justification — the ray travels along the normal because C is the centre of the sphere, so angle of incidence = 0° = angle of reflection. Quoting the textbook rule directly ("reflected back along the same path because the incident rays fall on the mirror along the normal") earns full marks. Don't just say "it follows laws of reflection" — explain why (the normal passes through C).