Unit 13 · Topic 13.2

13.2 Images Formed by Mirrors

Curved mirrors bend reflected rays toward or away from a focal point, so they form images that can be bigger, smaller, upside down or behind the mirror. You can find an image two ways: draw the principal rays, or use $\frac{1}{s_i} + \frac{1}{s_o} = \frac{1}{f}$ with the sign rules. Then describe the image as real or virtual, upright or inverted, and enlarged, reduced or the same size.

Key terms

concave mirror
convex mirror
focal point
real image
virtual image
magnification

Three kinds of mirror

A concave mirror curves inward, like the inside of a spoon. It is converging: rays that come in parallel to the principal axis (the line through the center of the mirror, perpendicular to it) all reflect through one point in front of the mirror, the focal point F.

A convex mirror bulges outward, like the back of a spoon or a store security mirror. It is diverging: parallel rays spread out after reflecting, as if they came from a focal point behind the mirror.

For a spherical mirror, the focal point sits on the axis about halfway between the mirror and its center of curvature C, the center of the sphere the mirror was cut from. So the focal length is $f = \frac{R}{2}$ , where R is the radius of curvature.

A plane (flat) mirror has its focal point infinitely far away. Its image is always as far behind the mirror as the object is in front, upright and the same size.

Real and virtual images

A real image forms where reflected rays actually meet. You can catch it on a screen or a sheet of paper. Only a concave mirror can make a real image of a real object, and the image is in front of the mirror.

A virtual image forms where reflected rays only seem to come from, when you trace them backward. The rays never actually pass through it, so a screen placed there shows nothing. Virtual images from a mirror are behind it.

Principal rays

To locate an image, draw at least two of these three rays from the top of the object. Where the reflected rays cross (or where their dashed extensions cross) is the top of the image.

A ray parallel to the axis reflects through F (concave), or reflects as if it came from F behind the mirror (convex).
A ray that hits the center of the mirror, where the axis meets it, reflects at an equal angle on the other side of the axis, like a flat mirror.
A ray headed through F (concave) or aimed toward F behind the mirror (convex) reflects parallel to the axis.

The mirror equation and sign rules

$\frac{1}{s_i} + \frac{1}{s_o} = \frac{1}{f}$

Here $s_o$ is the object distance and $s_i$ is the image distance, both measured from the mirror. The sign rules: $s_o$ is positive for a real object in front. f is positive for a concave mirror and negative for a convex one. A positive $s_i$ means a real image in front of the mirror; a negative $s_i$ means a virtual image behind it.

The size of the image comes from the magnification, $\lvert M \rvert = \left\lvert \frac{h_i}{h_o} \right\rvert = \left\lvert \frac{s_i}{s_o} \right\rvert$ . If $\lvert M \rvert > 1$ the image is enlarged; less than 1, reduced. For a single mirror, a real image is always inverted and a virtual image is always upright. (Some textbooks write $M = -\frac{s_i}{s_o}$ , where a negative M means inverted. It gives the same answers.)

Where the image ends up

Mirror and object position	Image type	Orientation	Size
Concave, object beyond C	Real, between F and C	Inverted	Reduced
Concave, object at C	Real, at C	Inverted	Same size
Concave, object between C and F	Real, beyond C	Inverted	Enlarged
Concave, object at F	No image (rays leave parallel)	None	None
Concave, object inside F	Virtual, behind mirror	Upright	Enlarged
Convex, any position	Virtual, behind mirror	Upright	Reduced
Plane, any position	Virtual, same distance behind	Upright	Same size

Worked examples

Try each one yourself first, then open the solution.

Example 1Calculator allowed
Real image from a concave mirror
A 2.0 cm tall candle stands 30 cm in front of a concave mirror with a focal length of 10 cm. Find the image location, its height and its type.
Show the solutionHide the solution
1. Step 1: Use the mirror equation with f = +10 cm (concave) and $s_o$ = +30 cm: $\dfrac{1}{s_i} = \dfrac{1}{10} - \dfrac{1}{30} = \dfrac{3 - 1}{30} = \dfrac{2}{30}$ , so $s_i$ = +15 cm.
2. Step 2: Positive $s_i$ means the image is real and in front of the mirror, so it's inverted.
3. Step 3: $\lvert M \rvert = \dfrac{15}{30} = 0.50$ , so $h_i$ = 0.50 × 2.0 cm = 1.0 cm.
4. Step 4: Check with the table: the object is beyond C (at 2f = 20 cm), so the image should be real, inverted, reduced and between F and C. It is.
Answer: Real, inverted image 15 cm in front of the mirror, 1.0 cm tall.
Example 2Calculator allowed
Object inside the focal point
You hold your face 6.0 cm from a concave makeup mirror with f = 10 cm. Where is the image, and how is it magnified?
Show the solutionHide the solution
1. Step 1: $\dfrac{1}{s_i} = \dfrac{1}{10} - \dfrac{1}{6.0} = \dfrac{3 - 5}{30} = -\dfrac{2}{30}$ , so $s_i$ = −15 cm.
2. Step 2: Negative $s_i$ means a virtual image 15 cm behind the mirror, so it's upright.
3. Step 3: $\lvert M \rvert = \dfrac{15}{6.0} = 2.5$ . Your face looks upright and 2.5 times larger, which is exactly what a makeup mirror is for.
Answer: Virtual, upright image 15 cm behind the mirror, magnified 2.5 times.
Example 3Calculator allowed
Convex mirror (sign trap)
A convex security mirror has a focal length of magnitude 20 cm. A shopper is 30 cm in front of it. Find the image distance and magnification.
Show the solutionHide the solution
1. Step 1: A convex mirror is diverging, so f is negative: f = −20 cm. Forgetting this sign is the classic mistake.
2. Step 2: $\dfrac{1}{s_i} = -\dfrac{1}{20} - \dfrac{1}{30} = -\dfrac{3 + 2}{60} = -\dfrac{5}{60}$ , so $s_i$ = −12 cm.
3. Step 3: The image is virtual, 12 cm behind the mirror, and upright, with $\lvert M \rvert = \dfrac{12}{30} = 0.40$ .
4. Step 4: A reduced image is what lets a convex mirror show a wide view of a whole store. If you had used f = +20 cm you'd get $s_i$ = +60 cm, a real image, which a convex mirror can never make.
Answer: Virtual, upright image 12 cm behind the mirror, 0.40 times the size.

Common mistakes

Using a positive focal length for a convex mirror. Diverging mirrors (and lenses) always have a negative f.
Saying a plane mirror's image is on the mirror's surface. It's as far behind the mirror as the object is in front.
Calling an image real just because you can see it. A real image is where rays actually meet and can be projected onto a screen; you can see virtual images too.
Drawing a ray through F that then reflects through F again. A ray through F reflects parallel to the axis, and a parallel ray reflects through F.

On the exam

Ray diagrams earn points only when they're precise: use a ruler, draw at least two principal rays with arrowheads, dash the extensions behind the mirror, and mark where the image forms.
Many questions ask how the image changes as an object moves toward the mirror. Use the table pattern: for a concave mirror the real image grows and moves away until the object reaches F, then the image becomes virtual and upright.

Connected topics

Videos

Converging (Concave) Mirrors Explained | AP Physics 2 - Unit 13 - Lesson 2
Allen Tsao The STEM CoachWatch on YouTube (opens in a new tab)
Ray Diagrams - Mirrors
Bozeman ScienceWatch on YouTube (opens in a new tab)
How to Draw Ray Diagrams for Concave and Convex Mirrors
The Physics UniverseWatch on YouTube (opens in a new tab)
Diverging (Convex) Mirrors Made Easy | AP Physics 2 - Unit 13 - Lesson 3
Allen Tsao The STEM CoachWatch on YouTube (opens in a new tab)
AP Physics 2 - Mirrors
Dan Fullerton (APlusPhysics)Watch on YouTube (opens in a new tab)
Spherical Mirrors & The Mirror Equation - Geometric Optics
The Organic Chemistry TutorWatch on YouTube (opens in a new tab)

Check yourself

4 questions on 13.2 Images Formed by Mirrors. Pick an answer to see if you got it, and why.

Question 1 of 4Calculator allowed

A person stands 2.0 m in front of a large plane mirror and walks straight toward it at 0.50 m/s. At what speed does the person's image approach the person?

Question 2 of 4Calculator allowed

A 1.5 m tall student stands 3.0 m in front of a vertical plane mirror. Which correctly describes the student's image?

Question 3 of 4Calculator allowed

An object is placed 30 cm in front of a concave mirror with a focal length of 10 cm. Where does the image form?

Question 4 of 4Calculator allowed

An object is placed 30 cm in front of a concave mirror whose focal length is 10 cm. The image forms 15 cm in front of the mirror. Which describes the image?

0 of 4 answered