Concave vs Convex Mirror: What Is the Difference?

The distinction, stated once

The AP Physics 2 course description gives both mirrors in the same form, so the two statements can be read side by side.

Concave. Essential knowledge 13.2.A.1: incident light rays parallel to the principal axis of a concave (converging) mirror will be reflected toward a common location, called the focal point.

Convex. Essential knowledge 13.2.A.2: incident light rays parallel to the principal axis of a convex (diverging) mirror will be reflected such that they appear to have originated from a common location behind the mirror, called the focal point.

Three words in those two sentences do all the work. The CED itself supplies the labels converging and diverging in parentheses, so the distinction is not something you have to infer from the shape. And the phrase behind the mirror is what makes the convex case different in kind rather than in degree: the concave mirror's focal point is a place light actually passes through, while the convex mirror's is a place light never reaches and only appears to come from.

Everything else follows. A concave mirror gathers rays, so it can bring them to a crossing point in front of itself, which is a real image. A convex mirror spreads rays apart, so nothing they do afterwards ever brings them together, and the only image it can form is virtual. The concave mirror is the one whose behaviour changes with object distance; the convex mirror does the same thing to everything.

A memory hook that survives contact with exam questions: a concave mirror is the one you could fill with water. Its reflecting surface is the inside of the bowl, facing the object. The convex mirror's reflecting surface is the outside of the bowl, bulging toward you.

The sign convention used on this page, declared before any number

The CED requires a sign convention and does not print one. Essential knowledge 13.2.A.7.i says the locations of a mirror's focal point, an object near the mirror, and the image of the object formed by the mirror follow sign conventions that are used to determine those locations relative to the mirror itself. The AP Physics 2 equation sheet then prints magnification with absolute value bars on every term,

so the printed relation gives sizes and no orientations. The signs have to come from somewhere, and that somewhere is a convention you state.

Here is the one this page uses, and it holds unchanged to the last line.

• Distances are measured from the reflecting surface of the mirror, which is what 13.2.A.7 specifies for the object distance.
• Object distance $s_o > 0$ for a real object. Every object on this page is real, so $s_o$ is positive everywhere below.
• Image distance $s_i > 0$ when the image forms in front of the mirror, on the side the light is on. $s_i < 0$ when it forms behind the mirror. Positive means real, negative means virtual.
• Focal length $f > 0$ for a concave mirror, whose focal point is in front. $f < 0$ for a convex mirror, whose focal point is behind, exactly as 13.2.A.2 describes.
• Magnification $M = -s_i/s_o$. Positive means upright, negative means inverted, and $\lvert M \rvert$ greater than one means enlarged.

The sign of $f$ is the only input that distinguishes the two mirrors. Once it is fixed, one equation does everything, and it is the relevant equation the CED gives for 13.2.A.7:

Rearranged for calculation, $1/s_i = 1/f - 1/s_o$. Put $f$ in with its sign, read the sign of the answer, and the image is classified. That is the whole method, and this page uses no other.

A note for the exam. Because the framework does not print a convention, write yours down in one line before substituting. A convention that quietly changes between parts of a question produces a set of answers that contradict each other, which is worse than a convention a grader would have set up differently.

Side by side

The row that decides most questions is does the image kind change with $s_o$, because it is the row that tells you how much work a question needs. For a convex mirror the answer is fixed before you compute anything: virtual, upright, reduced. For a concave mirror you have to compare the object distance with the focal length first, and only then does the classification follow.

The field-of-view row is the one that explains the everyday uses, and it is worth stating in the right direction. A convex mirror shows you more of the scene and shows all of it smaller. That is a single fact with two descriptions, not two separate advantages: the reduction is what makes room for the wider view.

The case that separates them: same three object distances, two mirrors

Take a concave mirror with $\lvert f \rvert = 15 \ \mathrm{cm}$ and a convex mirror with the same $\lvert f \rvert = 15 \ \mathrm{cm}$, so the only difference in the arithmetic is the sign of $f$. Put an object at $45 \ \mathrm{cm}$, then $30 \ \mathrm{cm}$, then $10 \ \mathrm{cm}$ in front of each.

Concave mirror, $f = +15 \ \mathrm{cm}$:

Convex mirror, $f = -15 \ \mathrm{cm}$:

One column in the convex table never changes. Every entry in its image column is the same three words, at every object distance, and the object distance only changes how close behind the mirror the image sits and how small it is. The concave table's image column changes twice over the same three distances.

Why the convex case cannot vary is worth proving rather than asserting, because it is a one-line argument. In $1/s_i = 1/f - 1/s_o$, a convex mirror makes the first term negative and a real object makes the second term subtract a positive quantity. Negative minus positive is negative, so $1/s_i < 0$ and therefore $s_i < 0$ for every positive $s_o$. A negative image distance is a virtual image, so a convex mirror can never form a real one from a real object, and $M = -s_i/s_o$ is then always positive, so the image is always upright.

The size claim follows the same way. Rearranging, $\lvert s_i \rvert = 1/(1/\lvert f \rvert + 1/s_o)$, which is smaller than $s_o$ for every positive $s_o$, so $\lvert M \rvert = \lvert s_i \rvert / s_o$ is always less than one and the image is always reduced. Three properties, one sign, no exceptions inside the course's scope.

The concave mirror's behaviour has a crossover instead, at $s_o = f$. Beyond the focal point, $1/f$ exceeds $1/s_o$, so $s_i$ comes out positive and the image is real and inverted. Inside the focal point the inequality reverses, $s_i$ is negative, and the image is virtual and upright. At exactly $s_o = f$, $1/s_i = 0$, no image forms, and the reflected rays leave parallel. The $s_o = 30 \ \mathrm{cm}$ row above is the other landmark: at $s_o = 2f$ the arithmetic gives $s_i = 2f$ and $M = -1$, a real inverted image the same size as the object.

Focal point, radius of curvature, and the plane-mirror limit

Essential knowledge 13.2.A.4 locates the focal point of a spherical mirror geometrically: it may be approximated as a point located on the principal axis of the mirror halfway between the surface of the mirror and the center of the mirror's radius of curvature.

Two things are worth reading carefully in that sentence. It says may be approximated, so it is stated as an approximation rather than an identity, and it locates the focal point halfway to the centre of curvature, which in magnitude means the focal length is half the radius of curvature. The CED does not print that as an equation, and neither does the AP Physics 2 equation sheet, which has no relation between focal length and radius of curvature anywhere in its waves, sound, and optics block. So if a question gives you a radius of curvature, use the geometric statement in 13.2.A.4 to halve it, and attach the sign from your convention: positive for concave, negative for convex.

The plane mirror is the third case the course covers, and it is a good consistency check on the whole convention. Essential knowledge 13.2.A.3 says the focal point of a plane mirror is an infinite distance from the mirror. So $1/f = 0$, and the shared equation collapses:

giving $s_i = -s_o$. Negative, so virtual, and equal in magnitude, so the image is as far behind the mirror as the object is in front. That is exactly what 13.2.A.7.ii states independently: the distance between the image formed and a plane mirror is equal to the distance between the object and the plane mirror. And $M = -s_i/s_o = +1$, so upright and the same size. Three separate CED statements agreeing with one substitution is how you know the signs are being used consistently.

A plane mirror is also the natural boundary between the two curved cases. Flatten a concave mirror and its focal length grows without limit; flatten a convex mirror and the same happens from the negative side. The plane mirror sits between them, converging nothing and diverging nothing, always giving the one image every mirror on this page's virtual column gives: upright, same size, behind the glass.

The Topic 13.2 boundary statement fixes the scope exactly here: AP Physics 2 limits the study of mirrors to plane mirrors, convex spherical mirrors, and concave spherical mirrors. Nothing else, and no other shape of curve.

When it costs a mark

Getting the sign of $f$ from the shape backwards. Concave is converging and positive; convex is diverging and negative. The CED hands you the converging and diverging labels in 13.2.A.1 and 13.2.A.2, so the shape-to-sign step does not have to be guessed.

Not declaring a convention and then changing it. 13.2.A.7.i requires a convention and the framework prints none. State yours once, in one line, and keep it for every part of the question.

Claiming a convex mirror can give a real image. It cannot, from a real object, and the sign argument above is the reason. If your working produces a positive $s_i$ for a convex mirror, the sign of $f$ went in as positive.

Claiming a concave mirror always gives a real image. It does not. An object inside the focal point gives a virtual, upright, enlarged image, which is exactly how a shaving mirror is used.

Reading the sheet's magnification as signed. The sheet prints $\lvert M \rvert$ with bars on every term. It cannot tell you upright from inverted. Get the sign from $M = -s_i/s_o$ or from a ray diagram.

Using the radius of curvature as the focal length. By 13.2.A.4 the focal point sits halfway between the mirror surface and the centre of curvature, so the focal length is half the radius in magnitude. Substituting $R$ where $f$ belongs doubles the focal length and moves every image.

Forgetting the final reciprocal. $1/s_i = 1/f - 1/s_o$ gives you the reciprocal of the image distance. A value of $-1/10$ per centimetre is not an image distance of $-0.1 \ \mathrm{cm}$; it is an image distance of $-10 \ \mathrm{cm}$.

Confusing enlarged with real. They are independent. The concave rows above include a real reduced image, a real same-size image, and a virtual enlarged one. Orientation comes from the sign of $M$ and size from its magnitude, and neither one determines the kind of image on its own.

When they behave alike, and why that lulls you

The two mirrors agree in three situations, and each one hides the difference for a different reason.

When the object is inside the concave mirror's focal point. Then the concave mirror also gives a virtual, upright image, so the words in the answer match the convex case exactly. The numbers do not: the concave mirror's virtual image is enlarged and the convex mirror's is always reduced. If a question only asks for real or virtual and upright or inverted, these two cases are indistinguishable.

When the object is very far away. Push $s_o$ toward very large values and $1/s_o$ becomes negligible, so $s_i$ approaches $f$ for both mirrors: the image forms essentially at the focal point, in front for the concave mirror and behind for the convex one. Both images are tiny, and the magnifications approach zero from opposite signs.

Both obey the same equation and the same law of reflection. There is no separate concave equation and no separate convex equation. $1/s_i + 1/s_o = 1/f$ handles both, and every ray in either diagram obeys $\theta_i = \theta_r$ from 13.1.B.2. The only thing that differs between the two calculations is one minus sign.

The distinction turns on exactly three questions, and they are the three that keep appearing on exams.

1. Which side is the focal point on? In front means concave and $f > 0$; behind means convex and $f < 0$.
2. Can this mirror produce a real image at all? Only the concave one, and only for an object beyond its focal point.
3. Does the answer depend on the object distance? For the convex mirror, only the size and position of the image; the kind and orientation are fixed. For the concave mirror, everything depends on it.

If you can answer those three, the pair is finished.

Where this sits on the AP exam

Mirrors are Topic 13.2, Images Formed by Mirrors, in Unit 13, Geometric Optics, which the CED weights at 12 to 15 percent of the multiple-choice section over a suggested 8 to 12 class periods. The topic has one learning objective, 13.2.A, describe the image formed by a mirror, and it is the only topic in Unit 13 that carries a boundary statement: AP Physics 2 limits the study of mirrors to plane mirrors, convex spherical mirrors, and concave spherical mirrors. That was checked against all four topic pages in the unit.

Ray diagrams are part of the requirement rather than an optional aid. Essential knowledge 13.2.A.9 says ray diagrams can be used to determine the location, type, size, and orientation of images formed by mirrors, and 13.2.A.9.i names the three principal rays typically used: the ray parallel to the principal axis, the ray that reflects at the center of the mirror where the principal axis intersects the mirror, and the ray that passes through the focal point of the mirror. Essential knowledge 13.2.A.9.ii then lists the choices a complete description makes: images formed by a mirror can be upright or inverted, virtual or real, and reduced, enlarged, or the same size as the object. Three choices, and a full-credit answer names all three.

The suggested skills the CED lists for Topic 13.2 are 1.A, create diagrams, tables, charts, or schematics to represent physical situations; 2.A, derive a symbolic expression from known quantities by selecting and following a logical mathematical pathway; 2.C, compare physical quantities between two or more scenarios or at different times and locations in a single scenario; and 3.C, justify or support a claim using evidence from experimental data, physical representations, or physical principles or laws. Skill 2.C is the reason the two-table comparison above is the shape of the question to rehearse, and 2.A is why a symbolic argument, like the proof that a convex mirror's image distance is always negative, is worth being able to write out.

For what real and virtual images actually are, and the same sign convention applied to lenses as well as mirrors, see real vs virtual image. For the mechanism, mirrors work by reflection while lenses work by refraction, see reflection vs refraction. Lenses themselves are Topic 13.4, where the convex lens plays the converging role that the concave mirror plays here, which is a naming reversal worth noticing before an exam rather than during one.