Topic 13.2: Images Formed by Mirrors

What Topic 13.2 requires

Topic 13.2 sits in Unit 13, Geometric Optics, which the CED weights at 12-15% of the multiple-choice section and estimates at about 8 to 12 class periods. Each topic in the unit carries one learning objective, and 13.2 carries the longest essential knowledge list.

13.2.A, describe the image formed by a mirror. Nine essential knowledge statements sit under it.

• 13.2.A.1 Incident rays parallel to the principal axis of a concave (converging) mirror are reflected toward a common location, the focal point.
• 13.2.A.2 Incident rays parallel to the principal axis of a convex (diverging) mirror are reflected such that they appear to have originated from a common location behind the mirror, the focal point.
• 13.2.A.3 The focal point of a plane mirror is an infinite distance from the mirror.
• 13.2.A.4 The focal point of a spherical mirror may be approximated as a point on the principal axis halfway between the surface of the mirror and the center of the mirror's radius of curvature.
• 13.2.A.5 A real image is formed when light rays emanating from a common point are reflected and then intersect at a common point.
• 13.2.A.6 A virtual image is formed when reflected rays diverge such that they appear to have originated from a common point.
• 13.2.A.7 The location of an image depends on the focal length and the object distance, with relevant equation $\dfrac{1}{s_i} + \dfrac{1}{s_o} = \dfrac{1}{f}$. 13.2.A.7.i adds that the focal point, object and image "follow sign conventions that are used to determine those locations relative to the mirror itself"; 13.2.A.7.ii that the image distance for a plane mirror equals the object distance.
• 13.2.A.8 Magnification is the ratio of image size to object size, with relevant equation $|M| = \left| \dfrac{h_i}{h_o} \right| = \left| \dfrac{s_i}{s_o} \right|$.
• 13.2.A.9 Ray diagrams determine the location, type, size and orientation of images. 13.2.A.9.i names the three principal rays; 13.2.A.9.ii states that images formed by a mirror can be upright or inverted, virtual or real, and reduced, enlarged, or the same size as the object.

The boundary statement is short enough to quote whole: "AP Physics 2 limits the study of mirrors to plane mirrors, convex spherical mirrors, and concave spherical mirrors." Three shapes, no others.

The CED lists four suggested skills: 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.

That list describes the questions: 1.A is the ray diagram, 2.A is why you should expect an expression rather than a number, 2.C the "now move the object closer" follow-up, 3.C the "explain why the image is virtual" prompt. Notice what is absent: 2.B, calculate an unknown quantity, is suggested for 13.1 and 13.4 but not here.

The sign convention, declared before anything else

Signs are the whole difficulty of this topic, so here is the convention this page uses, stated once and held to the end. A mirror sends light back the way it came, so everything real happens in front of the mirror, and that one fact sets every sign.

Read it as one rule: in front is positive, behind is negative. Behind the mirror is the region light never reaches, so anything located there is inferred by the eye rather than present.

The consequences then follow from the arithmetic instead of being asserted next to it.

• If the algebra returns $s_i > 0$, reflected rays are genuinely crossing in front of the mirror. That is the definition of a real image in 13.2.A.5, and a card held at that spot catches it.
• If the algebra returns $s_i < 0$, the meeting point is behind the mirror. Reflected rays are diverging and only their backward extensions meet, which is 13.2.A.6 word for word. A card there catches nothing.

So you never have to recall whether an arrangement gives a real or virtual image: solve for $s_i$ and read the sign.

Be precise about what the CED pins down and what it does not. Essential knowledge 13.2.A.7.i says only that these locations "follow sign conventions"; it prints no table of signs. Neither does the exam's own conventions list in the appendix: its nine bullets run from inertial frames and ideal wires through to the small-angle approximation for slit diffraction, and not one of them fixes an optics sign. What the equation sheet supplies is a hint in its symbol list, where $s$ is defined as position rather than distance. A position is a signed coordinate; a distance is not.

This is the convention the CED's own materials use when they work an example. The Topic 13.4 page traces it: the CED's thin-lens case study computes a negative image distance, places that image on the same side as the object, and labels it virtual.

Concave is converging, convex is diverging, and the sign of f carries it

The naming is a trap, and the CED defuses it by writing both words every time: "concave (converging) mirror" and "convex (diverging) mirror". Keep the pairing, because for lenses the words cross over and a convex lens converges. Carry "convex means diverging" from mirrors to lenses and every sign comes out backwards.

Concave, $f > 0$. Essential knowledge 13.2.A.1 says parallel rays are reflected toward a common location. Toward means they arrive. That point is in front of the mirror, in the positive region. Light really converges there, which is why a concave mirror can start a fire and a convex one cannot.

Convex, $f < 0$. Essential knowledge 13.2.A.2 places the focal point "behind the mirror" in its own words, and behind is the negative region. Nothing converges anywhere; the focal point is a place the rays only seem to come from.

Plane, $f$ infinite. Essential knowledge 13.2.A.3 puts the focal point an infinite distance away: a flat surface does not bend the bundle, so parallel in stays parallel out. In the equation that is $1/f = 0$.

Where the focal point sits on a curved mirror. Essential knowledge 13.2.A.4 places it halfway between the mirror surface and the center of curvature, so the magnitude of the focal length is half the radius:

Two warnings. The CED gives this in words only and it is not printed on the equation sheet, so if a question hands you a radius you halve it yourself. And note the word "may" in 13.2.A.4: the focal point may be approximated as that halfway point, which holds for rays near the axis. The CED flags the model's limits in the same sentence that gives you the model.

So a concave mirror of radius $R$ has $f = +R/2$ and a convex mirror of the same radius has $f = -R/2$. Problems quote $R$ as a plain positive length; the sign you attach from the shape.

Real and virtual, inverted and upright: how the signs produce the words

The CED's two definitions differ by one word. 13.2.A.5: rays from a common point are reflected and then intersect. 13.2.A.6: reflected rays diverge such that they appear to have originated from a common point. Real rays intersect; virtual rays only look like they came from somewhere. The sign of $s_i$ is the bookkeeping for that difference.

Orientation is where students guess. Essential knowledge 13.2.A.8 prints magnification as

and the absolute value bars are printed on the equation sheet too, not added here for safety. Both ratios are wrapped, so the printed relation returns a magnitude only. It answers enlarged, reduced, or same size. It cannot answer upright or inverted, because the sign that would carry that has been discarded before you see it.

Many textbooks define a signed $M = -s_i/s_o$ and read the minus as "inverted". The AP Physics 2 sheet does not print that version, so writing it in a free-response answer uses a convention your reference sheet did not give you.

Orientation comes instead from the two routes the CED does provide. The first is the diagram: 13.2.A.9 says ray diagrams determine "the location, type, size, and orientation" of the image. The second is a pairing that falls out of the geometry for a single mirror with a real object:

• A real image ($s_i > 0$) is inverted.
• A virtual image ($s_i < 0$) is upright.

That is not a separate convention to memorise, and one ray shows why. Take the ray striking the vertex, where the principal axis meets the surface. The axis is the normal there, so the ray reflects at an equal angle on the other side of it. If the image forms in front of the mirror, that reflected ray crossed the axis to get there and arrives on the opposite side from the object: inverted. If the rays instead diverge and you run them backwards, the extension never crosses the axis and the image lands on the same side: upright.

Size is a third, independent reading, and 13.2.A.9.ii lists all three possibilities.

So every mirror question has a fixed reading order: get $f$ with its sign, solve for $s_i$, read the sign for real or virtual, convert that to inverted or upright, then take $|M|$ for the size.

The three principal rays, exactly as the CED names them

Essential knowledge 13.2.A.9.i names them, and it is worth having the CED's list rather than a textbook's, because they are not always the same three: "The three principal rays are typically used to find the images formed by mirrors. The principal rays are 1) the ray parallel to the principal axis, 2) the ray that reflects at the center of the mirror where the principal axis intersects the mirror, and 3) the ray that passes through the focal point of the mirror."

Read ray 2 carefully. "The center of the mirror where the principal axis intersects the mirror" is the vertex, the middle of the reflecting surface. It is not the center of curvature, which sits a distance $R$ out in front. Many textbooks use the center-of-curvature ray, which reflects straight back on itself; that is a valid ray, but it is not one of the three the CED lists, and mixing up those two points is the most common way to draw a wrong diagram here.

Start every ray at the tip of the object, the only point you need to locate.

Concave mirror.

1. Parallel ray: travels parallel to the axis, reflects through the focal point. This is 13.2.A.1 used directly.
2. Vertex ray: travels to where the axis meets the mirror. The axis is the normal there, so the law of reflection from Topic 13.1 applies with the axis as the normal and the ray reflects at an equal angle on the far side.
3. Focal ray: passes through the focal point on the way in, reflects parallel to the axis. This is ray 1 reversed, which is why it works.

Convex mirror. The same three, with "through the focal point" becoming "along the line to the focal point behind the mirror", since there is none out in front to pass through. The parallel ray reflects so that its backward extension passes through that focal point; the vertex ray behaves as before; the focal ray aims at the focal point behind the mirror and reflects parallel to the axis.

Two habits worth building. Two rays locate an image, so the third is a check worth drawing, because an error shows up as three lines failing to meet at one point. And if the reflected rays diverge, do not stop: extend them backwards as dashed lines and mark the image where those meet. Dashed for virtual, solid for real, and skill 1.A is graded on that legibility.

Every case the CED asks for, in one table

The boundary statement limits mirrors to three shapes, so this table is complete for the topic. Throughout, $f$ is the signed focal length and the object is real and in front ($s_o > 0$). For a concave mirror the center of curvature C sits at $2f$.

Four things in it are worth saying out loud.

Only the concave mirror has a story. It moves through five behaviours as the object walks in, and the switch happens at the focal point: outside F the image is real and inverted, inside F it is virtual and upright. At the switch, $s_o = f$, we get $1/s_i = 1/f - 1/f = 0$ and the image distance runs to infinity, because the reflected rays leave parallel and never meet. That is the spotlight: put the bulb at the focal point and you get a beam.

Object and image swap places. Object beyond $2f$ gives an image between $f$ and $2f$; object between $f$ and $2f$ gives an image beyond $2f$. The equation is symmetric in $s_i$ and $s_o$, so the pair is reversible, which answers a comparison question without computing anything (skill 2.C).

The convex mirror has no story at all. One line proves it without cases: with $f$ negative, $1/s_i = 1/f - 1/s_o$ is a negative number minus a positive one, so it is negative for every positive $s_o$. Virtual and upright, always. A negative $f$ also makes $\lvert 1/s_i \rvert$ larger than $1/s_o$, so $\lvert s_i \rvert < s_o$ and $|M| < 1$ always.

Enlarged does not mean real. Two rows give enlarged images and they sit on opposite sides of the real-virtual divide. Size and type are independent readings, which is why the magnification relation carries absolute value bars: it was never meant to report the type.

The plane mirror is not a special case, it is the same equation

Plane mirrors are usually taught as a separate set of facts to memorise. Everything the CED says about them drops out of the equation you already have, which is the best evidence that the sign convention above is internally consistent.

Start from 13.2.A.3: the focal point is an infinite distance away, so $1/f = 0$ and the mirror equation becomes

which gives a single result:

Read the two halves separately. The magnitude says the image is as far behind the mirror as the object is in front, which is essential knowledge 13.2.A.7.ii: "The distance between the image formed and a plane mirror is equal to the distance between the object and the plane mirror." The CED states it as its own fact; the equation reproduces it without being told. The sign says the image is behind the mirror, so it is virtual and upright. That is why you cannot project a bathroom mirror's image onto paper, and why standing 1.5 m from a mirror puts your reflection 3.0 m away from you.

Magnification follows too:

Same size, always. Three CED facts about plane mirrors came out of one equation and one sign rule, with nothing memorised.

One thing the equation cannot capture is the unit's own essential question, "Why does a mirror flip words?" A plane mirror image is upright, so it does not flip top for bottom, and it does not really flip left for right either. What it reverses is front to back, along the axis pointing into the mirror. Hold up your right hand and its image is a left hand, because reversing the depth direction changes the handedness of the coordinate system. That is a three-dimensional effect, and no signed one-dimensional coordinate will show it to you.

Two-mirror arrangements, corner mirrors and periscopes all sit outside the boundary statement, which limits this topic to a single plane, convex spherical, or concave spherical mirror.

Traps, and how 13.2 gets tested

The suggested skills describe the question style, so pair each with the mistake it catches.

1.A, draw the diagram. The trap is the vertex ray. The CED's ray 2 reflects where the principal axis meets the mirror surface, not at the center of curvature. Draw the normal there as the axis itself and use equal angles.

2.A, derive an expression. Expect $s_o$ and $f$ as symbols and a request for $s_i$ or the magnification. Rearranging, $\dfrac{1}{s_i} = \dfrac{1}{f} - \dfrac{1}{s_o}$ gives $s_i = \dfrac{f s_o}{s_o - f}$, and dividing by $s_o$ gives the magnification ratio $\dfrac{f}{s_o - f}$. You have to combine the fractions before inverting. Inverting term by term, so that $1/s_i = 1/f - 1/s_o$ becomes $s_i = f - s_o$, is wrong and easy to do at speed.

2.C, compare two scenarios. The "object is moved closer" questions. The reversibility in the case table, and the fact that a convex mirror never changes behaviour, both beat recomputing.

3.C, justify a claim. These want the physics, not the arithmetic. A full answer names what the rays do, in the CED's terms: the reflected rays diverge and their extensions appear to originate from a common point behind the mirror, therefore the image is virtual. A sign on its own justifies nothing.

Four errors are worth naming directly.

• Losing the sign on $f$ for a convex mirror. It turns a virtual image into a real one and takes every later answer with it. Attach the sign the moment you read the word convex, before touching the algebra.
• Reading a magnification of 2 as "upright and twice as tall". The printed relation gives magnitude. Orientation comes from the sign of $s_i$ or from the diagram.
• Carrying this convention across to a lens. For a mirror, positive $s_i$ is on the same side as the object, because light comes back. For a lens it is on the far side, because light goes through. "Positive means real" survives the crossing; the geometry does not. Topic 13.4 sets it out for lenses.
• Assuming a separate mirror equation exists. It does not. The sheet prints $\dfrac{1}{s_i} + \dfrac{1}{s_o} = \dfrac{1}{f}$ once, in the Waves, Sound, and Optics group, and it serves mirrors and lenses alike. That group holds fifteen entries, and exactly two belong to this topic: the mirror equation and the magnification relation.

For the mechanism underneath all of this, the law of reflection and the difference between specular and diffuse surfaces, see Topic 13.1. For the same logic applied to refraction, see Topic 13.4. The Unit 13 overview puts the four topics in order, and the AP Physics 2 equation sheet shows what you have to work from.