Science (086) — AI-generated practice question

For a convex mirror, the focal length f is positive and the object distance u is negative (object is real, placed in front of the mirror).

From the mirror formula: $\dfrac{1}{v} = \dfrac{1}{f} - \dfrac{1}{u}$

Since f > 0 and u < 0, both terms $\dfrac{1}{f}$ and $-\dfrac{1}{u}$ are positive, so v > 0 (image is behind the mirror, virtual).

Magnification: $m = -\dfrac{v}{u}$

Since v > 0 and u < 0, $m = -\dfrac{(+)}{(-)} = +ve$ → m is never negative.

Also, for a convex mirror the image always forms between the pole and focus, so $|v| < |f|$ and $|v| < |u|$, meaning $|v| < |u|$, thus $m = \dfrac{v}{|u|} < 1$.

Therefore, 0 < m < 1 always for a convex mirror.

Source: Light – Reflection and Refraction, Chapter 9

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• Examiners expect you to apply sign convention explicitly: state signs of u, f, and deduce sign of v.
• Then use $m = -v/u$ and show why it is positive (erect image) and less than 1 (diminished image).
• The key insight: image distance |v| is always less than object distance |u| for a convex mirror, so the ratio stays between 0 and 1.
• Avoid vague statements — back every claim with the formula and signs.