Sign convention for spherical mirrors

The pole $P$ of the spherical mirror is treated as the origin, and the principal axis as the x-axis — exactly like plotting points on a graph. Four rules fix every sign:

• The object always sits to the left of the mirror, so incident light always travels left to right.
• All distances are measured from the pole, along or parallel to the principal axis.
• Distances measured in the same direction as the incident light (rightward, away from the object) are positive; distances measured against it (leftward, toward the object) are negative.
• Heights measured upward from the axis are positive; heights measured downward are negative.

Applied to a real setup, these four rules pin down every quantity in the mirror formula. The object distance $u$ is always negative, because the object always sits in front of the mirror, against the direction of incident light — there is no exception. For a concave mirror, the focus $F$ and centre of curvature $C$ lie in front of the mirror (same side as the object), so the focal length $f$ and radius of curvature $R$ are negative; for a convex mirror, $F$ and $C$ lie behind the mirror, so $f$ and $R$ are positive. The image distance $v$ takes whichever sign matches where the image actually forms — negative for a real image in front of the mirror, positive for a virtual image behind it. Object height $h$ is positive (object upright); image height $h'$ is positive if the image is erect and negative if inverted. These signed quantities plug directly into the two governing formulas:

where: $v$ is the signed image distance, $u$ is the signed object distance, and $f$ is the signed focal length, all measured from the pole $P$.

where: $m$ is magnification, $h'$ is signed image height, $h$ is signed object height, and $v, u$ are the same signed distances as above. A negative $m$ means an inverted, real image; a positive $m$ means an erect, virtual image.

The convention is deliberately built so that one formula, $1/v + 1/u = 1/f$, covers every combination of mirror type and object position without a separate case for each. Because $u$ is fixed as always negative, the algebra automatically "knows" the object is in front of the mirror — there is nothing left to decide by hand. Because $f$ flips sign with the mirror type, the same equation naturally pulls the image toward convergence for a concave mirror and toward divergence for a convex one, without needing a different formula for each shape. And because $v$ is left free to come out positive or negative, solving the equation tells you whether the image is real (negative, in front) or virtual (positive, behind) — the sign of the answer carries the physics, rather than a diagram having to be redrawn and re-inspected for every new problem. The magnification formula $m = -v/u$ inherits the same logic: since $u$ is always negative, the minus sign in front of $v/u$ flips again when $v$ is negative, making $m$ negative exactly when the image is real and inverted, and positive exactly when it is erect and virtual. This is why the convention is worth memorising as a system of rules rather than as isolated facts — every downstream sign falls out of the same four starting rules.

An object $2\ \text{cm}$ tall is placed $30\ \text{cm}$ in front of a concave mirror of focal length $10\ \text{cm}$. Find the image distance and the image height, and state whether the image is real or virtual, erect or inverted, and enlarged or diminished.

Step 1. Assign signed values. By convention, $u = -30\ \text{cm}$ (object always negative) and $f = -10\ \text{cm}$ (concave mirror). Object height $h = +2\ \text{cm}$ (upright).

Step 2. Solve the mirror formula for $v$, exactly as in the derivation above: $\dfrac{1}{v} = \dfrac{1}{f} - \dfrac{1}{u} = -\dfrac{1}{10} + \dfrac{1}{30} = -\dfrac{1}{15}$, giving $v = -15\ \text{cm}$.

Step 3. Find magnification: $m = -\dfrac{v}{u} = -\dfrac{-15}{-30} = -0.5$.

Step 4. Find image height from $m = h'/h$: $h' = m \times h = -0.5 \times 2 = -1\ \text{cm}$.

Step 5. Interpret every sign. $v = -15\ \text{cm}$ is negative, so the image is real, formed $15\ \text{cm}$ in front of the mirror. $m = -0.5$ is negative, so the image is inverted, and its magnitude is less than $1$, so it is diminished. $h' = -1\ \text{cm}$ confirms this: the image is $1\ \text{cm}$ tall, inverted below the axis.

The image is real, inverted, diminished, formed 15 cm in front of the mirror, with height 1 cm.

• Substituting magnitudes instead of signed values. Both $u$ and $f$ must carry their sign; dropping the negative sign for a concave mirror's $f$ gives a wrong, non-physical value of $v$.
• Assuming $u$ can ever be positive. The object is always placed in front of the mirror by convention, so $u$ is always negative — there is no scenario where it is positive.
• Mixing up which mirror gets which sign for $f$. Concave mirrors converge and take $f$ negative; convex mirrors diverge and take $f$ positive — reversing these flips every downstream answer.
• Reading the sign of $m$ without checking magnitude. A negative $m$ means inverted, but whether the image is enlarged or diminished depends on whether $|m|$ is greater than or less than $1$ — both checks are needed.

Ask the tutor: If assigning signs to $u$, $v$, and $f$ still feels shaky for a new mirror problem, ask the AI tutor to walk through two or three more sign-assignment examples before attempting full mirror-formula numericals.