13.2 Images Formed by Mirrors

Mirrors form images by reflecting light, and the type of image depends on the mirror's shape and where you place the object. Use the mirror equation $\frac{1}{s_o} + \frac{1}{s_i} = \frac{1}{f}$ and magnification $M = -s_i/s_o$ with the right sign conventions, then use ray diagrams to determine whether an image is real or virtual, upright or inverted, and enlarged or reduced.

Why This Matters for the AP Physics 2 Exam

Geometric optics is a solid chunk of the AP Physics 2 exam, and mirrors show up in both calculation and reasoning questions. You need to do more than plug numbers into the mirror equation: you have to predict and explain how an image changes when you move an object, justify why an image is real or virtual, and connect ray diagrams to the math.

This topic builds directly on reflection from the previous topic and uses the same kind of thinking you will apply to lenses later in the unit. Because the experimental design free-response question can pull from any part of the course, being able to describe how you would collect and linearize data, for example measuring image distance versus object distance, is useful practice here too.

Key Takeaways

• Concave (converging) mirrors reflect parallel rays toward a focal point in front of the mirror; convex (diverging) mirrors reflect parallel rays so they appear to come from a focal point behind the mirror.
• For spherical mirrors, the focal length is about half the radius of curvature, $f = R/2$, in the paraxial (small-angle) approximation.
• Real images form when reflected rays actually meet; virtual images form when reflected rays only appear to meet behind the mirror.
• Use $\frac{1}{s_o} + \frac{1}{s_i} = \frac{1}{f}$ with sign conventions: $f$ is positive for concave and negative for convex mirrors, and a negative $s_i$ means a virtual image behind the mirror.
• Magnification $|M| = |h_i/h_o| = |s_i/s_o|$ gives size; the signed form $M = -s_i/s_o$ tells you orientation (positive upright, negative inverted).
• The three principal rays (parallel, focal, and center/vertex) let you locate an image and classify it without doing any calculation.

Focal Points and Mirror Types

Concave (Converging) Mirrors

Concave mirrors curve inward like the inside of a bowl.

• Parallel rays striking the mirror reflect and converge at the focal point.
• The focal point sits in front of the mirror, on the same side as the incoming light.
• The distance from the mirror to the focal point is the focal length $f$.

These are called converging mirrors because they bring light together. Telescopes, makeup mirrors, and headlight reflectors are everyday applications of this concentrating effect.

Convex (Diverging) Mirrors

Convex mirrors curve outward like the outside of a sphere.

• Parallel rays reflect and diverge as if they came from a point behind the mirror.
• That point is the virtual focal point.
• Light never actually reaches behind the mirror, so the focal point there is virtual.

Convex mirrors spread light out, which gives a wider field of view. Security mirrors and vehicle side mirrors are common applications.

Plane Mirrors

Plane mirrors are flat with no curvature.

• Parallel rays stay parallel after reflection.
• The angle of incidence equals the angle of reflection for each ray.
• There is no finite focal point; the focal point is effectively at infinity.

Because of this, plane mirrors produce images the same size as the object.

Spherical Mirror Focal Length

For both concave and convex spherical mirrors, the focal point lies roughly halfway between the mirror surface and the center of curvature:

• $f = R/2$, where $R$ is the radius of curvature.
• $f$ is positive for concave mirrors and negative for convex mirrors.
• This holds well for rays close to the principal axis (paraxial), but breaks down for rays far from the axis, which causes spherical aberration.

Real vs Virtual Images

Whether an image is real or virtual depends on what the reflected rays actually do.

Real images form when reflected rays truly converge at a point:

• Light physically passes through each point of the image.
• A real image can be projected onto a screen.
• Real images are typically inverted.
• Only concave mirrors form real images, and only when the object is beyond the focal point.

Virtual images form when reflected rays only appear to come from a point:

• Light does not pass through the image.
• A virtual image cannot be projected onto a screen.
• Virtual images are typically upright.
• Plane mirrors, convex mirrors, and concave mirrors with the object inside the focal point all form virtual images.

Image Location and the Mirror Equation

The mirror equation links object distance ($s_o$), image distance ($s_i$), and focal length ($f$):

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

This works for all mirror types once you apply the sign conventions:

• Object distance ($s_o$): positive for a real object in front of the mirror.
• Image distance ($s_i$): positive for a real image in front of the mirror; negative for a virtual image behind the mirror.
• Focal length ($f$): positive for a concave (converging) mirror; negative for a convex (diverging) mirror.
• Plane mirrors: the focal point is at infinity. Under the same convention, an object in front has positive object distance and its virtual image behind the mirror has a negative image distance of equal magnitude, so $s_i = -s_o$. The image appears exactly as far behind the mirror as the object is in front.

Image Magnification

Magnification compares the size of the image to the size of the object. The required size relationship is:

$|M| = \left|\frac{h_i}{h_o}\right| = \left|\frac{s_i}{s_o}\right|$

Using the signed convention:

$M = \frac{h_i}{h_o} = -\frac{s_i}{s_o}$

so a positive magnification means an upright image and a negative magnification means an inverted image.

Where:

• $h_i$ is the image height
• $h_o$ is the object height
• $s_i$ is the image distance
• $s_o$ is the object distance

The formula gives you two things at once:

• $|M|$ tells you the size ratio between image and object.
• The sign of $M$ tells you orientation: positive upright, negative inverted.

For plane mirrors, $M = 1$, so the image is the same size and upright.

Ray Diagrams for Mirrors

Ray diagrams let you find and classify an image without heavy algebra. The three principal rays are:

1. Parallel ray: a ray parallel to the principal axis reflects through the focal point for a concave mirror, or reflects as if it came from the focal point for a convex mirror.
2. Vertex (central) ray: a ray aimed at the point where the principal axis meets the mirror reflects with equal angles on either side of the principal axis.
3. Focal ray: a ray passing through the focal point of a concave mirror, or aimed toward the focal point behind a convex mirror, reflects parallel to the principal axis.

The intersection of any two rays locates the image. For virtual images, extend the reflected rays backward to find where they appear to meet.

Ray diagrams tell you:

• Whether the image is real or virtual
• Whether the image is upright or inverted
• The relative size of the image
• The location of the image

Image Outcomes by Object Position

For a concave mirror, the image depends on where the object sits relative to the focal point ($F$) and the center of curvature ($C$):

1. Object beyond C: image is real, inverted, reduced, and forms between $C$ and $F$.
2. Object at C: image is real, inverted, same size, and forms at $C$.
3. Object between C and F: image is real, inverted, enlarged, and forms beyond $C$.
4. Object at F: reflected rays are parallel, so the image forms effectively at infinity (no image at a finite location).
5. Object between F and the mirror: image is virtual, upright, enlarged, and forms behind the mirror.

For convex mirrors, the image is always virtual, upright, and reduced, located behind the mirror, no matter where the object is.

For plane mirrors, the image is always virtual, upright, and the same size as the object.

How to Use This on the AP Physics 2 Exam

Problem Solving

• Write the mirror equation, plug in $s_o$ and $f$ with correct signs, then solve for $1/s_i$ before flipping to get $s_i$. Flipping too early is a common arithmetic slip.
• Check the sign of your answer: positive $s_i$ means a real image in front of the mirror, negative $s_i$ means a virtual image behind it.
• Use $M = -s_i/s_o$ to get both size and orientation, then find image height with $h_i = M \cdot h_o$.

Free Response

• When asked to explain rather than calculate, justify your image classification using ray behavior (do the reflected rays actually meet, or only appear to?).
• If a question asks how the image changes as you slide an object closer to the mirror, reason through the object-position cases instead of guessing.
• For an experimental design prompt, you can describe measuring image distance for several object distances and linearizing $\frac{1}{s_i}$ versus $\frac{1}{s_o}$ to find the focal length from the intercept.

Common Trap

• Mixing up which mirror gives real images. Only a concave mirror produces a real image, and only when the object is beyond the focal point.
• Forgetting that convex mirrors always give a reduced, upright, virtual image regardless of object position.

Practice Problem 1: Concave Mirror Image Formation

Solution

Use the mirror equation to find the image distance:

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

$\frac{1}{30 \text{ cm}} + \frac{1}{s_i} = \frac{1}{20 \text{ cm}}$

$\frac{1}{s_i} = \frac{1}{20 \text{ cm}} - \frac{1}{30 \text{ cm}} = \frac{3-2}{60 \text{ cm}} = \frac{1}{60 \text{ cm}}$

$s_i = 60 \text{ cm}$

The positive $s_i$ means the image is real and 60 cm in front of the mirror.

Now find the magnification:

$M = -\frac{s_i}{s_o} = -\frac{60 \text{ cm}}{30 \text{ cm}} = -2$

Then the image height from $h_i = M \cdot h_o$:

$h_i = (-2)(15 \text{ cm}) = -30 \text{ cm}$

The magnitude 30 cm means the image is twice as tall as the object, and the negative sign means it is inverted.

So the image is:

• Located 60 cm in front of the mirror
• Height $-30$ cm (30 cm tall, inverted)
• Real (can be projected on a screen)

Practice Problem 2: Plane Mirror Image Location

Solution

For a plane mirror, the image distance equals the object distance on the opposite side. The total distance between the person and the image is twice the distance from the person to the mirror.

Initial situation:

• Person is 2 m from the mirror
• Image is 2 m behind the mirror
• Total distance between person and image = 4 m

Final situation:

• Total distance between person and image = 4 m - 1.5 m = 2.5 m
• If the final person-to-mirror distance is $x$, then:
  • $2x = 2.5$ m
  • $x = 1.25$ m

The person needs to walk $2 \text{ m} - 1.25 \text{ m} = 0.75$ meters toward the mirror.

Common Misconceptions

• "Virtual images are not visible." You can see a virtual image just fine (think of yourself in a flat mirror); you just cannot project it onto a screen because no light actually passes through it.
• "All mirrors can magnify." Plane mirrors always give a same-size image, and convex mirrors always reduce. Only concave mirrors can enlarge an image.
• "A negative magnification means the image is smaller." The sign of $M$ tells you orientation, not size. A negative $M$ means inverted; its magnitude tells you the size ratio.
• "The focal point of a convex mirror is in front of it." For a convex mirror, the focal point is behind the mirror and is virtual, which is why $f$ is negative.
• "$f = R/2$ always holds exactly." This is the paraxial approximation. It works well for rays near the principal axis but loses accuracy for wide mirrors or rays far from the axis, which leads to spherical aberration.
• "A positive image distance means the image is behind the mirror." A positive $s_i$ is a real image in front of the mirror; a negative $s_i$ is the virtual image behind it.

Related AP Physics 2 Guides

• 13.1 Reflection
• 13.4 Images Formed by Lenses
• 13.3 Refraction