AP PHYSICS 2: ALGEBRA-BASED • GEOMETRIC OPTICS

Reflection

How light bounces off surfaces — the law that governs mirrors, imaging, and optical instruments.

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.

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.

1

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.
2

Angle of Incidence (θᵢ)

The angle between the incident ray and the normal at the point where the ray strikes the surface.
3

Angle of Reflection (θᵣ)

The angle between the reflected ray and the normal. The law of reflection states θᵢ = θᵣ.
4

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.
KEY TAKEAWAY
KEY TAKEAWAY

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.

Mathematical Framework

The Law of Reflection

LAW OF REFLECTION
θᵢ = θᵣ
θi = angle of incidence (measured from the normal), θr = 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
do = object distance (positive, in front of mirror), di = 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), do = object distance, di = image distance. The focal length relates to the radius of curvature by f = R/2.
MAGNIFICATION
M = hᵢ/hₒ = −dᵢ/dₒ
M = lateral magnification, hi = image height, ho = object height. If M > 0 the image is upright; if M < 0 it is inverted. |M| > 1 means the image is enlarged.
AP Exam Tip

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.
Specular vs. diffuse reflection comparison
PropertySpecular ReflectionDiffuse Reflection
Surface textureSmooth (irregularities ≪ λ)Rough (irregularities ≥ λ)
Reflected raysParallel bundle preservedScattered in many directions
Image formed?Yes — clear, defined imageNo — object is illuminated but not imaged
ExampleGlass mirror, still lakePaper, matte paint, unpolished wood
Law of reflectionθᵢ = θᵣ at every point (same normal)θᵢ = θᵣ at every point (varying normals)

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).

1
Step 1 — Find the Focal LengthThe 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 EquationUsing 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 MagnificationM = −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 ImageSince 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

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.

Comparison of plane, concave, and convex mirrors
PropertyPlane MirrorConcave MirrorConvex Mirror
Focal lengthf → ∞f > 0 (f = R/2)f < 0 (f = −R/2)
Image typeAlways virtualReal or virtual (depends on dₒ)Always virtual
Image orientationUprightInverted (real) or upright (virtual)Always upright
Magnification|M| = 1 always|M| can be > 1, = 1, or < 1|M| < 1 always
Common useBathroom mirrorsTelescopes, headlights, solar concentratorsSide-view car mirrors, security mirrors
KEY TAKEAWAY
KEY TAKEAWAY

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.

Geometric vs. wave treatment of reflection
AspectGeometric Optics (This Course)Wave / Physical Optics
Model of lightRays (straight lines)Electromagnetic waves
Reflection lawθᵢ = θᵣ (direction only)Fresnel equations (direction + amplitude + phase)
Accounts for polarization?NoYes — s- and p-polarization reflect differently
Partial reflection intensityNot predictedQuantified by reflectance R(θ)
Valid whenλ ≪ size of objectsAlways (more general)

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.

Practice Problems

1
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?
2
An object is placed 25 cm in front of a concave mirror with focal length 10 cm. What is the image distance?
3
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?
PROBLEM 4APPLIED
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.
PROBLEM 5CRITICAL 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.
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