Convex Mirror Ray Diagram: Always Virtual

A convex mirror is the curved mirror you see at blind corners in parking garages, the passenger-side mirror on your car ("objects may be closer than they appear"), and the security mirrors in convenience stores. It bulges outward — the reflective surface is on the outside of the curve. Because of this outward curvature, it does something very specific and very predictable: it always makes things look smaller, farther away, upright, and virtual.

If the concave mirror is the Swiss Army knife of mirrors (capable of producing real, virtual, magnified, diminished, inverted, and upright images depending on object position), the convex mirror is the opposite: a one-trick pony. But that one trick — providing a wide-angle, always-upright, always-diminished view — is so useful that convex mirrors are everywhere.

Why the focal length is negative

A convex mirror's reflective surface curves away from the incoming light. Parallel rays hitting it diverge after reflection, as if they came from a point behind the mirror. That point is the focal point $F$, and because it's behind the mirror (on the non-reflecting side), the focal length is assigned a negative value.

For our problem, $f = -10$ cm. The center of curvature $C$ is at $2f = -20$ cm (also behind the mirror).

Solving with the mirror equation (method 1)

$\dfrac{1}{f} = \dfrac{1}{d_{o}} + \dfrac{1}{d_{i}}$

$\dfrac{1}{d_{i}} = \dfrac{1}{f} - \dfrac{1}{d_{o}} = \dfrac{1}{-10} - \dfrac{1}{15} = -\dfrac{3}{30} - \dfrac{2}{30} = -\dfrac{5}{30} = -\dfrac{1}{6}$

$d_{i} = -6 \text{ cm}$

Negative $d_{i}$ → the image is behind the mirror (virtual).

$m = -\dfrac{d_{i}}{d_{o}} = -\dfrac{-6}{15} = +0.4$

Positive $m$ → upright. $|m| = 0.4$ → the image is 40% of the object's height. Diminished.

Why it's always virtual (proof by algebra)

For a convex mirror, $f < 0$, so $1/f$ is negative. And $d_{o} > 0$, so $1/d_{o}$ is positive:

$\dfrac{1}{d_{i}} = \underbrace{\dfrac{1}{f}}_{\text{negative}} - \underbrace{\dfrac{1}{d_{o}}}_{\text{positive}} = \text{more negative}$

$1/d_{i}$ is always negative → $d_{i}$ is always negative → the image is always virtual. Identical algebra to the concave (diverging) lens — diverging optics always produce virtual images (for real objects).

And $m = -d_{i}/d_{o} = -(- |d_{i}|)/d_{o} = |d_{i}|/d_{o}$, always positive (upright), and you can show $|d_{i}| < d_{o}$ so $|m| < 1$ (always diminished).

The three principal rays for a convex mirror (method 2)

Ray 1 — parallel → appears to come from F. A horizontal ray from the object tip hits the mirror and reflects outward. Tracing the reflected ray backward (behind the mirror), it appears to emanate from the focal point $F$ behind the mirror.

Ray 2 — aimed at C → reflects back on itself. A ray directed toward the center of curvature $C$ (behind the mirror) hits the mirror perpendicularly and bounces straight back.

Ray 3 — aimed at F → reflects parallel. A ray directed toward the focal point $F$ (behind the mirror) reflects back parallel to the principal axis.

All three reflected rays diverge. But their backward extensions (dashed lines behind the mirror) meet at a single point. That's the virtual image — smaller, upright, and behind the mirror.

More worked examples

Example 2: $d_{o} = 5$ cm, $f = -10$ cm.

$d_{i} = \dfrac{1}{-0.1 - 0.2} = \dfrac{1}{-0.3} = -3.33 \text{ cm}, \quad m = +0.667$

Close object: virtual image at $-3.33$ cm, 67% of original size. Even up close, the image never reaches full size.

Example 3: $d_{o} = 100$ cm, $f = -10$ cm.

$d_{i} = \dfrac{1}{-0.1 - 0.01} = \dfrac{1}{-0.11} = -9.09 \text{ cm}, \quad m = +0.091$

Far object: the image barely reaches 9% of the original size, pinned close to $F$ behind the mirror. As $d_{o} \to \infty$, $d_{i} \to f = -10$ cm and $m \to 0$. The image shrinks to a point right at $F$.

The field-of-view advantage

Here's why convex mirrors are on every car and in every parking garage: a flat mirror shows you exactly one angle — what's directly in front of it. A convex mirror, by shrinking everything, packs a much wider field of view into the same mirror size. The stronger the curvature (shorter $|f|$), the wider the field.

Your car's driver-side mirror is flat (accurate size, narrow field). Your passenger-side mirror is convex (smaller images, wider field). That's why it comes with the warning "objects in mirror are closer than they appear" — the convex curvature makes everything look farther away and smaller than it really is.

Concave vs convex mirror summary

• Concave (converging, $f > 0$): Can produce real or virtual images depending on object position. Five regimes. Used for magnification and focusing (telescopes, headlights, makeup mirrors).
• Convex (diverging, $f < 0$): Always virtual, always upright, always diminished. One regime. Used for wide-angle views (car mirrors, security mirrors, ATM monitors).

The analogy to lenses is exact:

• Concave mirror ↔ Convex lens (both converging, $f > 0$)
• Convex mirror ↔ Concave lens (both diverging, $f < 0$)

Common mistakes

• Thinking convex mirrors can make real images. They can't (for real objects). It's mathematically impossible — the algebra shows $d_{i} < 0$ always.
• Drawing $F$ and $C$ in front of the mirror. They're behind it. A convex mirror's focal point and center of curvature are on the non-reflecting side.
• Confusing "closer than they appear." The image in a convex mirror is closer to the mirror than the object is to the mirror ($|d_{i}| < d_{o}$), but the brain interprets the smaller image as being farther away. The warning is about this perceptual mismatch.

Try it in the visualization

Drag the object from 3 cm to 50 cm and notice how little the image moves — it's always trapped between the mirror and $F$. The image never gets bigger than the object ($|m| < 1$ always). Toggle the backward extensions to see the virtual image form behind the mirror. Compare with Problem 149 (concave mirror) by switching between them — same equation, completely different behavior.