Images — AP Physics 2 Study Guide

For: AP Physics 2 candidates sitting AP Physics 2.

Covers: Plane and spherical mirrors, thin converging/diverging lenses, the mirror/thin lens equation, magnification, real vs virtual images, ray tracing, and Cartesian sign conventions consistent with the AP Physics 2 CED.

You should already know: Law of reflection, Snell's law of refraction, basic geometry of curved surfaces.

A note on the practice questions: All worked questions in the "Practice Questions" section below are original problems written by us in the AP Physics 2 style for educational use. They are not reproductions of past College Board / Cambridge / IB papers and may differ in wording, numerical values, or context. Use them to practise the technique; cross-check with official mark schemes for grading conventions.

1. What Is Images?

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 images can be projected onto a screen, while virtual images can only be observed by looking through the optical element, as light never actually reaches the image location. Per the AP Physics 2 Course and Exam Description (CED), Unit 6 (Geometric and Physical Optics) makes up 20-25% of the total exam score, and the Images subtopic accounts for roughly 7-10% of the total exam weight. Images questions appear in both multiple choice (MCQ) and free response (FRQ) sections, often combined with ray drawing, calculation, and conceptual reasoning. This guide uses the standard Cartesian (real-is-positive) sign convention adopted by College Board for all AP Physics 2 problems, so all rules and examples align with exam expectations.

2. Image Formation by Mirrors

Mirrors form images via reflection; we divide mirrors into plane (flat) and spherical (curved) types. For plane mirrors, the relationship between object distance $d_o$ (distance from object to mirror, always positive for real objects) and image distance $d_i$ (distance from mirror to image) is simple: $d_i=-d_o$, meaning the image is the same distance behind the mirror as the object is in front. Magnification (ratio of image height to object height) for plane mirrors is $m=+1$, so the image is upright, same size as the object, and virtual.

For spherical mirrors, which have a constant radius of curvature $R$, the focal length (distance from mirror surface to focal point, where parallel rays converge) is $f=R/2$. The mirror equation, which relates $f$, $d_o$, and $d_i$, is: $$\frac{1}{f}=\frac{1}{d_o}+\frac{1}{d_i}$$ Sign convention: $f$ is positive for concave (converging) mirrors (curved inward toward the object) and negative for convex (diverging) mirrors (curved outward away from the object). A positive $d_i$ means the image is real, located in front of the mirror; negative $d_i$ means virtual, located behind. Magnification follows the same rule for all mirrors: $m=-\frac{d_i}{d_o}$, where positive $m$ means upright image, negative $m$ means inverted, and $|m|$ is the size ratio.

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 $f$: $f=R/2=16/2=8$ cm; convex mirrors have negative $f$, so $f=-8$ cm. We know $d_o=+12$ cm.
2. Rearrange mirror equation to solve for $d_i$: $\frac{1}{d_i}=\frac{1}{f}-\frac{1}{d_o}=\frac{1}{-8}-\frac{1}{12}=\frac{-3-2}{24}=-\frac{5}{24}$, so $d_i=-4.8$ cm.
3. Calculate magnification: $m=-\frac{d_i}{d_o}=-\frac{(-4.8)}{12}=+0.4$.
4. Calculate image height: $h^{\prime }=m\cdot h=0.4\cdot 3=1.2$ cm.
5. Description: The image is 4.8 cm behind the mirror, virtual, upright, and 1.2 cm tall (diminished relative to the object).

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.

3. Image Formation by Thin Lenses

Thin lenses form images via refraction, and follow nearly the same math 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: $m=-\frac{d_i}{d_o}$, with the same sign rules for orientation. 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. Converging lens means $f=+6$ cm, object distance $d_o=+4$ cm.
2. Solve for $d_i$: $\frac{1}{d_i}=\frac{1}{f}-\frac{1}{d_o}=\frac{1}{6}-\frac{1}{4}=\frac{2-3}{12}=-\frac{1}{12}$, so $d_i=-12$ 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.

4. 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, whose intersection (or 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 (for mirrors, this means traveling through the center of curvature, reflecting back along the original path).

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, and draw two focal points 5 cm on either side of the lens. Draw the upright object arrow 10 cm left of the lens.
2. Draw the three principal rays:
   • 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.
   • Focal ray heading toward the right (far-side) focal point: refracts parallel to the axis.
   • 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. $\frac{1}{d_i}=\frac{1}{-5}-\frac{1}{10}=-\frac{3}{10}$, so $d_i=-3.33$ cm. $m=-\frac{(-3.33)}{10}=+0.33$, which 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.

5. Common Pitfalls (and how to avoid them)

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

6. Practice Questions (AP Physics 2 Style)

Question 1 (Multiple Choice)

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 (so 4 m behind the mirror). How far behind the student's face does the image of the painting appear to be? A) 3 m B) 4 m C) 6 m D) 7 m

Worked Solution: For a plane mirror, the image is $d_o$ behind the mirror, where $d_o$ is the distance of the object in front of the mirror. The painting is $1\text{m}+3\text{m}=4\text{m}$ in front of the mirror, so the image is 4 m behind the mirror. The student's face is 1 m in front of the mirror, so the distance from the student's face to the image is $1\text{m}+4\text{m}=5$? Wait no, wait adjust question: Wait the question says the painting is 3 m behind the student's head, so distance from painting to mirror is 1 m (student to mirror) + 3 m (student to painting) = 4 m, image is 4 m behind mirror, so distance from student's face to image is 4 m (image to mirror) + 1 m (mirror to student) = 5? No, I messed up options. Let's correct the question: 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 (so 4 m in front of the mirror). 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. Okay, solution: The painting is 1 + 3 = 4 m in front of the mirror, so image is 4 m behind the mirror. The student's face is 1 m in front of the mirror, so total distance from student face to image is 1 m (face to mirror) + 4 m (mirror to image) = 5 m. Correct answer is B. Yep, that works. Wait let's write the solution correctly: Worked Solution: 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 $d_o=1\text{m}+3\text{m}=4\text{m}$. The image forms $d_i=-4\text{m}$, meaning 4 m behind the mirror. The distance from the student's face to the image is the distance from the student's face to the mirror plus the distance from the mirror to the image: $1\text{m}+4\text{m}=5\text{m}$. The correct answer is B.

Question 2 (Free Response)

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.

Worked Solution: (a) We know $d_o=12\text{cm}$, $d_i=10\text{m}=1000\text{cm}$. The image is projected onto a screen, so it is real, so $d_i$ is positive. Rearrange the thin lens equation: $$\frac{1}{f}=\frac{1}{d_o}+\frac{1}{d_i}=\frac{1}{12}+\frac{1}{1000}\approx 0.0833+0.001=0.0843{\text{cm}}^{-1}$$ $$f\approx \frac{1}{0.0843}\approx 11.86\text{cm}\approx 12\text{cm}$$

(b) Magnification $m=-\frac{d_i}{d_o}=-\frac{1000}{12}\approx -83.3$. Image width $w^{\prime }=m\cdot w=-83.3\cdot 5\text{cm}=-416.5\text{cm}\approx -4.2\text{m}$. Negative magnification means the image is inverted, positive $d_i$ means it is real, so the image is 4.2 m wide, inverted, real.

(c) From the thin lens equation, $1/d_o=1/f-1/d_i$. If $d_i$ decreases, $1/d_i$ increases, so $1/f-1/d_i$ decreases, meaning $d_o$ increases. The distance between the lens and the film must increase to keep the image in focus.

Question 3 (Application / Real-World Style)

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.

Worked Solution:

1. Magnification $m=\frac{h^{\prime }}{h}=\frac{-0.21\text{mm}}{10\text{mm}}=-0.021$ (negative because real images are inverted).
2. From $m=-d_i/d_o$, solve for $d_i$: $-0.021=-d_i/200\text{mm}⟹d_i=4.2\text{mm}$.
3. Use the mirror equation to find $f$: $\frac{1}{f}=\frac{1}{200\text{mm}}+\frac{1}{4.2\text{mm}}\approx 0.005+0.2381=0.2431{\text{mm}}^{-1}⟹f\approx 4.11\text{mm}$.
4. Radius of curvature $R=2f\approx 8.22\text{mm}$.

Interpretation: 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.

7. Quick Reference Cheatsheet

8. What's Next

Mastery of single-element image formation is the foundational prerequisite for all advanced optical system analysis in AP Physics 2. Next, you will apply the rules you learned here to multi-element optical systems, including compound microscopes and refracting/reflecting telescopes, where the image produced by one optical element becomes the object for the next. Without correctly calculating image position and magnification for single elements, you cannot correctly analyze compound systems, which are common on AP Physics 2 FRQs. This topic also feeds into the larger study of wave optics, where you will learn how diffraction limits the resolution of image-forming instruments, connecting geometric optics to physical wave behavior.

• Interference and diffraction
• Dispersion and polarization
• Applied optical systems