Reflection vs Refraction: What Is the Difference?

The distinction, stated once

Reflection keeps the light in the first medium. Essential knowledge 13.1.B.1 in AP Physics 2 says light that is incident on a surface can be reflected, and 13.1.B.2 gives the law of reflection: the angle between the incident ray and the normal, the line perpendicular to the surface, is equal to the angle between the reflected ray and the normal. The relevant equation is

The speed of the reflected light is unchanged, because it never left the medium it started in.

Refraction takes the light into the second medium. Essential knowledge 13.3.A.1 says refraction is the change in direction of a light ray as the ray passes from one medium into another, and 13.3.A.2 gives the cause: refraction is a result of the speed of light changing when light enters a new medium. The bending is a consequence, not the definition.

Those two paragraphs contain the whole distinction, and the cleanest way to hold it is to ask which side of the surface the light ends up on. Reflected light goes back. Refracted light goes through. One boundary, two rays, opposite sides.

What the definitions do not say is that you have to choose. Essential knowledge 14.3.A.1 states that a wave that travels from one medium to another can be transmitted or reflected, depending on the properties of the boundary separating the two media, and 14.3.A.1.i adds that a wave traveling from one medium to another will result in reflected and transmitted waves. Both happen. A window shows you the street outside by refraction and your own face by reflection at the same instant.

Side by side

The row worth reading twice is frequency, because it is the one both processes share and it is the reason the wavelength row says what it says. The frequency of a wave does not change when it travels from one medium to another, so if the speed changes then $\lambda = v/f$ forces the wavelength to change by the same factor. Light entering a slower medium keeps its colour, in the sense of frequency, and shortens its wavelength.

The row about angles is where marks are lost quietly. Both laws measure from the normal, not from the surface. An angle of incidence of $70^\circ$ from the surface is an angle of incidence of $20^\circ$ in both equations, and substituting the wrong one produces a plausible answer with no warning.

What the CED requires for each

Reflection is [Topic 13.1](/ap-physics-2/unit-13-geometric-optics/13-1-reflection), which also sets up the ray model the whole unit runs on. Objective 13.1.A asks you to describe light as a ray: 13.1.A.1 says a light ray is a straight line that is perpendicular to the wavefront of a light wave and points in the direction of travel of the wave; 13.1.A.1.i says light rays can be used to determine the behavior of light in geometric optics, where the wave nature of light can be neglected; 13.1.A.1.ii warns that rays are not sufficient to understand the spreading of light, and that in interference and diffraction the wave nature of the light is important; 13.1.A.1.iii notes that a laser is a common source of a single coherent, monochromatic beam of light that can be modeled as a ray; and 13.1.A.2 says ray diagrams depict the path of light before and after an interaction with matter.

Objective 13.1.B asks you to describe the reflection of light from a surface, and adds two kinds:

• 13.1.B.3: diffuse reflection is the reflection of light from a rough surface and results in light reflected in many different directions, because the line normal to the surface varies over the area over which the light is incident.
• 13.1.B.4: specular reflection is the reflection of light from a smooth surface and results in light uniformly reflected from the surface, because the line normal to the surface has an approximately constant direction over the area the light strikes.

Notice that the law of reflection is not suspended for a rough surface. Each patch still obeys $\theta_i = \theta_r$; what varies is the direction of the normal from patch to patch. That is the correct way to say why a matte wall does not show your reflection.

Refraction is [Topic 13.3](/ap-physics-2/unit-13-geometric-optics/13-3-refraction), one objective, 13.3.A, describe the refraction of light between two media. Beyond 13.3.A.1 and 13.3.A.2 above, 13.3.A.3 states that the index of refraction of a given medium is inversely proportional to the speed of light in the medium, with the relevant equation $n = c/v$, and 13.3.A.4 gives Snell's law, $n_1 \sin\theta_1 = n_2 \sin\theta_2$, relating the angles of incidence and refraction to the indices of the two media. Its three sub-points fix the direction of the bend:

• 13.3.A.4.i: travelling from higher index into lower index, the ray refracts away from the normal.
• 13.3.A.4.ii: travelling from lower index into higher index, the ray refracts toward the normal.
• 13.3.A.4.iii: incident along the normal, the transmitted ray is not refracted.

Both topics carry no boundary statement. Across the whole of Unit 13, Geometric Optics, exactly one topic has one, and it is Topic 13.2, limiting the study of mirrors to plane mirrors, convex spherical mirrors, and concave spherical mirrors. That was checked page by page across all four topics.

Both relations are on the AP Physics 2 equation sheet, in the waves, sound, and optics block: $n = c/v$ and $n_1 \sin\theta_1 = n_2 \sin\theta_2$, alongside $\lambda = v/f$. The law of reflection, $\theta_i = \theta_r$, is given in the CED as the relevant equation for 13.1.B.2, and the critical-angle relation discussed below is labelled in the CED as a derived equation rather than a relevant one, and it is not printed on the sheet.

The case that separates them: one boundary, two rays

Shine a laser from air onto the flat surface of a liquid at $40^\circ$ from the normal. Take the index of air as $n_1 = 1.00$ and of the liquid as $n_2 = 1.33$, both given. Two rays leave the point where the beam lands, and they are not symmetric.

The refracted ray bends toward the normal, so $28.9^\circ$ is smaller than $40^\circ$. The reflected ray keeps $40^\circ$ on the other side of the normal. Both are drawn on the same ray diagram, and a complete answer to "what happens to the beam" includes both.

Now reverse the trip, sending the beam up out of the liquid at $40^\circ$ from the normal. Snell's law now gives $\sin\theta_2 = (1.33)(\sin 40^\circ)/1.00 = 0.855$, so $\theta_2 = 58.7^\circ$: the ray bends away from the normal on the way out, exactly as 13.3.A.4.i requires, and the two trips are geometric reverses of each other. The reflected ray inside the liquid is still at $40^\circ$.

That reversibility is worth holding onto. Snell's law does not care which medium you call the first one; swap the labels and swap the angles and it is the same equation. What the labels do decide is whether the bend is toward the normal or away from it.

Total internal reflection: the one case where refraction stops

Push the exit angle further and something breaks. Essential knowledge 13.3.A.5 says total internal reflection may occur when light passes from one medium into another medium with a lower index of refraction, and 13.3.A.5.i says it occurs beyond a critical angle of incidence, with the CED's derived equation

Two further statements complete the picture. 13.3.A.5.ii says that for incident rays at the critical angle, the ray refracts at 90 degrees and travels along the surface of the material. 13.3.A.5.iii says that for incident rays beyond the critical angle, all light is reflected, and no light is transmitted into the other medium.

For the liquid-and-air boundary above, $\theta_{\mathrm{critical}} = \sin^{-1}(1.00/1.33) = 48.8^\circ$. So the $40^\circ$ ray above got out. A ray at $55^\circ$ does not: substituting into Snell's law gives $\sin\theta_2 = (1.33)(\sin 55^\circ) = 1.09$, and no angle has a sine greater than one. The algebra failing is the physics telling you there is no refracted ray.

This is the only situation in the course where the pair separates completely, so it is worth being precise about the two conditions the CED attaches:

1. The light must be going from higher index to lower. Total internal reflection cannot happen on the way into a denser medium, because then $n_2/n_1$ exceeds one and the inverse sine has no value.
2. The angle of incidence must be beyond the critical angle. At exactly the critical angle there is still a refracted ray, grazing along the surface at $90^\circ$.

Notice which quantity is unaffected. The reflected ray was always there, at $\theta_i = \theta_r$, at every angle from zero upward. Total internal reflection is not reflection switching on; it is refraction switching off, leaving reflection with all of the light.

What changes at a boundary and what does not

Three quantities, and only one of them survives the crossing untouched. The bookkeeping is worth doing once carefully because every refraction question is built on it.

Frequency does not change. Essential knowledge 14.3.A.1.iv states this directly: the frequency of a wave does not change when it travels from one medium to another. The source sets the frequency, and the boundary cannot renegotiate it.

Speed changes, and the index is the bookkeeping for it. From 13.3.A.3, $n = c/v$, so a medium with a larger index carries light more slowly. Rearranged, $v = c/n$, with $c = 3.00 \times 10^8 \ \mathrm{m/s}$ from the sheet's table of constants.

Wavelength changes, and it has to. The sheet prints $\lambda = v/f$. With $f$ fixed and $v$ divided by $n$, the wavelength is divided by $n$ as well. Light of wavelength $600 \ \mathrm{nm}$ in vacuum has a wavelength of $400 \ \mathrm{nm}$ inside a medium of index $1.50$, and it is still the same light.

That last point resolves a genuine confusion about colour. Colour tracks frequency, which is why a red laser stays red under water. It does not track wavelength inside the material, because that shortened. When a problem quotes a wavelength for light in a medium, check whether it means the vacuum wavelength or the in-medium one, because the two differ by a factor of $n$.

One more statement from Unit 14 is worth knowing, and it applies to the reflected wave rather than the transmitted one. 14.3.A.1.ii says a reflected wave is inverted if the transmitted wave travels into a medium in which the speed of the wave decreases, and 14.3.A.1.iii says it is not inverted if the transmitted wave travels into a medium in which the speed increases. That inversion is a phase statement, not a direction statement, and it is what thin film interference is built on. It does not affect any angle on this page.

When it costs a mark

Measuring angles from the surface instead of the normal. Both 13.1.B.2 and Snell's law are written for angles from the normal, and the CED spells out in 13.1.B.2 that the normal is the line perpendicular to the surface. A ray $20^\circ$ above a surface has an angle of incidence of $70^\circ$. Getting this backwards produces an answer that is wrong and looks fine.

Reporting only one of the two rays. At a boundary between transparent media, light is both reflected and transmitted (14.3.A.1.i). "The light refracts into the water" is an incomplete answer to what happens to the beam.

Bending the wrong way. Into a higher index, toward the normal; into a lower index, away from it. 13.3.A.4.i and 13.3.A.4.ii. A sketch with the refracted ray on the wrong side of the extended incident ray loses the diagram mark whatever the arithmetic says.

Bending the ray at normal incidence. 13.3.A.4.iii is explicit: a ray incident along the normal is not refracted. The speed still changes, and so does the wavelength, but the direction does not.

Saying the frequency changes on refraction. It does not, by 14.3.A.1.iv. Frequency is the quantity that carries through, and the wavelength is the one that adjusts.

Applying the critical angle in the wrong direction. Total internal reflection requires travel from higher index to lower. Computing $\sin^{-1}(n_2/n_1)$ with the larger index on top gives a quantity greater than one and no angle, which is the calculator telling you the geometry is impossible rather than that you mis-typed.

Assuming total internal reflection at exactly the critical angle. At the critical angle the refracted ray still exists, travelling along the surface at $90^\circ$ (13.3.A.5.ii). It is beyond the critical angle that all light is reflected (13.3.A.5.iii).

Forgetting that mirrors are reflection and lenses are refraction. Topic 13.2 builds images from reflection, Topic 13.4 builds them from refraction, and both use the same equation $1/s_i + 1/s_o = 1/f$. Getting the mechanism wrong in an explanation costs the justification mark even when the numbers come out right.

When they coincide, and why that lulls you

The two processes overlap in three situations, and each hides the distinction for a different reason.

At normal incidence, neither ray bends. The reflected ray goes straight back, the transmitted ray goes straight on, and both lie along the original line. Every angle in both equations is zero and both are satisfied trivially. A demonstration done at normal incidence teaches nothing about the difference.

When the two indices are equal, refraction disappears. Set $n_1 = n_2$ in Snell's law and it reduces to $\theta_1 = \theta_2$: the ray carries straight on. This is why a glass rod in a liquid of matching index becomes hard to see. There is no boundary as far as the light is concerned, which answers the CED's own Unit 13 essential question about how we can make things invisible.

Both are direction changes governed by the same normal, and both keep the frequency. If your description of what happened at a boundary is "the light changed direction and stayed the same colour", you have described both processes at once and distinguished neither.

The distinction turns on three questions, and they are the three that keep appearing on exams.

1. Which side of the surface is the ray on? This alone settles which process you are looking at.
2. Did the speed change? Only the refracted ray's did, and that is what makes its angle differ from the angle of incidence.
3. Can the ray get out at all? Only refraction can fail, and only when going into a lower index beyond the critical angle.

If you can answer those three, the pair is finished.

Where this sits on the AP exam

Both processes live in Unit 13, Geometric Optics, which the CED weights at 12 to 15 percent of the multiple-choice section over a suggested 8 to 12 class periods. Reflection is Topic 13.1 and refraction is Topic 13.3. The images each one builds get their own topics: mirrors in Topic 13.2 and lenses in Topic 13.4.

The suggested skills tell you the difference in how the two are examined. Topic 13.1 lists 1.A, create diagrams, tables, charts, or schematics; 2.B, calculate or estimate an unknown quantity with units; 2.C, compare physical quantities between two or more scenarios; and 3.B, apply an appropriate law, definition, theoretical relationship, or model to make a claim. Topic 13.3 lists 1.B, create quantitative graphs with appropriate scales and units, including plotting data; 2.B; 2.D, predict new values or factors of change using functional dependence between variables; 3.A, create experimental procedures appropriate for a given scientific question; and 3.B.

The presence of 1.B and 3.A on the refraction topic is a real signal: the CED's own sample activities for Unit 13 have students determining an index of refraction experimentally and finding the speed of light in water, and the unit's guidance says the third free-response question on the AP Physics 2 exam is the Experimental Design and Analysis question. Refraction is one of the places the course expects you to have designed a measurement, not just used a formula.

From here, the images built by reflection and refraction are the next step, and they share both a sign convention and an equation. See real vs virtual image for what the two kinds of image are and how the sign of the image distance tells them apart, and concave vs convex mirror for the reflection case in full. For the wave behaviour that the ray model deliberately ignores, 13.1.A.1.ii points to interference and diffraction in Unit 14, and the wave speed, frequency and wavelength guide covers the $\lambda = v/f$ bookkeeping that the change of medium relies on.