Reflection - AP Physics 2 | Varsity Tutors

SECTION 1

Historical Context & Motivation

The study of reflection is one of the oldest branches of physics, predating even the formal concept of a "law of nature." Ancient civilizations observed that polished metal surfaces and still water produced images, and Greek philosophers sought geometric explanations for this behavior. The law of reflection — deceptively simple in its statement — became one of the first quantitative principles in optics and remains a cornerstone of geometric optics today.

~300 BCE • Euclid's Optica

Euclid formalized the idea that light travels in straight lines and described the equality of incidence and reflection angles, establishing the geometric framework for optics.

~60 CE • Hero of Alexandria

Hero proved that the law of reflection follows from the principle that light takes the shortest path between two points via a reflecting surface, an early variational argument.

1621 • Snell's Law & Renewed Interest

Willebrord Snell's discovery of the refraction law prompted renewed study of how reflection and refraction work together at interfaces between media.

1662 • Fermat's Principle

Pierre de Fermat generalized Hero's argument into the principle of least time, from which the law of reflection can be derived as a special case.

The central question reflection addresses is straightforward: when electromagnetic radiation encounters a boundary between two media, what determines the direction of the wave that bounces back? Understanding this question leads directly to the design of mirrors, telescopes, fiber-optic systems, and everyday phenomena such as seeing your own image in a window.

SECTION 2

Core Principles & Definitions

Reflection occurs whenever a wave encounters a boundary and part of its energy is redirected back into the original medium. In geometric optics we model light as rays and track their directions before and after they strike a surface. All analysis rests on a few precisely defined geometric quantities.

Normal Line

An imaginary line drawn perpendicular to the reflecting surface at the point of incidence. All angles in reflection are measured from this line, not from the surface itself.

Angle of Incidence (θᵢ)

The angle between the incident ray and the normal at the point where the ray strikes the surface.

Angle of Reflection (θᵣ)

The angle between the reflected ray and the normal. The law of reflection states θᵢ = θᵣ.

Plane of Incidence

The plane containing the incident ray, the normal, and the reflected ray. All three are coplanar — this is the second part of the law of reflection often tested on the AP exam.

SECTION 3

Visual Explanation — The Law of Reflection

The incident ray (cyan) strikes the surface at the point of incidence (yellow dot). The dashed line is the normal. The reflected ray (pink) leaves at an equal angle on the opposite side of the normal. Both rays and the normal lie in the same plane.

In the diagram above, notice that both angles are measured from the normal — a frequent source of error on the AP exam. If a problem states that light hits a surface at 25° from the surface, the angle of incidence is actually 90° − 25° = 65° from the normal. The coplanarity condition means the three-dimensional problem reduces to two dimensions within the plane of incidence, simplifying ray tracing enormously.

SECTION 4

Mathematical Framework

The Law of Reflection

LAW OF REFLECTION

θᵢ = θᵣ

θ_i = angle of incidence (measured from the normal), θ_r = angle of reflection (measured from the normal). Both rays and the normal are coplanar.

This law holds for every type of reflection — specular, diffuse, or anything in between — because it applies at the local surface normal at each point. For a flat (plane) mirror, the normal is the same everywhere, so a parallel bundle of rays reflects as a parallel bundle, preserving the image. For a curved surface, each infinitesimal patch has its own normal, and the law still applies point by point.

Deriving Image Location for a Plane Mirror

PLANE MIRROR IMAGE DISTANCE

dᵢ = −dₒ

d_o = object distance (positive, in front of mirror), d_i = image distance (negative because the image is behind the mirror). The image is virtual, upright, and the same size as the object.

Curved Mirror Equation

MIRROR EQUATION

1/f = 1/dₒ + 1/dᵢ

f = focal length (positive for concave, negative for convex), d_o = object distance, d_i = image distance. The focal length relates to the radius of curvature by f = R/2.

MAGNIFICATION

M = hᵢ/hₒ = −dᵢ/dₒ

M = lateral magnification, h_i = image height, h_o = object height. If M > 0 the image is upright; if M < 0 it is inverted. |M| > 1 means the image is enlarged.

ℹ

AP Exam Tip

SECTION 5

Types of Reflection

The law of reflection (θᵢ = θᵣ) applies at every point on every surface, yet the macroscopic result depends on the surface's geometry. Two limiting cases — specular reflection and diffuse reflection — bracket a continuum that determines whether a surface acts as a mirror or simply scatters light.

Left: specular reflection from a smooth surface keeps parallel incident rays parallel after reflection, forming a clear image. Right: diffuse reflection from a rough surface scatters rays in many directions because the local normals vary, preventing a coherent image.

Property	Specular Reflection	Diffuse Reflection
Surface texture	Smooth (irregularities ≪ λ)	Rough (irregularities ≥ λ)
Reflected rays	Parallel bundle preserved	Scattered in many directions
Image formed?	Yes — clear, defined image	No — object is illuminated but not imaged
Example	Glass mirror, still lake	Paper, matte paint, unpolished wood
Law of reflection	θᵢ = θᵣ at every point (same normal)	θᵢ = θᵣ at every point (varying normals)

Specular vs. diffuse reflection comparison

Local incident/reflected angles 30° to normal match. A large local facet illustrates equal incidence and reflection angles. The companion changes facet slopes while preserving the incoming direction, so reflected directions differ without violating the same local law. This idealized ray model requires facets much larger than the wavelength.

Parallel incident rays hit facets with−15°,0°,15° slopes; reflected directions differ while each obeys local reflection law. Computed reflection at three local facet tilts

SECTION 6

Worked Example — Concave Mirror Imaging

A concave mirror has a radius of curvature R = 40.0 cm. An object 3.0 cm tall is placed 30.0 cm in front of the mirror. Find the image distance, the magnification, and describe the image (real/virtual, upright/inverted, enlarged/reduced).

Worked example

1
Step 1 — Find the Focal Length

The focal length of a spherical mirror is half the radius of curvature: f = R/2 = 40.0 cm / 2 = 20.0 cm. Because the mirror is concave, f is positive in the standard sign convention.

f = +20.0 cm
2
Step 2 — Apply the Mirror Equation

Using 1/f = 1/dₒ + 1/dᵢ, we substitute: 1/20.0 = 1/30.0 + 1/dᵢ. Solving for 1/dᵢ: 1/dᵢ = 1/20.0 − 1/30.0 = (3 − 2)/60 = 1/60. Therefore dᵢ = 60.0 cm.

dᵢ = +60.0 cm
3
Step 3 — Calculate Magnification

M = −dᵢ/dₒ = −60.0/30.0 = −2.0. The image height is hᵢ = M × hₒ = (−2.0)(3.0 cm) = −6.0 cm.

M = −2.0 ; hᵢ = −6.0 cm
4
Step 4 — Characterize the Image

Since dᵢ is positive, the image forms in front of the mirror and is real. The negative magnification tells us the image is inverted. Because |M| = 2.0 > 1, the image is enlarged (twice the object height).

Real, inverted, enlarged image at 60.0 cm from the mirror

A candle, screen with inverted candle projection, and concave mirror on an optical rail. A concave mirror forms a real image where reflected rays converge. When the object is inside the focal length, the reflected rays diverge and their backward extensions locate an upright virtual image.

Concave-mirror magnification. With f = 20 cm and object distance 30 cm, the image distance is 60 cm and magnification is −2. A 3 cm object therefore forms an inverted 6 cm image. With f = 20 cm and object distance 30 cm, the image distance is 60 cm and magnification is −2. A 3 cm object therefore forms an inverted 6 cm image.

Virtual image inside the focal length. An object 10 cm in front of a concave mirror with f = 20 cm has image distance −20 cm and magnification +2. Only backward extensions meet behind the mirror. An object 10 cm in front of a concave mirror with f = 20 cm has image distance −20 cm and magnification +2. Only backward extensions meet behind the mirror.

SECTION 7

Comparing Mirror Geometries

Mirrors come in three basic geometries — plane, concave, and convex — each producing distinct image characteristics. Selecting the right mirror for an application depends on whether you need a real, projectable image (concave) or a wide field of view with a virtual image (convex), or simply a faithful, same-size image (plane). The table below summarizes their properties as they are typically tested on the AP Physics 2 exam.

Property	Plane Mirror	Concave Mirror	Convex Mirror
Focal length	f → ∞	f > 0 (f = R/2)	f < 0 (f = −R/2)
Image type	Always virtual	Real or virtual (depends on dₒ)	Always virtual
Image orientation	Upright	Inverted (real) or upright (virtual)	Always upright
Magnification	\|M\| = 1 always	\|M\| can be > 1, = 1, or < 1	\|M\| < 1 always
Common use	Bathroom mirrors	Telescopes, headlights, solar concentrators	Side-view car mirrors, security mirrors

Comparison of plane, concave, and convex mirrors

SECTION 8

Connection to Wave Optics & Advanced Theory

Geometric optics treats light as rays, an approximation valid when the wavelength λ is much smaller than the dimensions of optical components. When surfaces or apertures approach the wavelength scale, wave phenomena — interference and diffraction — become significant, and the ray model breaks down. Reflection in the wave picture is governed by Maxwell's equations and the boundary conditions at an interface, which yield the Fresnel equations for reflectance and transmittance as functions of angle and polarization.

Aspect	Geometric Optics (This Course)	Wave / Physical Optics
Model of light	Rays (straight lines)	Electromagnetic waves
Reflection law	θᵢ = θᵣ (direction only)	Fresnel equations (direction + amplitude + phase)
Accounts for polarization?	No	Yes — s- and p-polarization reflect differently
Partial reflection intensity	Not predicted	Quantified by reflectance R(θ)
Valid when	λ ≪ size of objects	Always (more general)

Geometric vs. wave treatment of reflection

For AP Physics 2, you are expected to use the geometric (ray) model but should be aware that it is an approximation. The wave nature of light becomes important in the interference and diffraction units later in the course, and the concept of total internal reflection — where refracted rays vanish and all light is reflected — bridges both models.

SECTION 9

Practice Problems

Question 1: CONCEPTUAL

A beam of light strikes a plane mirror at an angle of 35° from the surface of the mirror. What is the angle between the reflected ray and the surface of the mirror?

A. 35°
B. 55°
C. 70°
D. 110°

Show answer (A)

The angle of incidence is measured from the normal, not the surface. If the light hits at 35° from the surface, then θᵢ = 90° − 35° = 55° from the normal. By the law of reflection, θᵣ = 55° from the normal. The angle between the reflected ray and the surface is therefore 90° − 55° = 35°. The reflected ray leaves at the same angle from the surface as the incident ray arrived — this symmetry is a direct consequence of θᵢ = θᵣ.

Question 2: BASIC CALCULATION

An object is placed 25 cm in front of a concave mirror with focal length 10 cm. What is the image distance?

A. 16.7 cm
B. 50.0 cm
C. −16.7 cm
D. 25.0 cm

Show answer (A)

Using the mirror equation: 1/f = 1/dₒ + 1/dᵢ → 1/10 = 1/25 + 1/dᵢ → 1/dᵢ = 1/10 − 1/25 = (5 − 2)/50 = 3/50. So dᵢ = 50/3 ≈ 16.7 cm. The positive value indicates a real image formed in front of the mirror. Choice A (16.7 cm) is correct. Choice B (50 cm) would result from incorrectly computing 1/dᵢ = 1/10 − 1/50. Choice C (−16.7 cm) flips the sign, which would imply a virtual image. Choice D (25 cm) confuses the object distance with the image distance.

Question 3: INTERMEDIATE

A convex mirror with focal length f = −15 cm produces an image that is 1/3 the size of the object. How far is the object from the mirror?

A. 30 cm
B. 45 cm
C. 15 cm
D. 10 cm

Show answer (A)

For a convex mirror the image is always virtual and upright, so M = +1/3. Since M = −dᵢ/dₒ, we get +1/3 = −dᵢ/dₒ → dᵢ = −dₒ/3. Substituting into 1/f = 1/dₒ + 1/dᵢ: 1/(−15) = 1/dₒ + 1/(−dₒ/3) = 1/dₒ − 3/dₒ = −2/dₒ. So dₒ = (−2)(−15) = 30 cm. Choice A (30 cm) is correct. Choice B (45 cm) results from using M = −1/3 instead of +1/3 for a convex mirror. Choice C (15 cm) confuses the focal length with the object distance. Choice D (10 cm) arises from an algebraic error in combining fractions.

Question 4: APPLIED

A student wants to determine the focal length of a concave mirror experimentally. Design a procedure the student could follow, identify the measurements to be taken, describe how the data should be analyzed, and state one assumption that must hold for the results to be valid.

Show answer

[1 pt] Procedure: Place a luminous object (e.g., a small lamp or candle) at a known distance dₒ from the concave mirror and adjust a screen on the same side as the object until a sharp image forms. Record dₒ and dᵢ for several different object positions (at least 5 trials with dₒ > f). [1 pt] Measurements: Measure dₒ (object-to-mirror distance) and dᵢ (mirror-to-screen distance where image is focused) using a meter stick for each trial. [1 pt] Analysis: Plot 1/dᵢ vs. 1/dₒ. The mirror equation 1/f = 1/dₒ + 1/dᵢ can be rewritten as 1/dᵢ = −1/dₒ + 1/f. The y-intercept of the best-fit line equals 1/f, so f = 1/(y-intercept). Alternatively, compute f for each trial and average. [1 pt] Assumption: The mirror must be a good approximation to a spherical (or parabolic) shape and the object must be near the principal axis (paraxial approximation) so that the equation 1/f = 1/dₒ + 1/dᵢ is valid. Spherical aberration must be negligible.

Question 5: CRITICAL THINKING

An object is placed between the focal point and the center of curvature of a concave mirror. (a) Use the mirror equation to derive a general expression for the image distance in terms of f and dₒ. (b) Show that the image is always real and beyond the center of curvature for f < dₒ < 2f. (c) Explain physically why the magnification magnitude |M| > 1 in this region. (d) Describe how the image changes as the object is moved from 2f toward f.

Show answer

[1 pt] (a) From 1/f = 1/dₒ + 1/dᵢ, solving for dᵢ: 1/dᵢ = 1/f − 1/dₒ = (dₒ − f)/(f × dₒ), so dᵢ = f × dₒ / (dₒ − f). [1 pt] (b) When f < dₒ < 2f, the denominator (dₒ − f) is positive, so dᵢ is positive → the image is real. We need dᵢ > 2f: f × dₒ / (dₒ − f) > 2f → dₒ / (dₒ − f) > 2 → dₒ > 2dₒ − 2f → 2f > dₒ. This is exactly the condition given, so the image is always beyond C. [1 pt] (c) |M| = dᵢ/dₒ. Since dᵢ > 2f > dₒ (from part b), we have dᵢ/dₒ > 1, so |M| > 1. Physically, the diverging rays from the object are only partially converged by the mirror before reaching the image point, requiring a longer path and thus producing an enlarged image. [1 pt] (d) At dₒ = 2f, the image forms at dᵢ = 2f with M = −1 (same size, inverted). As dₒ decreases toward f, dₒ − f → 0⁺, so dᵢ → +∞ and |M| → ∞. The image recedes, grows larger, and remains real and inverted until the object reaches f, at which point no image forms (rays emerge parallel).