Unit 13 · Topic 13.2 · about 30 minutes · 7 parts · 10 scored checks, 1 of them multiple choice, plus 2 from earlier lessons

AP Physics 2: Algebra-Based 13.2: Images Formed by Mirrors

Q: How do you find the magnification of a mirror?

|M| = |hiho| = |siso|. It gives the size; for one mirror and a real object, a real image is inverted and a virtual image is upright. If you see M = -siso, a negative M means inverted and a positive M upright; this agrees with reading orientation from the image type.

Built by Tanner Crow from College Board's course framework · Every answer key independently checked

Unit 13: 12 to 15% of the multiple-choice section

What you'll be able to do

Describe the image a plane, concave or convex mirror forms: find it with principal rays and with the mirror equation, and give its location, type, size and orientation.

Key vocabulary

Term	What it means	The exam trap
Principal axis	The line through the middle of a mirror and its center of curvature.	The principal axis is a line in front of and behind the mirror, not the line of the mirror's own surface.
Center of curvature	The center of the sphere that a spherical mirror is a piece of, written C.	C is a point in space on the principal axis, not a point on the mirror itself.
Magnification	How many times taller an image is than its object, the ratio of their heights.	A magnification below 1 is a smaller image, not a mistake: magnification does not have to be larger than 1.

Tap to open the picture full size

The security mirror worked in Watch it done.

Part 1 Think first Focusing on a reflection

Commit to a guess first. Being wrong now makes the right idea stick better.

Predict first

You stand 1.5 m in front of a flat bathroom mirror and take a photo of yourself in it. At what distance must the camera focus for your reflection to come out sharp?

In short. A mirror forms an image where reflected rays from each point of an object meet (a real image) or seem to come from (a virtual image), at a distance given by $\frac{1}{s_{i}} + \frac{1}{s_{o}} = \frac{1}{f}$ .

A concave mirror reflects rays parallel to its axis toward a focal point in front of it; a convex mirror spreads them as if from a focal point behind it; a plane mirror's focal point is infinitely far away.
For a spherical mirror, $f \approx \frac{R}{2}$ ; in this course's sign convention $f$ is negative for a convex mirror and $s_{i}$ is negative for a virtual image.
Magnification is a ratio of sizes, $| M | = | \frac{h_{i}}{h_{o}} | = | \frac{s_{i}}{s_{o}} |$ ; for one mirror, a real image is inverted and a virtual image upright.

Part 2 Learn it Focal points, images and the mirror equation

2a. Where parallel rays go

A spherical mirror is a piece of the surface of a sphere. The center of that sphere is the mirror's center of curvature, C, its radius is the radius of curvature, $R$ , and the principal axis runs through C and the middle of the mirror.

A concave, or converging, mirror curves in, like the inside of a spoon. Rays parallel to the principal axis reflect toward a common point in front of it, the focal point, F.
A convex, or diverging, mirror bulges out, like the back of a spoon. Rays parallel to the axis reflect and spread apart as if they came from a common point behind it, its focal point.
A plane mirror is flat. Parallel rays reflect and stay parallel, so its focal point is an infinite distance away.

For a spherical mirror, the focal point is approximately halfway between the mirror and its center of curvature, so its focal length is

f \approx \frac{R}{2}

It is an approximation that holds for rays striking near the principal axis, and every mirror problem here assumes such rays. AP Physics 2 limits mirrors to plane, concave spherical and convex spherical ones.

Quick check · Sort it

Each card describes a mirror. Tap a card, then the kind of mirror it describes.

Concave

Convex

Plane

2b. Real and virtual images, and three rays

Each point of an object sends out rays in all directions. After a mirror reflects them, either:

The reflected rays from one point meet again. Light really arrives there, so this is a real image: a screen there shows it.
The reflected rays spread apart, but traced backward they seem to come from a common point behind the mirror. No light reaches it, so this is a virtual image, seen only by looking into the mirror.

Any two of three principal rays from the top of an object find its image:

A ray parallel to the axis reflects through the focal point, or, from a convex mirror, as if from the focal point behind.
A ray striking the center of the mirror reflects at the same angle on the other side of the axis.
A ray through the focal point, or aimed at a convex mirror's focal point behind, reflects parallel to the axis.

Where the reflected rays, or their extensions behind the mirror, cross is the top of the image. Ray diagrams draw each reflection at the mirror's center line, the same near-axis approximation as $f \approx \frac{R}{2}$ .

Quick check · Fill the statement

Complete the three principal rays for a concave mirror. Tap a blank, then an answer.

A ray parallel to the principal axis reflects . A ray through the focal point reflects . A ray that strikes the center of the mirror reflects .

2c. The mirror equation and magnification

The mirror equation locates an image by calculation:

\frac{1}{s_{i}} + \frac{1}{s_{o}} = \frac{1}{f}

$s_{o}$ is the object's distance from the mirror and $s_{i}$ the image's. College Board's course framework says these locations follow a sign convention, measured from the mirror, but does not print one. This course uses the real-is-positive convention:

$f$ is positive for a concave mirror and negative for a convex mirror.
$s_{o}$ is positive for an object in front of the mirror.
$s_{i}$ is positive for a real image, in front of the mirror, and negative for a virtual image, behind it.

Magnification compares sizes:

| M | = | \frac{h_{i}}{h_{o}} | = | \frac{s_{i}}{s_{o}} |

The framework writes it as a size only, so read the orientation from the image type: for one mirror and a real object, a real image is inverted and a virtual image is upright. If you see $M = - \frac{s_{i}}{s_{o}}$ , a negative $M$ means inverted and a positive $M$ upright; this agrees with reading orientation from the image type.

A plane mirror has $\frac{1}{f} = 0$ , so $s_{i} = - s_{o}$ : its image is virtual, as far behind the mirror as the object is in front, and the same size.

Quick check · Calculate

A convex mirror has a focal length of magnitude $f$ . An object stands on the principal axis a distance $f$ in front of it, with rays close to the axis. Derive the image distance, with its sign in this course's convention (negative for a virtual image behind the mirror). It can be written as $s_{i} = k f$ , with $f$ the magnitude. What is $k$ ?

Your answer

Common mistakes

The mistake	Why it is wrong	Do this instead
Putting a plane mirror's image on the surface of the mirror.	Reflected rays from each point of the object seem to come from a point behind the glass, as far behind it as the object is in front.	Use $s_{i} = - s_{o}$ : the image is behind the mirror, at the object's distance.
Using a positive focal length for a convex mirror.	In this course's convention, a convex mirror's focal point is behind it, so its focal length is negative.	Write the sign with $f$ before you substitute: plus for concave, minus for convex.
Thinking a concave mirror always makes a real, inverted image.	An object closer to the mirror than the focal point sends rays that still spread apart after reflecting, so the image is virtual and upright.	Compare the object distance with the focal length before you decide the image type.
Thinking an image travels from the object to the mirror, or to a screen, as a picture.	Nothing travels as a picture. Each point of the object sends rays in many directions, and an image point is only where the rays from one object point meet again, or seem to come from.	Follow the rays from one point of the object at a time, and find where they meet.

Part 3 Watch it done A shopper in a security mirror

Predict first, then watch each step, then finish the problem yourself.

In the picture at the top of this lesson, a convex security mirror has a focal length of magnitude 0.60 m, and a shopper 1.8 m tall stands 3.0 m in front of it, on its principal axis. Taking the rays as close to the axis, where is her image, and how does its height compare with hers?

Predict first

Before you open the first step, predict: how tall will the shopper's image be?

Step 1 · Write the signs
A convex mirror has a negative focal length, so $f = - 0.60$ m. The shopper is in front of the mirror, so $s_{o} = + 3.0$ m.
Step 2 · Find the image distance
$\frac{1}{s_{i}} = \frac{1}{f} - \frac{1}{s_{o}} = \frac{1}{- 0.60 m} - \frac{1}{3.0 m} = - 1.67 m^{- 1} - 0.33 m^{- 1} = - 2.0 m^{- 1}$ , so $s_{i} = - 0.50$ m. The image is 0.50 m behind the mirror: negative, so virtual, and therefore upright.
Why this step?
Why is the image distance negative?
The rays a convex mirror reflects spread apart, so they never meet in front of it. Traced backward, they seem to come from a point behind the mirror, and in this course's convention a distance behind the mirror, where a virtual image forms, is negative.
Your step · Calculate

Finish it. A convex mirror has a focal length of -0.60 m, and a shopper 1.8 m tall stands 3.0 m in front of it. Her image forms 0.50 m behind the mirror. How many times as tall as her image is the shopper?

Your answer

Part 4 Try it Bringing an object closer

Predict first: as an object moves in from far away toward a concave mirror, does its image grow or shrink, and does it ever turn upright? Then match each position in the picture, and find one image distance with symbols alone.

A concave mirror on the right with its principal axis running left. The focal point F is marked 10 cm in front of the mirror and the center of curvature C 20 cm in front. Four object positions are marked on the axis: W at 30 cm, X at 20 cm (at C), Y at 15 cm, and Z at 5 cm, between F and the mirror. No rays are drawn.

Try it · Match

A concave mirror has a focal length of 10 cm. An object is placed at each of four positions in front of it: W at 30 cm, X at 20 cm (the center of curvature), Y at 15 cm and Z at 5 cm. Tap each position, then the image an object there forms.

Try it · Calculate

An object stands on the principal axis of a concave mirror with focal length $f$ , a distance $3 f$ in front of the mirror, with rays close to the axis. Derive the image distance; it can be written as $s_{i} = k f$ . What is $k$ ?

Your answer

Part 5 Explain it Draw it, then calculate it

On the exam this is handwritten, so answer on paper first. Nothing you type here leaves this page. Then check your answer against the model and each scoring point.

Free response is 50% of the exam score. This one is shaped like Question 2, Translation Between Representations.

Free-response practice · Question 2 style

A concave spherical mirror has a focal length of 10 cm. An object 4.0 cm tall stands upright on the principal axis, 15 cm in front of the mirror. Treat the rays as close to the principal axis.

(a) On paper, draw a ray diagram to scale, with at least two principal rays. Locate the image, and state whether it is real or virtual and whether it is upright or inverted. Do this by hand before you open the model.

(b) Use the mirror equation to calculate the image distance and the height of the image.

Your answer stays on this page. It is never sent or saved.

Model answer

(a) Compare your diagram with the model below. The ray parallel to the axis reflects through F; the ray to the center of the mirror reflects below the axis at the same angle; the ray through F reflects parallel to the axis. They cross 30 cm in front of the mirror and 8.0 cm below the axis. The image is real, because the reflected rays themselves meet there, and inverted.

(b) $\frac{1}{s_{i}} = \frac{1}{10 cm} - \frac{1}{15 cm} = \frac{1}{30 cm}$ , so $s_{i} = + 30$ cm: positive, so a real image in front of the mirror. $| M | = \frac{30 cm}{15 cm} = 2.0$ , so the image is $(2.0) (4.0 cm) = 8.0$ cm tall, and inverted because it is real.

(c) They agree: both give a real, inverted image 30 cm in front of the mirror, twice the object's height. A hand-drawn diagram can miss by a centimeter or so where two rays cross at a shallow angle. The equation gives the exact distance; the diagram shows at a glance that the image is real and inverted.

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One correct ray diagram. Your rays should cross at the same point, 30 cm out and 8.0 cm below the axis.

Score your own answer. This is a self-check, not a grade. Tick each point your answer makes.

Draws at least two principal rays correctly: parallel to the axis then through F, to the center of the mirror and back at the same angle, or through F then parallel.
Locates the image where the reflected rays cross, about 30 cm in front of the mirror and 8 cm below the axis, and labels it real and inverted.
Uses the mirror equation with $f = + 10$ cm and $s_{o} = + 15$ cm to get $s_{i} = + 30$ cm.
Calculates $| M | = 2.0$ and an image height of 8.0 cm.
States that the diagram and the calculation agree on the position, size, type and orientation, and puts any small difference down to drawing.

Part 6 Check yourself A dentist's mirror

Both case questions use the ray diagram.

A dentist holds a small concave mirror with a focal length of 2.0 cm so that a tooth 0.50 cm tall is 1.5 cm in front of it. Treat the rays as close to the principal axis. The ray diagram shows the tooth as the short upright arrow between F and the mirror, three principal rays from its tip, and each reflected ray traced backward through the mirror as a dashed line.

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Case question 1 · Multiple choice

Using the ray diagram of the dentist's mirror: which describes the image of the tooth?

Case question 2 · Calculate

Using the dentist's mirror: the mirror has a focal length of 2.0 cm, and the tooth is 1.5 cm in front of it. Find the image distance $s_{i}$ , with its sign in this course's convention (positive for a real image in front of the mirror, negative for a virtual image behind it), and compare it with the object distance $s_{o}$ . What is $\frac{s_{i}}{s_{o}}$ ?

Your answer

The rest stand on their own and do not use the case.

Check yourself · Decide and justify

A student walks straight toward a flat mirror at 1.0 m/s. She says that the distance between her and her image shrinks at 2.0 m/s. Is she right?

Which reason supports your answer?

Check yourself · Spot the error

A student locates the image formed by a convex mirror. Tap the first line that goes wrong.

Part 7 Keep it Bring back earlier ideas

Every principal ray obeys the law of reflection from the last lesson, and the magnetic field of a current from Unit 12 is still part of the course. Retrieving them now, between new ideas, is what makes them easy to find later.

From 13.1 Reflection · Calculate

A laser beam strikes a flat mirror. The beam makes an angle of 25° with the surface of the mirror. What is the angle of reflection?

Your answer °

Predict first: if you move twice as far from a long, straight wire, does its field become half as strong, a quarter as strong, or stay the same? Then sort the cards, each of which changes one thing about a wire or the point beside it.

From 12.3 Magnetism and Current-Carrying Wires · Sort it

Each card changes one thing and keeps everything else the same. Tap a card, then what happens to the size of the quantity it names.

It doubles

It halves

It stays the same

In three lines

A concave mirror converges parallel rays to a focal point in front of it, a convex mirror diverges them from one behind it, and a plane mirror's focal point is infinitely far away; for a spherical mirror $f \approx \frac{R}{2}$ .
Real images form where reflected rays meet; virtual images where they only seem to come from; three principal rays find either.
$\frac{1}{s_{i}} + \frac{1}{s_{o}} = \frac{1}{f}$ with $f$ negative for convex and $s_{i}$ negative for virtual images, and $| M | = | \frac{s_{i}}{s_{o}} |$ .

Exam strategy for this topic

Section I, Multiple Choice: 42 questions in 85 minutes.
Section II, Free Response: 4 questions in 95 minutes.
This lesson's Explain it is shaped like free-response question 2, Translation Between Representations.
A four-function, scientific, or graphing calculator is allowed on both sections of the exam.

When you draw a mirror ray diagram on paper, draw all three principal rays with a ruler even if two are enough. If the three do not cross at one point, one of them is drawn wrong, and you have caught it before the reader does.

Questions students ask

What is the focal point of a mirror?

For a concave mirror, the point in front of it where rays parallel to the principal axis meet after reflecting; for a convex mirror, the point behind it they seem to come from. For a spherical mirror it is about halfway to the center of curvature, $f \approx \frac{R}{2}$ .

What is the difference between a real and a virtual image?

A real image forms where reflected rays actually meet, and a screen there shows it. A virtual image forms where reflected rays only seem to come from, behind the mirror, and no screen can catch it.

What sign convention goes with the mirror equation?

College Board's framework requires a convention without printing one. This course uses real-is-positive: $f$ positive for concave and negative for convex, $s_{o}$ positive in front of the mirror, $s_{i}$ positive for a real image in front and negative for a virtual image behind.

How do you find the magnification of a mirror?

$| M | = | \frac{h_{i}}{h_{o}} | = | \frac{s_{i}}{s_{o}} |$ . It gives the size; for one mirror and a real object, a real image is inverted and a virtual image is upright. If you see $M = - \frac{s_{i}}{s_{o}}$ , a negative $M$ means inverted and a positive $M$ upright; this agrees with reading orientation from the image type.

Where is the image in a flat mirror?

Behind the mirror, as far behind it as the object is in front, upright and the same size. It is virtual: no light reaches the place it seems to be.

Beyond the AP Exam

College Board sets this limit on what the AP Exam asks about this topic. Anything past it is not on the exam.

AP Physics 2 limits the study of mirrors to plane mirrors, convex spherical mirrors, and concave spherical mirrors.

Course alignment, for teachers

AP Physics 2: Algebra-Based topic 13.2, Unit 13: Geometric Optics.

Skill 1.A: Create diagrams, tables, charts, or schematics to represent physical situations.
Skill 2.A: Derive a symbolic expression from known quantities by selecting and following a logical mathematical pathway.
Skill 2.C: Compare physical quantities between two or more scenarios or at different times and locations in a single scenario.
Skill 3.C: Justify or support a claim using evidence from experimental data, physical representations, or physical principles or laws.

This lesson maps all 13 essential knowledge statements of the topic to the part that teaches each one: 13 are checked by a scored question. The map is in the lesson file.