# Images

> AP Physics 2 · Geometric and Physical Optics, Unit 6
> Source: https://www.owlsprep.com/study/ap-physics-2-u6-images/

This guide covers image formation by plane/spherical mirrors and thin lenses, including the mirror/thin lens equations, magnification, ray tracing, and sign conventions aligned to the AP Physics 2 CED.

**Prerequisites:** Law of reflection; Snell's law of refraction; Basic curved surface geometry

## Learning objectives

- Distinguish between real and virtual images
- Apply the mirror and thin lens equations with correct AP sign conventions
- Perform ray tracing for mirrors and lenses
- Calculate magnification and describe image properties
- Identify and avoid common exam pitfalls

## Core Concepts of Image Formation

An image forms when light rays emitted or reflected from an object either converge to a single point (real image) or appear to diverge from a single point (virtual image).

**Real Image** — An image formed where light rays actually converge at the image location. Real images can be projected onto a screen.

*Example:* The image projected onto a movie theater screen

**Virtual Image** — An image formed where light rays only appear to diverge from the image location; no light actually reaches the image position. Virtual images cannot be projected onto a screen.

*Example:* The image you see of yourself in a plane mirror

Per the AP Physics 2 CED, Unit 6 (Geometric and Physical Optics) makes up 20-25% of the total exam score, and this subtopic accounts for roughly 7-10% of total exam weight, appearing in both MCQ and FRQ sections. This guide uses the standard Cartesian (real-is-positive) sign convention adopted by College Board for all AP Physics 2 problems.

## Image Formation by Mirrors

Mirrors form images via reflection, and are categorized as plane (flat) or spherical (curved). For plane mirrors, the object distance $d_o$ (always positive for real objects, the distance from the object to the mirror) relates to image distance $d_i$ (distance from the mirror to the image) as:

$$d_i = -d_o, \quad m = +1$$

Magnification $m$, the ratio of image height to object height, is $m=+1$ for plane mirrors, meaning the image is upright, same size as the object, and always virtual. For spherical mirrors with constant radius of curvature $R$, the focal length (distance from the mirror surface to the focal point, where parallel rays converge) is $f = R/2$. The mirror equation relating $f$, $d_o$, and $d_i$ is:

$$\frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i}$$

Sign convention for mirrors: $f$ is positive for concave (converging, curved inward toward the object) mirrors and negative for convex (diverging, curved outward away from the object) mirrors. A positive $d_i$ means a real image in front of the mirror; negative $d_i$ means a virtual image behind the mirror. Magnification for all mirrors follows:

$$m = -\frac{d_i}{d_o}$$

Positive $m$ indicates an upright image, negative $m$ indicates an inverted image, and $|m|$ gives the size ratio relative to the object.

**Worked example:** A 3 cm tall object is placed 12 cm in front of a convex (diverging) spherical mirror with radius of curvature 16 cm. Find the image position, height, and describe its properties.

1. Calculate the focal length. For a convex mirror, $f$ is negative, and we know $d_o = +12$ cm:

   $$f = \frac{R}{2} = \frac{16}{2} = 8 \text{ cm} \implies f = -8 \text{ cm}$$
2. Rearrange the mirror equation to solve for $d_i$:

   $$\frac{1}{d_i} = \frac{1}{f} - \frac{1}{d_o} = \frac{1}{-8} - \frac{1}{12} = \frac{-3 - 2}{24} = -\frac{5}{24} \implies d_i = -4.8 \text{ cm}$$
3. Calculate magnification:

   $$m = -\frac{d_i}{d_o} = -\frac{(-4.8)}{12} = +0.4$$
4. Calculate image height:

   $$h' = m \cdot h = 0.4 \cdot 3 = 1.2 \text{ cm}$$
5. Final description: The image is 4.8 cm behind the mirror (virtual), upright, and 1.2 cm tall (diminished).

> **Exam tip:** Always confirm the sign of $f$ before starting any calculation. For mirrors, curved inward = concave (positive f), curved outward = convex (negative f) — this never changes for real objects.

## Image Formation by Thin Lenses

Thin lenses form images via refraction, and follow nearly the same mathematics as mirrors, with only a small change to the sign interpretation of $d_i$. Lenses are divided into converging (convex, thicker at the center), which have positive $f$, and diverging (concave, thinner at the center), which have negative $f$, matching the sign convention for mirrors. The thin lens equation is identical in form to the mirror equation:

$$\frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i}$$

Magnification is also identical to that for mirrors:

$$m = -\frac{d_i}{d_o}$$

The key difference from mirrors is the interpretation of $d_i$: for lenses, a positive $d_i$ means the image is on the opposite side of the lens from the object (the side where light exits after refraction), which is a real image. A negative $d_i$ means the image is on the same side as the object, which is virtual. Converging lenses can form both real and virtual images depending on whether the object is outside or inside the focal length, while diverging lenses always form virtual, upright, diminished images for any real object.

**Worked example:** A magnifying glass is a converging lens with focal length 6 cm. If you hold the magnifying glass 4 cm from a flower to examine it, find the image position and describe the image.

1. A converging lens has positive focal length, so $f = +6$ cm, and $d_o = +4$ cm.
2. Solve for $d_i$:

   $$\frac{1}{d_i} = \frac{1}{f} - \frac{1}{d_o} = \frac{1}{6} - \frac{1}{4} = \frac{2 - 3}{12} = -\frac{1}{12} \implies d_i = -12 \text{ cm}$$
3. Calculate magnification:

   $$m = -\frac{d_i}{d_o} = -\frac{(-12)}{4} = +3$$
4. Interpretation: Negative $d_i$ means the image is on the same side of the lens as the flower, so it is virtual. Positive magnification means it is upright, and $m=3$ means it is 3 times larger than the actual flower, which matches the function of a magnifying glass.

> **Exam tip:** When asked to describe an image, AP graders require all three properties to earn full credit: 1) real or virtual, 2) upright or inverted, 3) magnified/diminished/ same size. Never leave out any of these three.

## Ray Tracing for Images

Ray tracing is a graphical method to locate images and confirm their properties, and is frequently required on AP Physics 2 FRQs. The method relies on three principal rays that follow simple rules; their intersection (or the intersection of their extensions) marks the top of the image. For both mirrors and lenses, the three principal rays are:

1. **Parallel ray**: Leaves the top of the object traveling parallel to the principal axis. After reflection/refraction, it passes through (or appears to diverge from) the focal point.
2. **Focal ray**: Leaves the top of the object passing through (or heading toward) the focal point, and exits parallel to the principal axis after reflection/refraction.
3. **Center ray**: Leaves the top of the object heading for the center of the mirror/lens, and travels straight through without changing direction.

If the actual rays intersect, the image is real; if only the extensions of the rays intersect, the image is virtual. Ray tracing can be used to check calculations from the lens/mirror equation, or to answer conceptual questions about image properties.

**Worked example:** Use ray tracing to locate the image of an object placed 10 cm in front of a diverging lens with focal length magnitude 5 cm, then confirm with the thin lens equation.

1. Draw the principal axis, mark the lens center, draw two focal points 5 cm on either side of the lens, and draw the upright object arrow 10 cm left of the lens.
2. Draw the three principal rays: (1) Parallel ray from the top of the object travels parallel to the axis, refracts so that it appears to diverge from the left (near-side) focal point. (2) Focal ray heading toward the right (far-side) focal point refracts parallel to the axis. (3) Center ray travels straight through the center of the lens without bending.
3. Extend all refracted rays back to the left of the lens. They intersect between the object and the lens, inside the focal length, at the top of the virtual image, which is upright and diminished.
4. Confirm with calculation: $f = -5$ cm, $d_o = 10$ cm:

   $$\frac{1}{d_i} = \frac{1}{-5} - \frac{1}{10} = -\frac{3}{10} \implies d_i = -3.33 \text{ cm}, \quad m = -\frac{(-3.33)}{10} = +0.33$$
5. This matches the ray tracing result of a virtual, upright, diminished image 3.3 cm left of the lens.

> **Exam tip:** Always draw arrows on your rays pointing in the direction of light travel. AP graders consistently deduct points for missing or reversed arrows on ray diagrams.

## AP Style Worked Practice Problems

**Worked example:** A student holds a plane mirror 1 m in front of their face, and looks at the image of a painting on the wall 3 m behind their head. How far behind the student's face does the image of the painting appear to be?
A) 4 m B) 5 m C) 7 m D) 8 m

1. For plane mirrors, the image distance equals the object distance from the mirror. The painting is 3 m behind the student's face, which is 1 m in front of the mirror, so the object distance of the painting from the mirror is:

   $$d_o = 1 \text{ m} + 3 \text{ m} = 4 \text{ m}$$
2. The image forms 4 m behind the mirror. The distance from the student's face to the image is the sum of the distance from the student's face to the mirror and the distance from the mirror to the image:

   $$1 \text{ m} + 4 \text{ m} = 5 \text{ m}$$
3. The correct answer is B.

**Worked example:** A small movie projector uses a single converging lens to project an image onto a screen 10 m away from the lens. The film (the object) is 5 cm wide, and is placed 12 cm from the lens.
(a) Calculate the focal length of the projector lens.
(b) Find the width of the image projected on the screen, and state if the image is upright or inverted, real or virtual.
(c) If the projector is moved closer to the screen, decreasing the image distance, how must the distance between the lens and the film change to keep the image in focus? Justify your answer.

1. (a) We know $d_o = 12$ cm, $d_i = 10$ m $= 1000$ cm. The image is projected onto a screen, so it is real, so $d_i$ is positive. Rearrange the thin lens equation:

   $$\frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i} = \frac{1}{12} + \frac{1}{1000} \approx 0.0843 \text{ cm}^{-1} \implies f \approx 11.9 \text{ cm}$$
2. (b) Calculate magnification and image width:

   $$m = -\frac{d_i}{d_o} = -\frac{1000}{12} \approx -83.3, \quad w' = m \cdot w = -83.3 \cdot 5 \text{ cm} \approx -4.2 \text{ m}$$
3. Negative magnification means the image is inverted, positive $d_i$ means it is real. The image is 4.2 m wide, inverted, and real.
4. (c) From the rearranged thin lens equation: $\frac{1}{d_o} = \frac{1}{f} - \frac{1}{d_i}$. If $d_i$ decreases, $\frac{1}{d_i}$ increases, so $\frac{1}{f} - \frac{1}{d_i}$ decreases, meaning $d_o$ increases. The distance between the lens and the film must increase to keep the image in focus.

**Worked example:** An ophthalmologist measures the radius of curvature of a patient's cornea, which acts as a converging mirror when reflecting light from a test source. A 10 mm tall test object is placed 200 mm in front of the cornea, and the real image of the test object is measured to be 0.21 mm tall. Calculate the radius of curvature of the patient's cornea, and state if it is steeper or flatter than the typical value of 8 mm.

1. Magnification is the ratio of image height to object height. Real images are inverted, so $h'$ is negative:

   $$m = \frac{h'}{h} = \frac{-0.21 \text{ mm}}{10 \text{ mm}} = -0.021$$
2. Solve for $d_i$ using the magnification formula:

   $$m = -\frac{d_i}{d_o} \implies -0.021 = -\frac{d_i}{200 \text{ mm}} \implies d_i = 4.2 \text{ mm}$$
3. Use the mirror equation to find $f$:

   $$\frac{1}{f} = \frac{1}{200 \text{ mm}} + \frac{1}{4.2 \text{ mm}} \approx 0.2431 \text{ mm}^{-1} \implies f \approx 4.11 \text{ mm}$$
4. Radius of curvature is $R = 2f$, so:

   $$R = 2(4.11 \text{ mm}) \approx 8.22 \text{ mm}$$
5. This patient's cornea has a radius of ~8.2 mm, which is slightly larger than the typical 8 mm, so it is slightly flatter than an average cornea.

## Common pitfalls

- **Wrong:** Assigning a negative focal length to a converging lens or concave mirror
  - Why it fails: Students mix up the sign rules for converging/diverging elements across lenses and mirrors
  - Correct: In the AP convention, all converging elements (concave mirrors, convex lenses) have positive f, all diverging elements (convex mirrors, concave lenses) have negative f, this holds for both mirrors and lenses.
- **Wrong:** Claiming a negative $d_i$ for a lens means the image is on the opposite side from the object
  - Why it fails: Students carry over the mirror sign interpretation to lenses
  - Correct: For mirrors: positive $d_i$ = in front (same side as object, real), negative $d_i$ = behind (opposite side, virtual). For lenses: positive $d_i$ = behind (opposite side from object, real), negative $d_i$ = in front (same side, virtual). Write this down at the start of any FRQ if unsure.
- **Wrong:** Forgetting the negative sign in magnification, so calling an inverted image upright
  - Why it fails: Students only remember the magnitude of magnification and drop the negative sign during calculation
  - Correct: Always carry the sign of $d_i$ through the magnification calculation. A quick check: all real images from a single mirror or lens are inverted, so $m$ should be negative.
- **Wrong:** Drawing a parallel ray for a diverging lens that actually passes through the far-side focal point
  - Why it fails: Students assume all parallel rays pass through a focal point, regardless of lens type
  - Correct: For diverging elements, the parallel ray only appears to come from the near-side focal point. Draw solid lines for actual light paths, dashed lines for extensions to virtual image points.
- **Wrong:** Using $R = f/2$ instead of $f = R/2$ for spherical mirrors
  - Why it fails: Students rush and reverse the relationship between focal length and radius of curvature
  - Correct: The focal point is halfway between the mirror and the center of curvature, so $f$ is always half of $R$.

## Cheatsheet

| Category | Formula / Rule | Notes |
| --- | --- | --- |
| Plane Mirror | $d_i = -d_o$, $m = +1$ | Always virtual, upright, same-size |
| Spherical Mirror Focal Length | $f = \frac{R}{2}$ | $f>0$ = concave (converging), $f<0$ = convex (diverging) |
| Mirror Equation | $\frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i}$ | $d_o>0$ for real objects; $d_i>0$ = real in front, $d_i<0$ = virtual behind |
| Thin Lens Focal Length | $f>0$ = convex (converging), $f<0$ = concave (diverging) | Same f sign convention as mirrors |
| Thin Lens Equation | $\frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i}$ | $d_i>0$ = real opposite object side, $d_i<0$ = virtual same side |
| Magnification (all elements) | $m = -\frac{d_i}{d_o}$, $h' = m h$ | $m>0$ = upright, $m<0$ = inverted; $\|m\|>1$ = magnified |
| Parallel Ray Rule | Parallel to principal axis → passes through/appears from focal point | First ray to draw for any diagram |
| Center Ray Rule | Travels straight through center of lens / curvature of mirror | Second ray to confirm image position |

## What's next

Mastery of single-element image formation is the foundational prerequisite for all advanced optical systems, including compound lenses, optical instruments like telescopes and microscopes, and further topics in wave optics such as interference and diffraction. Understanding image properties and sign conventions for geometric optics also prepares you to solve conceptual problems about the human eye and vision correction, which are common AP Physics 2 exam topics. Building a solid foundation in image formation will help you tackle more complex optical problems with confidence on exam day.

- [Unit 6 Overview](https://www.owlsprep.com/study/ap-physics-2-u6-overview/)
- [Interference and Diffraction](https://www.owlsprep.com/study/ap-physics-2-u6-interference-and-diffraction/)
- [Quantum, Atomic, and Nuclear Physics Overview](https://www.owlsprep.com/study/ap-physics-2-u7-overview/)

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